This is the question I am trying to solve. "Suppose that n items are being tested simultaneously, the items are independent, and the length of life of each item has the exponential distribution with parameter β. Determine the expected length of time until three items have failed"
I approached the problem like this. Since the sampled items are independent, and we are looking for time until the third failure, we can model this as gamma distribution with parameter alpha = 3 and beta = beta. Hence the Expected value is 3 / β.
But when I checked the solution, the correct answer is
Could you kindly explain why it cant be considered as gamma distribution but instead we need to find the expectation of sum of time until successive failures.
The problem is in Probablity and Statistics 4th ed Problem 11 on sec 5.7. Page 325
Thankyou.
Best Answer
The gamma distribution with parameters $\ 3,\beta\ $ is the distribution of the sum of three $\ \beta$-exponentially distributed independent random variables, so its mean is the expected value of time taken to test three items to failure sequentially. But this isn't the problem you're trying to solve, in which there are $\ n\ $ items being tested simultaneously. This means that when the first failure occurs, all the other tests are still running, so the times when any of the items in those other tests fail can't be earlier than the time when the first failure occurred.
Thus, if $\ f_1\ $ is the index of the item which fails first, then $$ X_{f_1}\le X_i\ \ \text{ for all }\ \ i\ , $$ or, equivalently, $$ X_{f_1}=\min\big(X_1,X_2,\dots,X_n\big)\ . $$
Now if $\ X_{f_1}=\min\big(X_1,X_2,\dots,X_n\big)\ $, then $\ X_{f_1}>x\ $ if and only if $\ X_i>x\ $ for all $\ i=1,2,\dots,n\ $. Therefore, \begin{align} \mathbb{P}\big(X_{f_1}>x\big)&=\mathbb{P}\big(X_1>x, X_2>x,\dots,X_n>x\big)\\ &=\prod_{i=1}^n\mathbb{P}\big(X_i>x\big)\\ &=e^{-n\beta x}\ , \end{align} and the cumulative distribution function of $\ X_{f_1}\ $ is given by \begin{align} \mathbb{P}\big(X_{f_1}\le x\big)&=1-\mathbb{P}\big(X_{f_1}>x\big)\\ &=1-e^{-n\beta x}\ . \end{align} Thus, $\ X_{f_1}\ $ is $\ n\beta$-exponentially distributed and therefore has mean $\ \frac{1}{n\beta}\ $.
Likewise, if $\ f_2\ $ and $\ f_3\ $ are the indices of the second and third items to fail, then $\ X_{f_1}\le X_{f_2}\le X_{f_3}\le X_i\ $ for all $\ i\not\in\big\{f_1,f_2\big\}\ $, and the expected time until the third failure (which is the quantity you're asked to find) is $\ \mathbb{E}\big(X_{f_3}\big)\ $.
By a derivation similar to that of the distribution of $\ X_{f_1}\ $, we get the conditional cumulative distribution of $\ X_{f_2}\ $ given $\ X_{f_1}\ $ to be given by $$ \mathbb{P}\big(X_{f_2}\le x\,\big|\, X_{f_1}=z\big)=1-e^{-(n-1)\beta(x-z)}\ . $$ Therefore \begin{align} \mathbb{P}\big(X_{f_2}-X_{f_1}\le x\,\big|\, X_{f_1}=z\big)&=\mathbb{P}\big(X_{f_2}-z\le x\,\big|\, X_{f_1}=z\big)\\ &=\mathbb{P}\big(X_{f_2}\le x+z\,\big|\, X_{f_1}=z\big)\\ &=1-e^{-(n-1)\beta x}\ . \end{align} That is, the conditional distribution of $\ X_{f_2}-X_{f_1}\ $ given $\ X_{f_1}\ $ is exponential with parameter $\ (n-1)\beta\ $, so its conditional expectation given $\ X_{f_1}\ $ is $$ \mathbb{E}\big(X_{f_2}-X_{f_1}\,\big|\, X_{f_1}=z\big)=\frac{1}{(n-1)\beta}\ , $$ and integrating this with respect to the distribution of $\ X_{f_1}\ $ gives $$ \mathbb{E}\big(X_{f_2}-X_{f_1}\big)=\frac{1}{(n-1)\beta}\ . $$ Therefore \begin{align} \mathbb{E}\big(X_{f_2}\big)&=\mathbb{E}\left(\big(X_{f_2}-X_{f_1}\big)+X_{f_1}\right)\\ &=\frac{1}{(n-1)\beta}+\frac{1}{n\beta}\ . \end{align}
Essentially the same argument will give \begin{align} \mathbb{E}\big(X_{f_3}-X_{f_2}\,\big|\, X_{f_2}=z\big)&=\frac{1}{(n-2)\beta}\ ,\\ \mathbb{E}\big(X_{f_3}-X_{f_2}\big)&=\frac{1}{(n-2)\beta}\ ,\text{ and}\\ \mathbb{E}\big(X_{f_3}\big)&=\frac{1}{(n-2)\beta}+\frac{1}{(n-1)\beta}+\frac{1}{n\beta}\ . \end{align}