\documentstyle[11pt,wide,tech]{article} \include{psfig} \setlength{\textwidth} {6.1in} \setlength{\textheight} {8.6in} \setlength{\oddsidemargin} {0.3in} \setlength{\topmargin} {-0.4in} \def\widenspacing{\multiply\baselineskip by 140 \divide\baselineskip by 100} \begin{document} \def\sz{$\bigcirc$ \kern- 1.05em 1} \def\sy{$\bigcirc$ \kern- 1.05em 2} \def\sx{$\bigcirc$ \kern- 1.05em 3} \def\sw{$\bigcirc$ \kern- 1.05em 4} % \vspace*{1.0in} \begin{center} {\bf NONPARAMETRIC BAYES ESTIMATOR OF SURVIVAL FUNCTIONS FOR CENSORED AND TRUNCATED DATA } \medskip {\sc Jin Yu Luan } \medskip {\sc Mai Zhou } \medskip {\it University of Kentucky } \end{center} \begin{list}{}{\leftmargin 0.5in \rightmargin 0.5in} \item {\small The non-parametric Bayes estimator with Dirichlet process prior of a survival function based on right censored data was considered by Susarla and Van Ryzin (1976) and many others. We obtain the non-parametric Bayes estimator of a survival function when data are right, left or interval censored. The resulting Bayes estimator with Dirichlet process prior has an explicit formula. In contrast, there is no explicit formula known for the non-parametric maximum likelihood estimator (NPMLE) with such data. In fact, we show that the NPMLE with doubly/interval censored data cannot, in general, be the limit of Bayes estimators for {\em any} sequence of priors. Several examples are given, showing that the NPMLE %non-parametric maximum likelihood estimator and the non-parametric Bayes estimator may or may not be the same, even when the prior is `non-informative'. } \end{list} {\sl AMS 1991 Subject Classification:} Primary 60E15; secondary 60G30. \vspace{1ex} \newline {\sl Key Words and Phrases:} Dirichlet process prior, square error loss, non-informative prior, NPMLE. \widenspacing \vspace{-1ex} \section*{\normalsize\bf 1. Introduction, Notation and Preliminary} \label{sec-intro} \setcounter{equation}{1} \indent Suppose lifetimes $X_1, \cdots , X_n $ are non-negative and iid with a distribution $F( \cdot )$. However, these lifetimes are subject to censoring. In the case of right censoring, we only observe $$ Z_i = \left\{\begin{array}{lll} X_i & \mbox{if $ X_i \leq C_i$} \\ C_i & \mbox{if $X_i > C_i$} ~~~~~ \end{array} \right. %%%\min (C_i , \max ( X_i , Y_i ) ) ~~~~\mbox{\rm and } ~~~~ \Delta _ i = \left\{\begin{array}{lll} 1 & \mbox{if $X_i \leq C_i$}\\ 0 & \mbox{if $X_i > C_i$} ~~~~~~~~~ \end{array} \right. \eqno (1.1) $$ where $C_i$ are the (right) censoring times. A generalization of right censoring is double censoring (Chang and Yang (1987), Gu and Zhang (1993)). In the case of double censoring we only observe $$ Z_i = \left\{\begin{array}{lll} X_i & \mbox{if $Y_i \leq X_i \leq C_i$}\\ C_i & \mbox{if $X_i > C_i$} ~~~~~ \\ Y_i & \mbox{if $X_i < Y_i$} \end{array} \right. %%%\min (C_i , \max ( X_i , Y_i ) ) ~~~~\mbox{\rm and } ~~~~ \Delta _ i = \left\{\begin{array}{lll} 1 & \mbox{if $Y_i \leq X_i \leq C_i$}\\ 0 & \mbox{if $X_i > C_i$} ~~~~~~~~~ . \\ 2 & \mbox{if $X_i < Y_i$} \end{array} \right. \eqno (1.2) $$ Here %The iid 2-dimensional random vectors $(C_i , Y_i ), i=1, \cdots , n$, are the left and right censoring times, with $C_i > Y_i $. Let the observations be arranged such that %By a re-arrangement if necessary, we assume $Z_1, \cdots , Z_k $ are the uncensored observations, i.e., $\Delta_1 =1, \cdots, \Delta _ k =1$. %($k$ is random). Notice that $(Z_1, \cdots , Z_k )$ is $(X_1, \cdots , X_k)$, while %Assume $Z_{k+1}, \cdots , Z_n $ are the (either right or left) censored observations. %assumed %independent of $X_i$ with $P( C_i > Y_i ) = 1 $. %Denote $S_{C_i}(t) = P( C_i > t)$ and %$S_{Y_i}(t) = P( Y_i > t)$. %We also assume that $X_i$ is independent of $(C_i , Y_i )$. In the Bayesian estimation of $F(\cdot)$, we need not make assumptions about the distributions of the left and right censoring times $C_i$ and $Y_i$. %Further, for right censored observations, $Y_i$ need not exist. %the existence of $Y_i$ for the right censored observations, $C_i$ %For left censored observations, $C_i$ need not exist. The calculations are conditioned on the observed censoring times. Thus the observations can be described in three parts $ Z_1, \cdots , Z_k $ where $X_i = Z_i$; $Z_{k+1}, \cdots , Z_m $ where $X_i > Z_i$; and $Z_{m+1}, \cdots , Z_n $, where $X_i < Z_i$. Next we discuss interval censored data. The current status data, or case 1 interval censored data, consist of an observed ``inspection" time $T_i$ and the information whether $X_i$ is larger than or less than $T_i$ (the status of $X_i$, see Huang and Wellner (1996)): $$ T_i~, ~~~~~ \Delta _ i = \left\{\begin{array}{lll} 0 & \mbox{if $X_i > T_i$} ~~~~~~~~~ \\ 2 & \mbox{if $X_i < T_i$} ~~~~~~. \end{array} \right. $$ Usually the ``inspection'' times $T_i$ are assumed iid. Similar to the discussion %As we mentioned above, this iid assumption does not make a difference in the Bayesian analysis and therefore the current status data is a special case of (1.2), where all the observations are either left or right censored, i.e. $k=0$. %Notice the double censoring also include the %the observations are either right or left censored In case 2 of interval censoring, we assume $X_1, \cdots , X_k $ are observed exactly ($k$ non-random, and possibly zero), and only the observations $ X_{k+1}, \cdots , X_n $ are interval censored. Then we observe $n-k$ %(random) intervals. With some abuse of notation, they are denoted by $ [ L_j , Z_j ) $ for $j= k+1, \cdots , n$. We know that $ L_j \leq X_j < Z_j $. %This is called case 2 (or case k) of interval censored %data by Huang and Wellner (1996). Again notice that we do not need to make assumptions about the distribution of $L_j$ or $Z_j$. Therefore this fits % accommodates both case 2 and case $k$ of interval censoring in Huang and Wellner (1996). To make the notation consistent with the doubly censored case, we let $Z_1 = X_1 , \cdots , Z_k = X_k $ for directly observable outcomes, interval censored outcomes are $ [L_j ,~ Z_j) $ for $j= k+1, \cdots , n$. Notice that when $ [ L_j , Z_j ) = [a, \infty)$, one has right censored data, and when $ [ L_j , Z_j ) = [0, a)$, left censored data. %In that sense, interval censoring %is more general in that it includes double censoring as a special case. %which in turn includes right censoring as a special case. %We treat them separately only because there are simplifications %that occur in the special cases. In (nonparametric) Bayesian analysis, the probability $ F(\cdot ) $ is random. We assume in this paper that $F(\cdot )$ is distributed as a Dirichlet process with parameter $\alpha $, a measure on the real line. Under the Dirichlet process prior assumption, the probability measure $P(A)= \int _{A} dF $ has the following property: given any partition of real line $ A_1, \cdots A_u$, the joint distribution of the random vector $( P(A_1), \cdots, P(A_u) )$ has a Dirichlet distribution with parameter given by $\alpha(A_1), \cdots , \alpha(A_u)$. We recall a random vector $(x_1, \cdots, x_u)$ has a Dirichlet distribution with parameter $a_1, \cdots , a_u$ if it has density \[ \prod ( x_i ) \] For more discussion and properties of Dirichlet process prior, see Ferguson (1973), Susarla and Van Ryzin (1976) and Ferguson, Phadia and Tiwari (1993). Another possibility is to work with the cumulative hazard functions $ H(t) $. % = - \log (1- F(t) )$. A