I have a question regarding problem 12(b) and (c) of chapter 5 of M.Ross "Introduction to probability models". The question is as follows:

If $X_1, X_2, X_3$ are independent exponential random variables with rates $\lambda_i$, $i = 1,2,3$, find

(b) $P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3)$;

(c) $E[\max X_i \mid X_1 < X_2 < X_3]$.

For (b), my first thought was that $P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3) = P(X_1 < X_2)$ since from my point of few, $\max(X_1,X_2,X_3) = X_3$ does not have to influence $X_1 < X_2$. According to the answer book of Ross: $P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3) = \frac{P(X_1 < X_2 < X_3)}{P(X_1 < X_2 < X_3) + P(X_2 < X_1 < X_3)}$. Can anyone explain me what mistake I made with my initial reasoning?

For (c) I have no clue how to reason to get the correct answer.


(b) The following might help to understand your mistake.

Let $U_{1},U_{2},U_{3}$ be independent random variables where $P\left(U_{1}=1\right)=P\left(U_{1}=3\right)=\frac{1}{2}$ and $P\left(U_{2}=2\right)=1=P\left(U_{3}=2\right)$.

Then it is evident that: $$P\left(U_{1}<U_{2}\mid\max\left(U_{1},U_{2},U_{3}\right)=U_{3}\right)=1\neq\frac{1}{2}=P\left(U_{1}<U_{2}\right)$$

(Almost) degenerated random variables can be very helpful to examine questions like: "is my intuition correct here?"


Under condition $X_{1}<X_{2}<X_{3}$ you are dealing with the original PDF divided by probability $P\left(X_{1}<X_{2}<X_{3}\right)$.

To be worked out is the integral:$$\frac{\frac{1}{\lambda_{1}\lambda_{2}\lambda_{3}}\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}ze^{-\lambda_{1}x-\lambda_{2}y-\lambda_{3}z}dzdydx}{\frac{1}{\lambda_{1}\lambda_{2}\lambda_{3}}\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}e^{-\lambda_{1}x-\lambda_{2}y-\lambda_{3}z}dzdydx}$$

Note that the denominator equals $P(X_1<X_2<X_3)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.