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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.

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(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?"


(c)

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 $P(X_1<X_2<X_3)$.

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