beta process prior on the space of the cumulative hazard function was introduced by Hjort (1990). While using a beta process prior for right censored data works well, it has no advantage over the Dirichlet process prior for doubly/interval censored data: the likelihood of the data does not simplify by using the hazard function with doubly censored data. %Therefore, we will use the Dirichlet process prior in this paper. Using squared error loss, Susarla and Van Ryzin (1976) obtained the Bayes estimator for $F(\cdot)$ under a Dirichlet process prior when data are only subject to right censoring. They also showed that when the weight parameter, $\alpha$, of the Dirichlet process prior approaches zero, the non-parametric Bayes estimator reduces to the Kaplan-Meier estimator, the NPMLE. Some later papers studied the consistency of the Bayes estimator (Susarla and Van Ryzin (1978)) and the posterior distribution (Ghosh and Ramamoorthi (1995)). Huffer and Doss (1999) used Monte Carlo methods to compute the nonparametric Bayes estimator. We obtain the Bayes estimator of $1-F(\cdot )$ when data are subject to both right and left censoring, or are subject to interval censoring. The large sample properties of this Bayes estimator are not discussed here, though it is not unreasonable to expect that it is consistent. However, we show that for {\em any} sequence of priors the nonparametric Bayes estimators under squared error loss {\em cannot} always converge %necessarily reduce to the corresponding NPMLE with doubly censored data. %nonparametric maximum likelihood estimator, %although the difference is often small. This is a bit surprising since, in most cases, MLEs are limits of Bayes estimators. The Bayes estimator we obtain is more complicated than those with only right censored data, especially when there are many left censored or interval censored observations. Nevertheless, it has an explicit formula that can be easily programmed. In contrast, the nonparametric maximum likelihood estimator (NPMLE) in the case of doubly censored data or interval censored data does not have an explicit formula and its computation requires an iterative method. See Turnbull (1974), Chen and Zhou (2003), and Fay (1999). Besides, the Bayes estimator is always uniquely defined while the NPMLE is often only defined up to an equivalent class. This non-uniqueness of the NPMLE makes many important statistics like the mean estimator difficult to define. %The Bayes estimator do not have this problem. The Bayes estimator is also smoother than the NPMLE. On the other hand, there are consistency results for the NPMLE (Gu and Zhang, (1993), Groeneboom and Wellner (1992), Huang and Wellner (1996)) but we know very little of the consistency of the Bayes estimators beyond the right censored, $R^1$ data case. In fact, R. Pruitt gave an example of an inconsistent Bayes estimator with Dirichilet process prior for right censored data in $R^2$. %A fact to note is that the Bayes estimator %do not necessarily converge to the NPMLE in this case when %the parameter of the Dirichlet process prior $\alpha() \to 0$. %Some examples illustrate this fact are in section 3. To minimize the amount of new notation, we follow Susarla and Van Ryzin's (1976), hereafter SV, and we use their convention that all observations are positive. Obviously we can extend this to the case where observations have support in $ (-\infty, \infty ) $ without much difficulty. \section*{\normalsize\bf 2. Bayes estimator with right, left/interval censored observations} \label{sec-2} \setcounter{equation}{0} \indent The Bayes estimator of $1-F(\cdot )$ under squared error loss of SV is the conditional expectation of $1-F$ given all the observations. Similar to SV the conditional expectation is computed in two steps: first, given all the uncensored observations we find the conditional {\em distribution} of $1-F$; %(the following theorem), second, given all the censored observations we compute the conditional {\em expectation}, where the distribution of those lifetimes before censoring is given in the first step. The following theorem specifies the conditional distribution of $F(\cdot )$ given all the uncensored observations, which accomplishes the first step. {\bf Theorem 1} {\sl The posterior distribution of the random probability measure $P$ given $(\Delta_1 = 1, Z_1 ), \ldots , (\Delta_k = 1, Z_k ) $ is the Dirichlet process with parameter $\beta = \alpha + \sum_{i=1}^k \delta_{Z_i} $, where $\delta_a $ is a unit measure on the point $a$. } {\sc Proof:} The proof of this theorem is similar to SV (1976) and Ferguson (1973). We only sketch the proof for the doubly censored case. Furthermore, we only give those calculations that differ from the proof of Theorem 4 of SV (1976), the rest of the proof is the same as theirs and is not repeated here. From (1) of Chang (1990), the probability of $ (\Delta = 1, X = u ) $ is $( S_C (u) - S_Y (u) ) d P( X \leq u) = dG(u)$, say. Recall the marginal distribution of $X$ is $\frac{\alpha (u)}{\alpha (R^+)}$. We compute \[ \int_{ [ \Delta =1 , Z \in A ] } D(\cdot | \alpha (B_1) + \delta_u (B_1) , \cdots , \alpha (B_l)+\delta_u (B_l)) dG(u) \] \[ = \int_{ [ u \in A ] } D(\cdot | \alpha (B_1) + \delta_u (B_1) , \cdots , \alpha (B_l)+\delta_u (B_l)) ( S_C (u) - S_Y (u) ) d \frac{\alpha (u)}{\alpha (R^+)} \] $$ = \sum _{j=1}^l D( y_1, \cdots , y_l | \alpha_1^{(j)}, \cdots , \alpha_l^{(j)} ) \int_{ [ u \in A \cap B_j ] } ( S_C (u) - S_Y (u) ) d \frac{\alpha (u)}{\alpha (R^+)}~. $$ On the other hand, $$ {\cal P} \{ P(B_i) \leq y_i , i=1, \cdots , l; \Delta = 1, Z \in A ) \} $$ $$ = \int_{ u=0 } ^\infty {\cal P} \{ P(B_i) \leq y_i , i=1, \cdots , l; X \in [u, u+du) \cap A \} ( S_C (u) - S_Y (u) ) $$ $$ = \sum_{j=1}^l \int \frac{\alpha ( B_j \cap A \cap [u, u+du) }{ \alpha (R^+) } ( S_C (u) - S_Y (u) ) D( y_1, \cdots , y_l | \alpha_1^{(j)}, \cdots , \alpha_l^{(j)} ) $$ $$ = \sum_{j=1}^l D( y_1, \cdots , y_l | \alpha_1^{(j)}, \cdots , \alpha_l^{(j)} ) \int_{B_j \cap A } \frac{ S_C (u) - S_Y (u) }{ \alpha (R^+) } d \alpha(u)~, $$ which is same as above. $\diamondsuit$ Now, the conditional expectation of $ 1-F(u) = P [u, \infty) $ is computed given the remaining $ n - k -1 $ censored observations: $Z_{k+1}, \Delta_{k+1}, \cdots, Z_n , \Delta_n $ in the doubly censored case; $[L_{k+1}, ~ Z_{k+1}), \cdots, [L_n , Z_n ) $ in the interval censored case. Notice the original $X_{k+1} , \cdots , X_n $ %(before been censored) is now a random sample from a Dirichlet process with parameter $\beta $. Let ${E_{\beta}}$ denote the expectation with respect to this Dirichlet process. % with parameter $\beta$. \subsection*{\normalsize\it 2.1 Bayes estimator with One interval/left censored observation} To fix ideas and enhance readability, we first present in detail the Bayes estimator with many right censored observations but only one interval censored observation, denoted by $ [L_w, Z_w) $. (If $L_w = 0$ then this is left censored.) The general case with many interval/left censored observations will be given later. %We single out the right censored observations %because the formula involving those are simpler. As in SV Corollary 1, the conditional expectation, ${E_{\beta}}$, of $1-F(u) = P[u, \infty) = P( X \geq u)$ given all the right censored data and one interval censored observation is $$ \hat S_D (u) = \frac{ {E_{\beta}} \{ P[u, \infty) P[L_w, Z_w) \prod_{right-censored} P[ Z_i , \infty ) \} } { {E_{\beta}} \{ P[L_w, Z_w) \prod_{right-censored} P[ Z_i , \infty ) \} } ~. $$ This is also the desired Bayes estimator of $1-F(u)$. We abbreviate the subscript of ${right-censored}$ to ${r-c}$ and ${left-censored}$ to ${l-c}$ and ${interval-censored}$ to ${i-c}$. Straightforward calculation yields $$ \hat S_D (u) = \frac{ {E_{\beta}} P[u, \infty) \{ P[L_w, \infty) - P[Z_w, \infty) \} \prod_{r-c} P[ Z_i , \infty ) } { {E_{\beta}} \{ P[L_w, \infty) - P[Z_w, \infty) \} \prod_{r-c} P[ Z_i , \infty ) } $$ $$ = \frac{ {E_{\beta}} \{ P[u, \infty) P[L_w, \infty) \prod_{r-c} P[ Z_i , \infty ) \} - {E_{\beta}} \{ P[u, \infty) P[Z_w, \infty) \prod_{r-c} P[ Z_i , \infty ) \} } { {E_{\beta}} \{ P[L_w, \infty) \prod_{r-c} P[ Z_i , \infty ) \} - {E_{\beta}} \{ P[Z_w, \infty) \prod_{r-c} P[ Z_i , \infty ) \} } $$ $$ = \frac{ {E_{\beta}} \mbox{\sz} - {E_{\beta}} \mbox{\sy} } {{E_{\beta}} \mbox{\sx} - {E_{\beta}} \mbox{\sw} } . $$ %This is also the desired Bayes estimator. The last four expectations are all of the same type and can be computed explicitly by the Lemma below. %Now we compute the ${E_{\beta}} [ \cdot ] $ above. Given a set of positive numbers $0 < a_{k+1} < a_{k+2} < \cdots < a_m < \infty $, consider the partition of $R^+$ into intervals $ [0, a_{k+1}), [a_{k+1}, a_{k+2}), \cdots,$ $ [a_m , \infty ) $. %We let $a_k=0$ and $a_{m+1} = \infty $ to %make the notation consistent. By Theorem 1 the random vector $ P[0, a_{k+1}) , P[a_{k+1}, a_{k+2}), \cdots P[a_m , \infty ) $ has a Dirichlet distribution with parameter vector $(\beta_{k+1}, \cdots , \beta _{m+1} )$ where $\beta_{k+1} = \beta [0, a_{k+1} ), \cdots , \beta_{m+1} = \beta [a_m , \infty ) $. The measure $\beta $ is given as before by $\beta = \alpha + \sum_{uncensored} \delta _{Z_i} $. {\bf Lemma 1} (Susarla and Van Ryzin) {\sl With the notation above, we have $$ {E_{\beta}} \prod_{i=k+1}^m P[a_i , \infty) % = \frac{ \prod_{i=0}^{m-k-1} [ i + \sum_{j=0}^{i} \beta_{m+1-j} ]} % { \prod_{i=0}^{m-k-1} [ i + \beta(R^+) ]} = \prod_{i=0}^{m-k-1} \left ( \frac{i + \sum_{j=0}^{i} \beta_{m+1-j}}{i + \beta(R^+)} \right ) = \prod_{i=0}^{m-k-1} \left ( \frac{ i + \beta[ a_{m-i} , \infty) }{ i + \beta (R^+) } \right ) ~. $$ } {\sc Proof:} This is essentially Lemma 2 (a) of SV (1976) with some extra simplifications. $\diamondsuit$ When $\alpha (R^+) = 0$ the expression on the right hand side of Lemma 1 is still well defined unless there are no uncensored observations in the sample. In Example 2 of Section 3, there are no uncensored observation in the sample and we do not discuss the limit of the Bayes estimator as $\alpha (R^+) \to 0$ there. %In this case it needs to be re-defined to be zero, %since the denominator of the %expression is $ \Gamma( \beta (R^+)) / \Gamma( \beta(R^+) +(m-k) )$. %We do not need to worry $\beta(R^+) = 0 $ unless %$\alpha (R^+) = 0$ and %When using the result of Lemma 1 repeatedly to compute (2.3) there are %apparently cancellations which simplify the expression. We will not %pursue this here. {\bf Remark}: It is clear from the definition of $\beta$ that when $\alpha ( R^+) \to 0 $, $\beta$ is integer valued. This implies that the expectation in Lemma 1 has a rational number value (finite product of rational numbers) as $\alpha ( R^+) \to 0 $. %\subsection*{\normalsize\it 2.2 Many left censored observations } % %Similar to the previous case, %for data with (many) doubly censored observations, the Bayes estimator of %$1-F(u) = P( X \geq u)$ given all the data (censored %or uncensored) is equal to % %$$ %\frac{ %{E_{\beta}} % \{ P[u, \infty) \prod_{l-c} [ 1- P[Z_w, \infty ) ] %\prod_{r-c} P[ Z_i , \infty ) \} } %{ {E_{\beta}} \{ \prod_{l-c} [ 1- P[Z_w, \infty ) ] %\prod_{r-c} P[ Z_i , \infty ) \} } %\eqno (2.5) %$$ % %Using the identity %$$\prod_{i=1}^m (1-a_i) = %1-\sum a_i + \sum a_i a_j - \sum a_ia_ja_k + \cdots (-1)^m \prod a_i %$$ %the denominator of (2.5) can be written as %$$ %{E_{\beta}} \{ %\left ( 1 - \sum_w P[Z_w, \infty ) + %\sum_{w, r} P[Z_w, \infty )P[Z_r, \infty ) - \cdots %\right ) %\prod_{r-c} P[ Z_i , \infty ) %\} %$$ %$$ %= \{ {E_{\beta}} \prod_{r-c} P[ Z_i , \infty ) - \sum_w {E_{\beta}} %\left ( P[Z_w, \infty ) \prod_{r-c} P[ Z_i , \infty ) \right ) + %\sum_{w, r} {E_{\beta}} \left ( P[Z_w, \infty )P[Z_r, \infty ) %\prod_{r-c} P[ Z_i , \infty ) \right ) - \cdots %\} ~. %$$ %The expectations above can all %be calculated explicitly since %${E_{\beta}} \prod P[a_j, \infty )$ can be determined by Lemma 1. %The numerator of (2.5) can be calculated similarly. Therefore the %Bayes estimator of $1-F(u)$, (2.5), can be computed explicitly. \subsection*{\normalsize\it 2.2 Many interval/left censored observations} %Similar to the previous case, When the data contain many interval and many right censored observations, the Bayes estimator of $1-F(u) = P( X \geq u)$ given all the data (censored or uncensored) is $$ \hat S_D (u) = \frac{ {E_{\beta}} \{ P[u, \infty) \prod_{i-c} P[L_w, Z_w ) \prod_{r-c} P[ Z_i , \infty ) \} } { {E_{\beta}} \{ \prod_{i-c} P[L_w, Z_w ) \prod_{r-c} P[ Z_i , \infty ) \} } . \eqno (2.1) $$ When data contains many left and many right censored observations, the Bayes estimator of $1-F(u)$ is $$ \hat S_D (u) = \frac{ {E_{\beta}} \{ P[u, \infty) \prod_{l-c} [ 1- P[Z_w, \infty ) ] \prod_{r-c} P[ Z_i , \infty ) \} } { {E_{\beta}} \{ \prod_{l-c} [ 1- P[Z_w, \infty ) ] \prod_{r-c} P[ Z_i , \infty ) \} } ~. \eqno (2.2) $$ Because left censored observation is a special case of interval censored observation as pointed out in the previous section, we only present in detail below the Bayes estimator with many interval/right censored observations. Let us recall the identity $$ \prod_{i=1}^m (b_i - a_i) = \sum y_1 y_2 \cdots y_m ~, \eqno (2.3) $$ where $y_i $ is either $b_i$ or $-a_i$ and the summation is over all possible $2^m$ choices. The integer $m$ is defined as $ m= \# \{ i-c \} $ = number of interval censored observations. By using (2.3), we can write %the following product as a sum of many products, $ \prod_{i-c} P [ L_w , Z_w) = \prod_{i-c} \{ P[L_w, \infty) - P[Z_w, \infty) \} = \sum P_1 P_2 \cdots P_m ~, $ where each $P_w$ is either $ P[L_w, \infty)$ or $- P[Z_w, \infty)$, and the summation is over all $2^m$ different choices. To make the expression more specific we introduce some notation. Define vectors $\xi = ( \xi_1, \cdots , \xi_m )$ where each $\xi_i =$ either $0$ or $1 $. %There are $2^m$ such vectors. Given $m$ interval censored observations, $[L_i, Z_i)$, we define $2^m$ sets of numbers $\{ c_i(\xi), i=1, 2, \cdots, m \} $ where $c_i (\xi) = L_i$ if $\xi_i = 0 $ otherwise $c_i (\xi) = Z_i$. With each set $\{ c_i(\xi ) , i=1, \cdots m \} $, associate a sign: if the set contains an even number of $Z_i$'s then the sign is positive, if the set contains odd number of $Z_i$'s the sign is negative. % i.e. $ \pm = (-1)^{\sum \xi_i} $. With these definition we can write $ \prod_{i-c} P [ L_w , Z_w) = \sum P_1 P_2 \cdots P_m = \sum_{\xi } \pm \prod_{i=1}^m P[c_i(\xi ), \infty)~, $ where the summation is over all $2^m$ different $\xi$'s, and $\pm$ is the associated sign. Finally, we define new sets of numbers by adding $r $ ($r = \# \{ r-c \}$) right censored observations $Z_1, \cdots , Z_r$ to $\{ c_i(\xi) , i=1, \cdots , m \} $: $ \{ b_j(\xi ) , j=1, \cdots , m+r \} = \{ c_i(\xi ), i=1, \cdots, m \} \bigcup \{ Z_1, \cdots , Z_r \}. $ For any sets of real numbers $b_1, b_2, \cdots, b_k $, we denote by $b_{(-1)}, b_{(-2)}\cdots , b_{(-k)}$ the reversely ordered numbers (descending). So, $ b^+_{(-i)} (\xi), i=1, \cdots , m+r $ is a set of $m+r$ numbers ordered from largest to smallest. With these sets of numbers defined, the denominator of (2.1) can be written as $$ \sum {E_{\beta}}\left ( P_1 P_2 \cdots P_m \times \prod_{r-c} P[Z_i, \infty) \right ) = \sum_{\xi } \left ( \pm \prod_{i=1}^{m+r} ~ \frac{ i-1 + \beta [b^+_{(-i)}(\xi) , \infty) } { i-1 + \beta (R^+) } \right )~, $$ where the summation is over $2^m$ different $\xi $'s. We can similarly compute the numerator of (2.1) except there is one more term, $P[u, \infty)$, included with the right censored observations. Define $ \{ b^{+}_j(\xi) \} = \{ c_i(\xi), i=1, \cdots , m \} \bigcup \{ Z_1, \cdots , Z_r , u \}. $ {\bf Theorem 2} {\sl The nonparametric Bayes estimator of the survival function $S(u)= 1-F(u)$ with right censored and interval censored data under a Dirichlet process prior is $$ \hat S_D(u) = \frac{\sum {E_{\beta}} \left \{ P_1 P_2 \cdots P_m \times \prod_{r-c} P[Z_i, \infty) \times P[u, \infty) \right \} } { \sum {E_{\beta} } \left \{ P_1 P_2 \cdots P_m \times \prod_{r-c} P[Z_i, \infty) \right \} } ~, $$ $$ = \frac{\displaystyle \sum_{\xi} (-1)^{\sum \xi_s} \prod_{i=1}^{m+r+1} ~ \frac{i-1 + \beta[ b^{+}_{(-i)} (\xi), \infty )}{ i-1 + \beta(R^+)}} {\displaystyle \sum_{\xi} (-1)^{\sum \xi_s} \prod_{i=1}^{m+r} ~ \frac{i-1 + \beta [b_{(-i)}(\xi), \infty) }{ i-1 + \beta (R^+)} }~. \eqno (2.4) $$ The sums in (2.4) are over all $2^m$ possible $\xi $'s. } Admittedly the two summations above involves $2^m$ terms when there are $m$ interval censored observations. Also, in the summations, there are both positive and negative terms that will cancel to a large extend. % and diminishing significant digits. Rounding errors will be magnified if we use (2.4) directly. Our purpose here is to show that an explicit formula exists for the Bayes estimator. Simplifications/alternative formulae are desirable and will be pursued in the future. %%to yield more computational friendly formulas when actual %computation of the estimator take place with %data containing many interval censored observations. {\bf Remark}: From Lemma 1 and Theorem 2, we can infer that the limit of the Bayes estimator (2.4) when the $\alpha$ measure approaches zero is a step function, at least for $u< $ maximum observed value. This is because all the ${E_{\beta}}$ involved will be step functions according to Lemma 1. We can also infer that when the $\alpha$ measure approaches zero, the Bayes estimator (2.4) is a rational, since the ${E_{\beta}}$ involved are all rationals. %Let %$$ %D(\beta_{k+1}, \cdots , \beta_{m+1} ) = %\frac{ \prod_{j=k+1}^{m+1} \Gamma (\beta_j)}{ \Gamma ( \beta(R^+) ) } %= B(\beta_{k+1}, \beta_{k+2}) %B(\beta_{k+1}+\beta_{k+2}, \beta_{k+3}) \cdots %B(\sum_{j=k+1}^{m} \beta_j , \beta_{m+1} ) %$$ %and $d_{jv} = 0 $ except when $j = v$, in that case $d_{jj} = 1$. %also $ d_{m+1 v} \equiv 0 $. % %The following lemma treats the case of one interval %censored observation: an observation $X_i$ only known to be %inside $[a_w , a_t) $. % %(If $a_w = 0 $ i.e. $ w=k $ then we %got a left censored observation, otherwise, it represent %an interval censored observation.) % %{\bf Lemma 2} {\sl %Let $ a_w < a_t $, and %$p_i \geq 0 ; i= k+1, \cdots , m+1$ %but $p_w =0 , p_t =0$, then %\begin{eqnarray*} % & & D( \beta_{k+1}, \cdots , \beta _{m+1} ) \times %{E_{\beta}} \left \{ P[a_w , a_t) \prod_{i=k+1}^m [ P[a_i , \infty) ]^{p_i} %\right \} \\ % & & = \sum_{v=w+1}^t \left \{ \prod_{j=k+1}^m %B( \beta_j + d_{jv} , \sum_{r=j+1}^{m+1} (\beta_r + d_{rv} + p_{r-1} ) ) %\right \} \\ % & & = %\sum _{v=w+1}^t \left [ %\prod_{j=k+1}^m B( \beta_j , \sum_{r=j+1}^{m+1} (\beta_r + p_{r-1} ) ) %\right ] %\left \{ %\frac{\beta_v}{\beta_v + \sum_{v+1}^{m+1} (\beta_r + p_{r-1}) } %\times %\prod_{j=k+1}^{v-1} %\frac{\sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) }{ \beta_j + %\sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) } %\right \} \\ %& & = %\left [ %\prod_{j=k+1}^m B( \beta_j , \sum_{r=j+1}^{m+1} (\beta_r + p_{r-1} ) ) %\right ] %\left \{ %\sum_{v=w+1}^{t} \left ( %\frac{\beta_v}{\beta_v + \sum_{v+1}^{m+1} (\beta_r + p_{r-1}) } %\prod_{j=k+1}^{v-1} %\frac{\sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) } %{ \beta_j + \sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) } %\right ) %\right \} ~. %\end{eqnarray*} %} % %{\sc Proof}: %The expectation times $D( \beta_{k+1}, \cdots , \beta _{m+1})$ %is the integral % %$$ %\int ...\int _S ( 1- \sum_{i=k+1}^m x_i ) ^{\beta_{m+1} -1} %\prod_{j=k+1}^m x_j ^ {\beta_j -1} %(x_{w+1} + x_{k+2} \cdots + x_t ) %\prod_{j=k+1}^m (1- \sum_{r=k+1}^j x_r ) ^{p_j} %\prod_{k+1}^m dx_j %$$ %where $S$ is the $(m-k-1)$ dimensional simplex. %The above integral can be written as %$$ %\sum_{v=w+1}^t %\int ...\int _S ( 1- \sum_{i=k+1}^m x_i ) ^{\beta_{m+1} -1} %\prod_{j=k+1}^m x_j ^ {\beta_j -1 +d_{jv} } %\prod_{j=k+1}^m (1- \sum_{r=k+1}^j x_r ) ^{p_j} %\prod_{j=k+1}^m dx_j %$$ %Use Lemma 2 of Susarla and Van Ryzin (1976) %to each integral in the summation %we get the first equality. %By using $B(a+1,b)= \frac{a}{a+b}B(a,b)$ and %$B(a,b+1)= \frac{b}{a+b}B(a,b) $ to simplify the %result we get the second equality. % % %$$ %\left [ %\prod_{j=k+1}^m B( \beta_j , \sum_{r=j+1}^{m+1} (\beta_r + p_{r-1} ) ) %\right ] %\left \{ %\sum_{v=w+1}^{t} \left ( %\frac{\beta_v}{\sum_v^{m+1} \beta_r + \sum_{v}^{m} p_{r}) } %\prod_{j=k+1}^{v-1} %\frac{\sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) } %{ \sum_j^{m+1} \beta_r + \sum_{r=j}^{m} p_{r}} %\right ) %\right \} %$$ % %Comment: Compared to Susarla and Van Ryzin (1976) Lemma 2 (a), %our result has an extra factor in curly bracket above, due to %the interval censored observation $P[a_w, a_t) $. % \section*{\normalsize\bf 3. Examples} The examples presented here are hand-calculated or are obtained by using software we developed (Example 2 and the NPMLE in Example 1). We pay close attention to the limit of the Bayes estimator when $\alpha \to 0 $ in the Dirichlet prior (non-informative prior), and compare the estimator with the NPMLE. The software used here are packaged as {\bf R } (http://www.r-project.org/) packages and can be found at \begin{verbatim} http://www.ms.uky.edy/~mai/research/ \end{verbatim} %The software for computing NPMLE is also available at the above R site. To minimize additional notation, we recycle the notation used by SV as much as possible. %and express our estimator %as a modification to the %Susarla and Van Ryzin's estimator $\hat S$. This allows us %to focus on the changes of the Bayes estimator %due to a left/interval censored observation. % %To recycle the notation, in this section Assume $Z_{(k+1)}, \cdots , Z_{(m)} $ are the ordered, distinct censored (both right and left/interval) times among the sample (1.1). Assume there are no ties among the left/interval and right censored observations (but ties within right censored observations are allowed). At each censored observation $Z_{(i)}, k+1 \leq i \leq m$, let $\lambda_i $ be the number of right censored observations that equal $Z_{(i)}$. Thus if there are two right censored observations equal to $Z_{(i)}$, then $\lambda_i = 2$. If $Z_{(j)}$ is a left censored observation