Homework 2 . Due Sept. 16, 2016 

1. Given i.i.d. right censored observations (T_1, \delta_1) ... (T_n, \delta_n). Further assume random independent censorship
   model and our interest is in the distribution of the lifetimes (not censor times).

   Assume a piecewise exponential model for the lifetimes with the hazard function given by 

  h(t) = \lambda_1 I[0 <= t < K_1] + \lambda_2 I[K_1 <= t < K_2] + \lambda_3 I[K_2 <= t < K_3] + \lambda_4 I[K_3 <= t < \infty].

  We assume the cut points 0 < K_1 < K_2 < K_3 < \infty are given and the 4 lambda's are parameters (must be positive).


 (a)  Find the MLE of lambda_1, lambda_2, lambda_3 and lambda_4 (when they exist. Under what condition some MLE may not exist?)

 (b)  Find the observed 4x4 Fisher information matrix and its inverse matrix. (assume all MLE in (a) exist)


 (c) what is the approx. variance of \hat \lambda_3  ?


2. For purely discrete r.v. X, define its cumulative hazard function H(x) as we did in class, (also as below)
   find and verify the inverse formula. i.e. from H(x) to CDF F(x).

    Define H(t) = \int_{-\infty}^t \frac{dF(s)}{1-F(s-)}  then we have the inverse formular

     1-F(t) = \prod_{s \leq t} [ 1- dH(s) ]

   (you may assume a distribution that have just 5 jumps, if that makes your life easier.)
 


3. For the "myel" data in our textbook, fit all the available parametric models to the data set use
   SAS proc lifereg and/or R function survreg
   [exponential; 2-parameter weibull, weibull with shape fixed at 2, 1, 0.5; gamma, etc].
   
   and plot the hazard functions of the fitted models in one plot. 

   A final comment: you are to use the variable "dur" in the "myel" as survival times when working on problem 3, you should also 
   use "status" in addition to "dur". But we do not use the variables "treat" and "renal" at this stage.


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 When a CDF F(t) is discrete, we know the corresponding H(t) is also discrete.

   Determine which inequality below hold and show it: For a given t that 1 > F(t) > 0

       (a) true or False:  1-F(t) > exp[ - H(t) ]


       (b) true or false:  1-F(t) < exp[ - H(t) ]

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