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Chapter 3: Problem 6

Let \(X_{1}, X_{2}\), and \(X_{3}\) be iid random variables, each with pdf \(f(x)=e^{-x}\), \(0y)=1-P\left(X_{i}>y, i=1,2,3\right)\). (b) Find the distribution of \(Y=\operatorname{maximum}\left(X_{1}, X_{2}, X_{3}\right)\).

Short Answer

Expert verified
The distribution of \(Y=minimum(X1, X2, X3)\) is \(f_Y(y)= 3e^{-y}[1-(1-e^{-y})^2]\), and the distribution of \(Y=maximum(X1, X2, X3)\) is \(f_Y(y)=3[1-e^{-y}]^2e^{-y}\).

Step by step solution

01

Calculate Probability that Y is less than or equal to y

First, based on the probability rule for the minimum of multiple independent random variables, calculate \(P(Y \leq y)\) using \(P(Y \leq y)=1-P(Y>y)=1-P\left(X_{i}>y, i=1,2,3\right)\). Given the pdf of \(X_{i}\) as \(f(x)=e^{-x}\), we can calculate \(P\left(X_{i}>y\right) = 1 - \int_{0}^{y} e^{-x} dx\). This should be computed for \(i=1,2,3\).
02

Compute the distribution of the minimum

Next, compute the distribution of Y (the minimum) by multiplying the three individual probabilities together, considering independence. That is: \(F_Y(y) =1 - [1 - (1-e^{-y})]^3 \). Take the derivative of \(F_Y(y)\) with respect to \(y\) to find the pdf, \(f_Y(y)\).
03

Calculate Probability that Y is greater than y

Now, based on the probability rule for the maximum of multiple independent random variables, calculate \(P(Y > y)\) using \(P(Y > y)=P\left(X_{i}>y, i=1,2,3\right)\). We already know \(P\left(X_{i}>y\right) =1 - \int_{0}^{y} e^{-x} dx \).
04

Compute the distribution of the maximum

Next, compute the distribution of Y (the maximum) by multiplying the three individual probabilities together, considering independence. That is: \(F_Y(y) =[1 - (1-e^{-y})]^3 \). Take the derivative of \(F_Y(y)\) with respect to \(y\) to find the pdf, \(f_Y(y)\).

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