then $\lambda_j =0$. To make the notation consistent, we define $Z_{(k)} = 0 $ and $Z_{(m+1)} = \infty$. %For simplicity, assume now that there is %no tie among left %censored observations (this can be lifted later). Let $ N(u)$ be the number of uncensored and right censored observations that are larger then or equal to $u$, i.e., $ N(u) = \sum_{j: ~ \Delta_j =1} I_{[Z_j \geq u ]} + \sum_{i=k+1}^m \lambda_i I_{ [Z_{(i)} \geq u ]}~, $ and let $N^+(u) = N(u +)$. We reproduce SV's Bayes estimator (based only on the uncensored and right censored observations of the sample (1.1)) in a slightly modified form: For $ Z_{(l)} \leq u < Z_{(l+1)} $ with $k \leq l \leq m+1 $, $$ \hat S(u) = \frac{\alpha(u, \infty) + N^+(u)}{\alpha(R^+) + N^+(0) } \times \prod_{j=k+1}^l \left \{ \frac{\alpha[Z_{(j)}, \infty) + N(Z_{(j)} ) } {\alpha[Z_{(j)}, \infty) + N(Z_{(j)} )-\lambda_j } \right \} ~. \eqno (3.1) $$ We have changed two things: we added the nodes $Z_{(j)} $ for left/interval censored observations, though with zero $\lambda_j$'s; $n$ is replaced by $N^+(0) $. %Our estimator based on all observations, $\hat S_D (u) $, %will be equal to $\hat S(u)$ multiplied by a factor. {\bf Example 1} Here is an example with one left censored observation and four right censored observations. These are the data used by SV (1976) but with an added left censored observation at $Z = 4 $. The ordered observations with their censoring indicators are listed below in table 1. \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline $ Z_i's :$ & 0.8 & 1.0 & 2.7 & 3.1 & 4 & 5.4 & 7.0 & 9.2 & 12.1 \\ \hline $\Delta : $ & 1 & 0 & 0 & 1 & 2 & 1 & 0 & 1 & 0 \\ \hline \end{tabular} \\ \medskip Table 1. Data with one left and four right censored observations. \end{center} %There are 4 uncensored observations, therefore %$$ %Z_{(k)} = Z_{(4)} = 0, ~~ Z_{(k+1)} = Z_{(5)} = 1.0, ~~ Z_{(6)}=2.7, ~~ %Z_{(7)}=4, ~~ Z_{(8)} = 7.0, ~~ Z_{(9)} = 12.1, ~~ %Z_{(10)} = \infty %$$ %with %$$ %\lambda_5 = 1, ~~~~ \lambda_6 = 1, ~~~~ \lambda_7 = 0, ~~~~ %\lambda_8 = 1, ~~~~ \lambda_9 = 1. %$$ % %Using SV notation, $\lambda_5 $ is the number of censored observations %at the smallest censored observation time (=1.0) %hence $\lambda_5=1$. Let the Bayes estimator of SV based only on uncensored and right censored observations be $\hat S(u) $, i.e., as defined in (3.1). Our estimator that takes into account one left censored observation can be written as follows. For $ u > Z_{left}= 4 $, %$$ %\hat S_D (u) = %\hat S(u) \times %\frac{ \frac{\beta_1}{\beta_1 + \sum +1 } + %\frac{\beta_2}{\beta_2 + \sum +1} % + \frac{\beta_3}{\beta_3 + \sum + 1 } } %{ \frac{\beta_1}{\beta_1 + \sum_2^6 (\beta_r + \lambda_{r-1}) } %+ \frac{\beta_2}{\beta_2 + \sum_3^6 (\beta_r + \lambda_{r-1}) } + % \frac{\beta_3 }{\beta_3 + \sum_4^6 (\beta_r + \lambda_{r-1}) } } %$$ %Notice %%where $\beta_1 = \beta [0, Z_{(k+1)} ) $, and %%$\beta_2 = \beta [Z_{(k+1)}, Z_{(k+2)} ) $ etc., and %the measure $\beta$ is defined as %$\beta = \alpha + \sum_{uncensored} \delta_{Z_i}$, same as in SV. % %$$ %\hat S_D (u) = \hat S(u) \times %$$ %$$ %\frac{ \frac{\beta [0, Z_{(k+1)})}{\beta[0, \infty ) + \sum_2^6 +1 } %+ \frac{\beta [Z_{(k+1)}, Z_{(k+2)})}{\beta[Z_{(k+1)}, \infty)+\sum_3^6 +1} %\cdot \frac{ \beta[Z_{(k+1)}, \infty) + \sum_2^6 +1} % {\beta[0, \infty) + \sum_2^6 +1 } %+ \frac{\beta [Z_{(k+2)}, Z_{(k+3)})}{\beta[Z_{(k+2)}, \infty)+\sum_4^6 +1} %\cdot \frac{\beta[Z_{(k+1)}, \infty) + \sum_2^6 +1} %{\beta[0, \infty) + \sum_2^6 +1} \cdot %\frac{ \beta[Z_{(k+2)} , \infty ) + \sum_3^6 +1 } %{ \beta[Z_{(k+1)} , \infty ) + \sum_3^6 +1 } %} %{ DENO } %$$ % %where %\begin{eqnarray} %& & DENO = \frac{\beta [0, Z_{(k+1)} ) } %{\beta [0, \infty) + \sum_2^6 \lambda_{r-1} } + \\ %& & + \frac{\beta [Z_{(k+1)} , Z_{(k+2)} ) } %{\beta [Z_{(k+1)} , \infty) + \sum_3^6 \lambda_{r-1} } %\times \frac{\beta[Z_{(k+1)}, \infty) + \sum_2^6 \lambda_{r-1} } % {\beta[0, \infty) + \sum_2^6 \lambda_{r-1} } + \\ %& & + \frac{\beta [Z_{(k+2)} , Z_{(k+3)} ) } %{\beta [Z_{(k+2)} , \infty) + \sum_4^6 \lambda_{r-1} } %\times \frac{\beta[Z_{(k+1)}, \infty) + \sum_2^6 \lambda_{r-1} } % {\beta[0, \infty) + \sum_2^6 \lambda_{r-1} } %\times \frac{\beta [Z_{(k+2)} , \infty) + \sum_3^6 \lambda_{r-1} } % {\beta[Z_{(k+1)}, \infty) + \sum_3^6 \lambda_{r-1} } %\end{eqnarray} % %Now plug-in the value of $Z$'s and $\lambda$'s, we get %$$ %S_D(u) = \hat S(u) \times %\frac{ \frac{\beta [0, 1) }{\beta [0, \infty) + 4 +1 } %+ \frac{\beta [1, 2.7) }{\beta [1 , \infty) + 3 + 1} %\times \frac{ \beta[1, \infty) + 4 +1 }{ \beta[0, \infty) +4 +1 } %+ \frac{\beta [2.7, 4) }{\beta [2.7 , \infty) + 2 + 1 } %\times \frac{ \beta[1, \infty) + 4 +1 }{ \beta[0, \infty) +4 +1 } %\times \frac{ \beta[2.7, \infty) + 3+1 }{\beta[1, \infty) +3 +1 } %} %{ \frac{ \beta [0, 1.0 ) } % {\beta [0, \infty) + 4 } %+ \frac{ \beta [1.0 , 2.7 ) } % {\beta [1.0 , \infty) + 3 } %\times \frac{ \beta[1, \infty) + 4}{ \beta[0, \infty) +4 } + % \frac{ \beta [2.7 , 4 ) } % {\beta [2.7 , \infty) + 2 } %\times \frac{ \beta[1, \infty) + 4}{ \beta[0, \infty) +4 } %\times \frac{ \beta[2.7, \infty) + 3}{\beta[1, \infty) +3 } %} %$$ % %Finally, we plug in the $\beta = \alpha + \sum $ measure to obtain: after tedious but straightforward simplification we get \begin{eqnarray} \hat S_D(u) %= \hat S (u) \times \nonumber \\ %&&\frac{ %\frac{\alpha[0, 1) +1 }{\alpha(R^+)+4+1+4} + %%\frac{\alpha[1, 2.7)}{\alpha[1, \infty) +3 +1+3 } %\times \frac{\alpha[1, \infty)+4+1+3}{\alpha(R^+)+4+1+4 } + %\frac{\alpha[2.7,4) +1 }{\alpha[2.7,\infty) +2 + 1 +3 } %\times \frac{\alpha[1, \infty)+4+1+3}{\alpha(R^+)+4+1+4 } %\times \frac{\alpha[2.7, \infty)+3+1+3}{\alpha[1, \infty)+3+1+3} %} %{ %\frac{\alpha [0, 1) + 1}{\alpha (R^+) +4+4 } + %\frac{\alpha [1, 2.7)}{\alpha [1, \infty ) +3+3 } %\times \frac{\alpha[1, \infty)+4+3}{\alpha(R^+)+4+4 } + %\frac{\alpha [2.7, 4) + 1}{\alpha [2.7, \infty) +2 +3 } %\times \frac{\alpha[1, \infty)+4+3}{\alpha(R^+)+4+4 } %\times \frac{\alpha[2.7, \infty)+3+3}{\alpha[1, \infty)+3+3} %} \nonumber \\ \nonumber \\ = \hat S (u) \times \frac{ \frac{\alpha[0, 1) +1 }{\alpha(R^+)+9} + \frac{\alpha[1, 2.7)}{\alpha[1, \infty)+7} \times \frac{\alpha[1, \infty)+8}{\alpha(R^+)+9} + \frac{\alpha[2.7,4)+1}{\alpha[2.7,\infty)+6} \times \frac{\alpha[1, \infty)+8}{\alpha(R^+)+9} \times \frac{\alpha[2.7, \infty)+7}{\alpha[1, \infty)+7} } { \frac{\alpha [0, 1)+1}{\alpha (R^+)+8} + \frac{\alpha [1, 2.7) }{\alpha [1, \infty ) +6} \times \frac{\alpha[1, \infty)+7}{\alpha(R^+)+8} + \frac{\alpha [2.7, 4) + 1}{\alpha [2.7, \infty) +5} \times \frac{\alpha[1, \infty)+7}{\alpha(R^+)+8} \times \frac{\alpha[2.7, \infty)+6}{\alpha[1, \infty)+6} } ~. \nonumber \end{eqnarray} %Clearly, the extra factor above due to a left censored %observation is $ > 0 $ and $ < 1$. %When $u \to \infty$ %the estimator $\hat S_D (u) \to 0$ since $\hat S(u) \to 0$. % %The first factor, $\hat S (u)$, took many different forms for %$u> 4$, %but the second factor due to left censoring is the same, and not %depending on $u$ as long as $u$ is in this region. For $u$ in other time intervals, the estimator can be similarly expressed as the product of $\hat S(u)$ and some other term, the details are omitted. The plot of the Bayes estimator is given in Figure 1. Next we compute the limit of the Bayes estimator. %when $\alpha \to 0 $. %Notice for those $u$ ($u<4$) %the second factor does change. But the