Question

(b) Suppose that xi, . . . ,Xn are a random sample of lifetimes for individuals diagnosed with a certain disease. Assume a model with λ, x>0, where k is fixed and known Interest is in the parameter Ç P(X > 25A) which gives the probability that an individual will survive more than 25 years with the disease. It can be shown that the cumulative distribution Determine the MLE of

Answer 1

$f(x;\lambda)=\frac{k}{\lambda}\left(\frac{x}{\lambda} \right )^{k-1}\exp\left(-\left(\frac{x}{\lambda} \right )^k\right),~x>0\\\\ The~likelihood~function~of~\lambda=L(\lambda)=k^n\left(\prod_{i=1}^nx_i \right )^{n-1}(1/\lambda)^{nk}\exp\left(-\sum_{i=1}^n\left(\frac{x_i}{\lambda} \right )^k\right)\\ \log L(\lambda)=l(\lambda)=n\log k&plus;(n-1)\sum_{i=1}^n\log x_i-nk\log \lambda-\sum_{i=1}^n\left(\frac{x_i}{\lambda} \right )^k\\ or,~\frac{\mathrm{d} l(\lambda)}{\mathrm{d} \lambda}=-\frac{nk}{\lambda}&plus;\frac{k}{\lambda^{k&plus;1}}\sum_{i=1}^nx_i^k=0~or~\hat{\lambda}=\left(\frac{1}{n}\sum_{i=1}^nx_i^k \right )^{\frac{1}{k}}\\ \left.\begin{matrix} \frac{\mathrm{d}^2 l(\lambda)}{\mathrm{d} \lambda^2} \end{matrix}\right|_{\lambda=\hat{\lambda}}=\frac{nk}{\hat{\lambda}^2}-\frac{k(k&plus;1)n}{\hat{\lambda}^2}<0\\$

$Hence~MLE~of~\lambda=\hat{\lambda}=\left(\frac{1}{n}\sum_{i=1}^nX_i^k \right )^{\frac{1}{k}}\\ \zeta =P(X>25)=\int_{25}^\infty\frac{k}{\lambda}\left(\frac{x}{\lambda} \right )^{k-1}\exp\left(-\left(\frac{x}{\lambda} \right )^k\right)dx\\ =\int_{\left(\frac{25}{\lambda} \right )^{k}}^\infty \exp(-t)dt~~~~take,~\left(\frac{x}{\lambda} \right )^k=t\Rightarrow \frac{k}{\lambda}\left(\frac{x}{\lambda} \right )^{k-1}dx=dt\\ =\exp\left(-\left(\frac{25}{\lambda} \right )^{k} \right ).\\ We~know~if~T~is~the~MLE~of~\theta~and~\xi(\theta)~is~one~to~one~function~of~\theta,~\\ then~\xi(T)~is~the~MLE~of~\xi(\theta).\\ Since~\exp\left(-\left(\frac{25}{\lambda} \right )^{k} \right )~is~one~to~one~function~of~\lambda,~(\lambda>0)~so~\exp\left(-\left(\frac{25}{\hat{\lambda}} \right )^{k} \right )~is\\ MLE~of~\zeta ~where~\hat{\lambda}=\left(\frac{1}{n}\sum_{i=1}^nX_i^k \right )^{\frac{1}{k}}.$

$Note~that~expression~of~F(x;\lambda)~is~not~correct,\\ Correct~expression~is\\ F(x;\lambda)=1-e^{-(x/\lambda)^k}$