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Chapter 2: Problem 9

Let \(X_{1}, X_{2}, X_{3}\) be iid with common pdf \(f(x)=\exp (-x), 0

Short Answer

Expert verified
(a) The conditional probability \(P\left(X_{1}<X_{2} \mid X_{1}<2 X_{2}\right)\) equals to 0.5, (b) The conditional probability \(P\left(X_{1}<X_{2}<X_{3} \mid X_{3}<1\right)\) equals to 1/6.

Step by step solution

01

Understanding the given problem

The given variables \(X_{1}, X_{2}, X_{3}\) are iid (independently identically distributed), each having a common PDF \(f(x)=\exp (-x)\), for \(0<x<\infty\), which describes an exponential distribution. Part (a) is asking for the probability \(X_{1}<X_{2}\) given \(X_{1}<2 X_{2}\), and part (b) is asking for the probability that \(X_{1}<X_{2}<X_{3}\) given that \(X_{3}<1\) .
02

Calculate Part (a)

Since \(X_1, X_2\) are independent, the probability that \(X_1 < X_2\) given \(X_1 < 2X_2\) simplifies to \(P(X_1 < X_2)\), which due to symmetry (since both \(X_1, X_2\) are identically distributed) equals 0.5.
03

Calculate Part (b)

To perform calculations for part (b), observe that the joint density of \(X_{1}, X_{2}, X_{3}\) is the product of the marginal densities due to independence, which is the cube of the exponential distribution given, since all variables are identical and independent. We need to integrate this joint density over the quadruple-interval \(0<X_{1}<X_{2}<X_{3}<1\). This will give the conditional probability \(P\left(X_{1}<X_{2}<X_{3} \mid X_{3}<1\right)\). But due to symmetry (all variables are identical and independent), this condition can be thought of as choosing 3 items from a set of 3, therefore the answer is \(1/3!\).

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