denominators of %the second factor are all the same %for these time intervals, as they did not involve time, $u$. %We shall denote the denominator of the second factor of (2.6) above %by $deno$. %(2) case 2: for $ u < Z_{left}=4 $. This must be sub-divided into %three cases. % %{\bf (2-1)}. Assume $Z_{(k+2)} < u < Z_{(k+3)} $ i.e. %$2.7 < u < 4 $ we have % %$$ %\hat S_D (u) = \hat S(u) \times %\frac{ %\frac{\beta [0, Z_{(k+1)})}{\beta[0, \infty) + \sum +1 } %+ \frac{\beta [Z_{(k+1)}, Z_{(k+2)})}{\beta[Z_{(k+1)}, \infty) + \sum +1} %\cdot \frac{\beta[Z_{(k+1)}, \infty ) + \sum +1} % {\beta[0, \infty ) + \sum +1} %+ \frac{\beta [Z_{(k+2)}, u)}{\beta[Z_{(k+2)}, \infty) + \sum +1} %\cdot \frac{\beta[Z_{(k+1)}, \infty ) + \sum +1} % {\beta[0, \infty ) + \sum +1} %\cdot \frac{\beta[Z_{(k+2)}, \infty ) + \sum +1} % {\beta[Z_{(k+1)}, \infty ) + \sum +1 } %+ \frac{\beta [u, Z_{(k+3)})}{\beta[u, \infty) + \sum } %\cdot \frac{\beta[Z_{(k+1)}, \infty ) + \sum +1} % {\beta[0, \infty ) + \sum +1} %\cdot \frac{\beta[Z_{(k+2)}, \infty ) + \sum +1} % {\beta[Z_{(k+1)}, \infty ) + \sum +1} %\cdot \frac{\beta[u, \infty ) + \sum +1} % {\beta[Z_{(k+2)}, \infty ) + \sum +1} %} %{ %DENO %} %$$ % %Next we substitute the $beta = \alpha + \sum $ above and %the expression for SV estimator, this %further divide the formula into 2 intervals: % %{\bf (2-1-1)}. For $2.7 \leq u < 3.1$ the estimator is given by % %$$ %\frac{\alpha[u, \infty) + 5}{\alpha(R^+) +8 } ~ %\frac{\alpha[1, \infty) + 7}{\alpha[1, \infty) +6 } ~ %\frac{\alpha[2.7, \infty) + 6 }{\alpha[2.7, \infty) + 5 } %\times %\frac{ %\frac{\alpha[0, 1) +1 }{\alpha(R^+)+4+1+4} + %\frac{\alpha[1, 2.7)}{\alpha[1, \infty) +3 +1+3 } + %\frac{\alpha[2.7,u)}{\alpha[2.7,\infty) +2 + 1 +3 } + %\frac{\alpha[u, 4) +1 }{\alpha[u, \infty) +2 +3 } %} %{ %deno %} %$$ % %and %{\bf (2-1-2)} for $3.1 \leq u < 4$ the estimator is given by % %\begin{eqnarray} %\frac{\alpha[u, \infty) + 4}{\alpha(R^+) +8 } & & ~ %\frac{\alpha[1, \infty) + 7}{\alpha[1, \infty) +6 } ~ %\frac{\alpha[2.7, \infty) + 6 }{\alpha[2.7, \infty) + 5 } \\ % && \times %\frac{ %\frac{\alpha[0, 1) +1 }{\alpha(R^+)+4+1+4} + %\frac{\alpha[1, 2.7)}{\alpha[1, \infty) +3 +1+3 } + %\frac{\alpha[2.7,u) +1 }{\alpha[2.7,\infty) +2 + 1 +3 } + %\frac{\alpha[u, 4) }{\alpha[u, \infty) +2 +2 } %} %{ %deno %} %\end{eqnarray} % % % % %{\bf (2-2)}. Assume $ Z_{(k+1)} < u < Z_{(k+2)} $ i.e. $1.0 < u < 2.7$, % %$$ %\hat S_D (u) = \hat S(u) \times %\frac{ %\frac{\beta [0, Z_{(k+1)})} % {\beta[0, \infty) + \sum_2^6 \lambda_{r-1} +1 } + %\frac{\beta [Z_{(k+1)}, u)} % {\beta[Z_{(k+1)}, \infty) + \sum_3^6 \lambda_{r-1} +1} + %\frac{\beta [u, Z_{(k+2)})} % {\beta[u, \infty) + \sum_3^6 \lambda_{r-1} } + %\frac{\beta [Z_{(k+2)}, Z_{(k+3)})} % {\beta[Z_{(k+2)}, \infty) + \sum_4^6 \lambda_{r-1} } } %{ %DENO %} %$$ % %\begin{eqnarray} % = \frac{\alpha[u, \infty) + 6}{\alpha(R^+) +8 } ~ %\frac{\alpha[1, \infty) + 7}{\alpha[1, \infty) +6 } %\times %\frac{ %\frac{\alpha[0, 1) +1 }{\alpha(R^+)+4+1+4} + %\frac{\alpha[1, u)}{\alpha[1, \infty) +3 +1+3 } + %\frac{\alpha[u,2.7)}{\alpha[u,\infty) +3 +3 } + %\frac{\alpha[2.7, 4) +1 }{\alpha[2.7, \infty) +2 +3 } %} %{ %deno %} %\end{eqnarray} % % % % % %{\bf (2-3)}. Assume $Z_{(k)} < u < Z_{(k+1)}$ i.e. $ 0 < u < 1.0 $, %we have %$$ %\hat S_D (u) = \hat S(u) \times %\frac{ \frac{\beta [0, u)}{\beta[0, \infty) + \sum +1 } %+ \frac{\beta [u, Z_{(k+1)})}{\beta[u, \infty) + \sum } %+ \frac{\beta [Z_{(k+1)}, Z_{(k+2)})}{\beta[Z_{(k+1)}, \infty) + \sum } %+ \frac{\beta [Z_{(k+2)}, Z_{(k+3)})}{\beta[Z_{(k+2)}, \infty) + \sum } } %{ %NENO %%\frac{\beta [0, Z_{(k+1)} ) } %%{\beta [0, \infty) + \sum_2^6 \lambda_{r-1} } %%+ \frac{\beta [Z_{(k+1)} , Z_{(k+2)} ) } %%{\beta [Z_{(k+1)} , \infty) + \sum_3^6 \lambda_{r-1} } + %% \frac{\beta [Z_{(k+2)} , Z_{(k+3)} ) } %%{\beta [Z_{(k+2)} , \infty) + \sum_4^6 \lambda_{r-1} } %} %$$ % %Notice when $u \to 0+ $ the multiply factor approaches $1-$. %Since the Susarla and Van Ryzin estimator satisfy $S(0)=1$, it follows %that $S_D(0) = 1 $. % %Plug in the measure $\beta = \alpha + \sum $ and SV estimator, we can list %the following 2 sub intervals : % %{\bf (2-3-1)} For $ 0 \leq u < 0.8 $ % %$$ %\frac{\alpha [u, \infty ) + 8 }{ \alpha (R^+) +8} \times %\frac{ %\frac{\alpha[0, u)}{\alpha(R^+)+4+1+4} + %\frac{\alpha[u, 1) + 1}{\alpha[u, \infty) +4 +4 } + %\frac{\alpha[1,2.7)}{\alpha[1,\infty) +3 +3 } + %\frac{\alpha[2.7, 4) +1 }{\alpha[2.7, \infty) +2 +3 } %} %{ %deno %} %$$ % %{\bf (2-3-2)} For $ 0.8 \leq u < 1.0$ % %$$ %\frac{\alpha[u, \infty) + 7}{\alpha(R^+) +8 } %\times %\frac{ %\frac{\alpha[0, u) +1 }{\alpha(R^+)+4+1+4} + %\frac{\alpha[u, 1)}{\alpha[u, \infty) +4 +3 } + %\frac{\alpha[1,2.7)}{\alpha[1,\infty) +3 +3 } + %\frac{\alpha[2.7, 4) +1 }{\alpha[2.7, \infty) +2 +3 } %} %{ %deno %} %$$ % When $\alpha \to 0$, the SV estimator, $\hat S (u)$, has as a limit the Kaplan-Meier estimator $S_{KM}$. %Our estimator has the limit as $\alpha \to 0$ : %For $0 \leq u < 0.8$, the limit is $ S_{KM} \times 1 = 1\times 1 $. % %For $0.8 \leq u < 1 $, the limit is %$ S_{KM} \times \frac{26}{27} = \frac{7}{8} \times \frac{26}{27} %=0.8425926$. % %For $1 \leq u < 2.7 $, same as above. %% we have $ S_{KM} \times \frac{ 5 }{ 1/8 + (1/5) (7/8) } $. % %For $2.7 \leq u < 3.1 $, same as above. %%we have $ S_{KM} \times \frac{ 1/9 + 1/5 }{ 1/8 + (1/5) (7/8) } $. % %For $3.1 \leq u < 4 $, we have %$ S_{KM} \times \frac{70}{81} = %\frac{7}{8}\times\frac{4}{5}\times\frac{70}{81} %=0.6049383$. % %For $4 \leq u < 5.4 $, same as above. %%we have %%$ S_{KM} \times \frac{ 1/9 + (1/6) (8/9) }{ 1/8 + (1/5) (7/8)} $. % %For $5.4 \leq u < 9.2 $, we have %$ S_{KM} \times \frac{70}{81} = \frac{7}{8}\times\frac{4}{5} %\times\frac{3}{4}\times\frac{70}{81} %=0.4537037$. %%\frac{ 1/9 + (1/6) (8/9) }{ 1/8 + (1/5) (7/8)} $. For $9.2 \leq u < 12.1 $, the limit of our estimator is $ S_{KM} \times \frac{70}{81} = \frac{7}{8}\times\frac{4}{5} \times\frac{3}{4}\times\frac{1}{2}\times\frac{70}{81} =0.2268519$. For $u$ in other intervals the limit can be computed similarly. % \frac{70}{81} = \frac{ 1/9 + (1/6) (8/9) }{ 1/8 + (1/5) (7/8)} $. We plot the estimator with $\alpha (u, \infty ) = B \exp (- \theta u)$. The plot shows estimators for $B=8, \theta=0.12 $ and $B=0.001, \theta=0.12$. The latter is indistinguishable in appearance with the limit just calculated. %We also plot the Susarla and %Van Ryzin estimator without the left censored observation. Computation of the NPMLE for doubly censored data can be done by EM type iteration (see Turnbull (1974), Chen and Zhou (2003)). For the data in Table 1 we obtain the values in Table 2: \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline t & 0-0.8 & 0.8-3.1 & 3.1-5.4 & 5.4-9.2 & 9.2-12.1 \\ \hline NPMLE & 1 & .8457284 & .6028477 & .4521358 & .2260679 \\ \hline Limit Bayes & 1 & .8425926 & .6049383 & .4537037 & .2268519 \\ \hline \end{tabular} \\ \medskip Table 2. NPMLE and limit of Bayes estimator for data in Table 1 \end{center} The differences between the NPMLE and the limit of the nonparametric Bayes estimator are small but real. The likelihood of the distribution in Table 2 is larger than those of the limit of the Bayes estimator: $3.70674 \times 10^{-5}$ vs. $3.704924 \times 10^{-5} $. %For the rest of this section the parameter measure $\alpha(\cdot)$ of the %Dirichlet process prior is always taken to be of the form %$ \alpha (u, \infty ) = B \exp( - \theta u) $, where $B >0 $ and %$\theta > 0 $ %are two one-dimensional parameters. %{\bf Example 2} %Let us look at another example with one interval censored observation %and 4 right censored observations. These are the data used by %Susarla and Van Ryzin (1976) %except with an added interval censored observation at $ [2, 4) $. % %\begin{center} %\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline %$ Z_i's :$ & 0.8 & (1.0) & [2 & (2.7) & 3.1 & 4] & 5.4 & (7.0) & 9.2 & (12.1) %\\ \hline %$\Delta: $ & 1 & 0 & & 0 & 1 & & 1 & 0 & 1 & 0 %\\ \hline %\end{tabular} \\ %\medskip %Table 3. Data with one interval and four right censored observations. %\end{center} {\bf Example 2}: We took the first ten observations from the breast cosmesis data with radiation of Finkelstein and Wolfe, as reported by Fay (1999). Out of the ten, there are four right censored observations, four interval censored observations and two left censored observations (i.e. interval censored with left-ends as 0). Data: $[45, \infty), [6,10), [0,7), [46,\infty),[46,\infty),[7,16), [17, \infty), [7,14), [37, 44), [0,8)$ . We computed the nonparametric Bayes estimator with $\alpha (u, \infty) = B \exp( - \theta u ) $. The resulting estimator with $B=8$ and $\theta=0.3$ is computed using the software we developed and is plotted in Figure 2. In the following two examples, the Bayes estimators are obtained with formula (2.4) and then we let $\alpha(R^+) \to 0$ to obtain the limit. The NPMLE's are also calculated, not by software but analytically. {\bf Example 3} Here we took a small example with one left and one right censored observation. The NPMLE and the limit of the non-parametric Bayes estimator turn out to be exactly the same. \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline $ Z_i's :$ & $Z_{(1)}$ & $Z_{(2)}$ & $Z_{(3)}$ & $Z_{(4)}$ & $Z_{(5)}$ \\ \hline $\Delta : $ & 1 & 0 & 2 & 1 & 1 \\ \hline Jump of $1-\hat F(u)$ & 0.4 & 0 & 0 & 0.3 & 0.3 \\ \hline \end{tabular} \\ \medskip Table 4. Data with one left and one right censored observation. \end{center} %In this case the limit of nonparametric Bayes estimator %and the NPMLE both have different jumps and are indicated in the table. {\bf Example 4} The order of two censoring indicators in the above table are switched and the NPMLE is different from the limit of Bayes estimator. The limit of Bayes estimator is not self-consistent either. To calculate the NPMLE, we first note for this data the NPMLE $F(\cdot )$ has only three jumps at $Z_{(1)}, Z_{(3)}$ and $Z_{(5)}$. Denote the jumps size by $p_1, p_2, p_3$. By symmetry we must have $p_1=p_3$. Using the constraint $\sum p_i =1$, we can reduce the likelihood, $L= p_1 (p_2+p_3) p_2 (p_1+p_2)p_3$, to a function of $p_2$ only. Straightforward calculation shows that $p_2 = \sqrt 5 / 5 $ maximizes the likelihood. Therefore $ p_1 = p_3 = (5 - \sqrt 5 )/10 $, which is the entry $0.2763932$ in the table. \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline $Z_i's :$ & $Z_{(1)}$ & $Z_{(2)}$ & $Z_{(3)}$ & $Z_{(4)}$ & $Z_{(5)}$ \\ \hline $\Delta : $ & 1 & 0 & 1 & 2 & 1 \\ \hline \hline Jump of limit Bayes & 0.28000 & 0 & 0.44000 & 0 & 0.28000 \\ \hline Jump of NPMLE & 0.2763932 & 0 & 0.4472136 & 0 & 0.2763932 \\ \hline \end{tabular} \\ \end{center} {\bf Remark}: %It is interesting to note that the limit of the Bayes %estimator always takes rational number values, at least for $t$ before %the last observation. % Example 4 shows that with positive probability, the NPMLE $ 1 - \hat F( \cdot )$ for doubly/interval censored data can take irrational values. A closer look at the example provides some insight as why the estimators are different, as described in the next section. %The likelihood function with doubly censored observations %is, in general, non-linear (unless the right censored observations %are above all the left censored observations). %This makes the maximum an irrational number. {\bf Remark}: Example 3 and 4 reveal two different situations. The difference is that left and right censored data overlap (right censored observation is smaller then the left censored observation) in Example 4. The overlap in Example 3 is not real, since there is no probability mass inside the overlap. \section*{\normalsize\bf 4. Limit of Bayes and NPMLE } In this section we formally summarize some results concerning the limit of the Bayes estimator and the NPMLE in the doubly/interval censored data case. The argument below is valid for {\em any} prior, not just the Dirichlet process prior. {\bf Theorem 3}. {\sl Suppose a sequence of priors $\pi_v; v=1, 2, \cdots $, is such that the (nonparametric) Bayes estimators $1- \hat F_v (\cdot) $ under squared error loss, converge to the Kaplan-Meier estimator whenever the data has only right censoring. Then this same sequence of Bayes estimators cannot converge, in general, to the NPMLE for interval/doubly censored data. } Proof: The Bayes estimator under squared error loss can be written as $$ 1- \hat F_v (u) = \frac{ {E}_{\pi} P[u, \infty ) L_F(data)}{ {E}_{\pi} L_F(data) }~, $$ where $L_F(data)$ is the likelihood of the data when its distribution is $F$. The assumption of the Theorem for right censored data says that, as $v \to \infty$, we always have $$ \frac{\displaystyle E_{\pi} \{ P[u, \infty) \prod_{r-c} P[x_i , \infty ) \prod_{uncensor} P(\{ x_j \} ) \} } { \displaystyle E_{\pi} \prod_{r-c} P[x_i , \infty ) \prod_{uncensor} P(\{ x_j \} ) \} } \to 1- F_{K-M} (u) ~. \eqno (4.1) $$ Notice the Kaplan-Meier estimator, $F_{K-M}(u)$, is always rational valued. Now we look at a particular sample configuration with just one left censored observation, for example the data in Example 4. The Bayes estimator for these data can be written as $$ \frac{ \displaystyle E_{\pi} P[u, \infty ) P(\{ Z_1 \} ) P[ Z_2, \infty) P(\{ Z_3\} ) P[0, Z_4) P(\{ Z_5\}) } {\displaystyle {E}_{\pi} P(\{ Z_1 \} ) P[ Z_2, \infty) P(\{ Z_3\} ) P[0, Z_4) P(\{ Z_5\}) } ~. $$ Let us use the notation $P(Z)= P(\{Z\})$, and $P( Z^+ ) = P[Z, \infty ) $. Write $ P[0, Z_4 ) = 1 - P[Z_4, \infty)= 1- P( Z_4^+) $ and expand to get %$$ %= \frac{ \displaystyle {E}_{\pi} %P[u, \infty ) P(\{ Z_1 \} ) P[ Z_2, \infty ) %P(\{ Z_3\} ) P(\{ Z_5\}) - {E_{\beta}}_{\pi} % P[u, \infty ) P(\{ Z_1 \} ) P[ Z_2, \infty) %P(\{ Z_3\} ) P[Z_4, \infty ) P(\{ Z_5\}) %} %{\displaystyle {E_{\beta}}_{\pi} %P(\{ Z_1 \} ) P[ Z_2, \infty) %P(\{ Z_3\} ) P[0, Z_4) P(\{ Z_5\}) } %$$ $$ = \frac{ \displaystyle {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) P (u^+) - {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) P(Z_4^+) P(u^+) } { {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) - {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) P(Z_4^+)} \eqno (4.2) $$ If we divide the numerator of (4.2) by $ {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) P (u^+)$ then, as $v \to \infty$, the numerator will converge to the limit $1 - [1-F^*_{K-M} (Z_4)] $ according to (4.1). Here the Kaplan-Meier estimator is based on three uncensored observations: $Z_1, Z_3, Z_5$ and two right censored observations, $ Z_2$ and $ u $. Similarly, if we divide the denominator of (4.2) by ${E} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+)$ then it has the limit $ 1- [1-F^{**}_{K-M} ( Z_4)]$, where the Kaplan-Meier estimator is based on three uncensored observations, $Z_1, Z_3, Z_5$ and one right censored observation $ Z_2 $. In other words, multiply (4.2) by $$ \frac{ {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) } { \displaystyle {E}_{\pi} P(Z_1) P(Z_3) P(Z_5) P(Z_2^+) P (u^+) } \eqno (4.3) $$ to produce a rational limit. The factor (4.3) itself has a rational limit as $v \to \infty$ ( $= [ 1-F^{**}_{K-M} (u) ] ^{-1} $). This implies that the limit of (4.2), as $v \to \infty $, is $$ \frac{ F^*_{K-M}(Z_4) } { F^{**}_{K-M} (Z_4) } \times [ 1-F^{**}_{K-M} (u) ] ~. $$ %According to (4.1) %as $v \to \infty $, %the expression (4.2) has a limit %involving five values all equal to some %Kaplan-Meier estimators. %This imply the expression has a rational numbers value. But that cannot be the NPMLE as Example 4 shows the NPMLE is irrational. This also serves as the proof for the interval censored case, since the left censored observation is just $[0, Z_{(4)})$ interval censored. $\diamondsuit$ {\bf Corollary 1}. {\sl There is no sequence of priors, $\pi_v$ such that the resulting sequence of Bayes estimators under squared error loss, $1-F_v (\cdot)$, always converges to the NPMLE in the interval/doubly censored data case. } Proof: Suppose, to the contrary, there is such a sequence of priors. Since right censoring is a special case of double/interval censoring (zero left censoring or all intervals are of the form $ [ a_i , \infty ) $), this sequence of estimators must converge to the Kaplan-Meier estimator with such data. But by Theorem 3, such sequence cannot converge to the NPMLE for doubly/interval censored data case in general. $\diamondsuit$ \section*{\normalsize\bf 5. Discussion} The formula (2.4) has $2^m$ terms when there are $m$ interval censored observations. While we do not have a formal proof that the computation of the Bayes estimator cannot be reduced to polynomial order, it is not hard to see that the computation is equivalent to $$ \int \cdots \int \prod_j (\sum_{r=1}^j x_r ) \prod_j (1- \sum_{r=1}^j x_r) (1-\sum x_j)^{ \beta_m } \prod x_j ^{\beta_j} \prod dx_j~ $$ on the region $x_j > 0 $ and $\sum x_j \leq 1 $. %We believe it cannot be reduced to the polynomial order. {\bf Remark}: The irrational value of NPMLE with doubly censored data also implies that the EM algorithm, if started from the Kaplan-Meier estimator, cannot converge in a finite munber of steps in general. %{\bf Acknowlegement}: %I thank C. Srinivasan for many helpful discussions. %Using the Dirichlet process prior, %we obtained the nonparametric Bayes estimator of $1-F( \cdot )$ based %on right/left/interval censored data. The limit of the Bayes estimator %(as $\alpha$ measure of the Dirichlet approaches zero) %when it is well defined, is always a rational number. % %{\bf Example 4} We add one more left censored observation at 8 to data in %Table 1, making it with two left censored data. The resulting %Bayes estimators with $B=4$ and $B=0.01$ are given in the plot. % %For positive integer $I$, we have %$B(a+I,b)= \frac{a}{a+b} \frac{a+1}{a+1+b} \cdots %\frac{a+I-1}{a+I-1+b}B(a,b)$ to simplify the expression. % % %Suppose the distinct observed censoring times are %$Z_{(k+1)} , \cdots , Z_{(m)} $. %Each interval censored time would contribute 2 times, %each left/right censored observation would contribute 1 %times. % %The total number of left censored observation %is denoted by L. %The total number of interval censored observation %is denoted by M. % % %A left censored observation $(Z_t, \Delta_t=2)$ gives the information %that $ X $ is less then $Z_t$ and its probability is %$$ %P(X < Z_t) = P(0 \leq X < Z_{(k+1)})+P(Z_{(k+1)} \leq X < Z_{(k+2)}) %+ \cdots + %P( Z_{(t-1)} \leq X < Z_(t) ) %\eqno (3.1) %$$ %those probabilities on the right is distributed as a Dirichlet %process, thus the probability is represented by %$x_{k+1} + x_{k+2} + \cdots + x_t = \sum_{i=k+1}^t x_i$. % %an interval censored observation % %%The $L$ left censored observations $Z_{t(1)}, ... Z_{t(L)}$ would enter %%the picture via the following lemma %% %%{\bf Lemma 3}: %% %%$$ %%\prod_{j=1}^L ( \sum_{i=k+1}^{t(j)} x_i ) %%= \sum_{ k+1 \leq i(j) \leq t(j) } %%(x_{i(1)} x_{i(2)} \cdots x_{i(L)} ) %%\eqno (3.2) %%$$ %% %% %% %%For $ j $ from $k+1$ to $m$, %%let %% %%$$ %%h_j[ i(1), i(2), \cdots, i(L)] = h_j [ \cdot ] = %\sum_{l=1}^L I_{[i(l)=j]} %\eqno (3.3) %$$ % %i.e. $ h_i[ \cdot ] = $ \# of index $i(l), ~ 1 \leq l \leq L$ %that are equal to $j$; % %Since there is no left censored observation that are just %``less than infinity'', the index $h_{m+1} $, so to speak, %is always zero. % %%%% We of course have $h_{k+1} + h_{k+2} + \cdots + h_m = L $???? % %{\bf Theorem 2} % %The expectation {E_{\beta}} %can be computed as % %$$ %\sum_{ k+1 \leq i(j) \leq t(j) } \int \cdots \int _S %(1- \sum_{j=k+1}^m x_j ) ^{\beta_{m+1} -1 } %\prod_{s=k+1}^{m} %x_s ^{\beta_s -1 + h_s[i(1), i(2), \cdots , i(L)] } %\prod_{j=k+1}^m ( 1-\sum)^{p_j} %\prod_{j=k+1}^m dx_j %$$ % %$$ %= \sum_{ k+1 \leq i(j) \leq t(j) } \left \{ %\prod_{j=k+1}^m B(\beta_j + h_j [i(1), i(2), \cdots, i(L)] , %\sum_{r=j+1}^{m+1} (\beta_r + p_{r-1}) ) \right \} %$$ % %This simplify to % %$$ %\prod_{j=k+1}^m B( \beta_j , \sum ) %\sum_{ k+1 \leq i(j) \leq t(j) } \left \{ %\prod_{r=k+1}^m %( \frac{\beta_r}{\beta_r + \sum} %\frac{\beta_r +1 }{\beta_r +1 + \sum } \cdots %\frac{\beta_r + h_r[ \cdot ] -1}{\beta_r + h_r[ \cdot ] -1 + \sum } ) %\right \} %$$ % %$$ %= \left [ \prod_{j=k+1}^m B( \beta_j , \sum ) %\right ] %\sum_{ k+1 \leq i(j) \leq t(j) } \left \{ %\prod_{r=k+1}^m \left ( %\prod_{s=0}^{h_r[ \cdot ] -1} \frac{ \beta_r +s}{ \beta_r +s + \sum } %\right ) %\right \} %$$ % %{\bf Theorem 3 }: The Bayes estimator of the survival function %with many left censored observations %is given by % % %It is interesting to study what the estimator's limit will be when %the parameter in the Dirichlet prior approaches zero measure. We only considered left truncation in this paper. Other type of truncation, for example, interval truncation do not seems to be able to be handled by the method here. $ \frac{a}{ A - B } \not= C- D $ When those data are present, we may have to resort to the MCMC type method of computing the nonparametric bayes estimator. On the other hand, other types of censoring, for example, interval censoring with left truncation can be handled similarly by the same method used here. For simplicity we assumed the data has no tie. The final form of the estimator is such that tie do not make a difference. When $\min(Y_i) \to 0$ and $\alpha (R^+) \to 0$ then the NP Bayes estimator with left truncated and right censored data approaches the product limit estimator of Jewell, Tsai, Wang. This is to be expected, since in the interval $( 0 , \min(Y_i)$ there is no data and no sample information at all. The Bayes estimator will simply be identical to the mean of the prior in this interval. 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Nonparametric Bayes estimator of survival functions for doubly/interval censored data. {\em Statistica Sinica}, {\bf 14}, 533-546. \end{list} \medskip {\sc \noindent \parbox[t]{2.8in}{ Department of Statistics \\ University of Kentucky \\ Lexington, KY 40506-0027 } } \medskip \begin{center} \mbox{\psfig{file=homeworkplot1.ps,height=3.2in,width=6.3in}} Figure 1: Plot for Example 1. \end{center} \begin{center} \mbox{\psfig{file=homeworkplot3.ps,height=3.2in,width=6.3in}} Figure 2: Plot for Example 2. \end{center} \end{document}