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Chapter 9: Problem 5

Let \(X_{1}\) and \(X_{2}\) be two independent random variables. Let \(X_{1}\) and \(Y=\) \(X_{1}+X_{2}\) be \(\chi^{2}\left(r_{1}, \theta_{1}\right)\) and \(\chi^{2}(r, \theta)\), respectively. Here \(r_{1}

Short Answer

Expert verified
Based on the properties of chi-square distribution and the given conditions of the problem, it is proven that the variable \(X_{2}\) follows a \(\chi^{2}\left(r-r_{1}, \theta-\theta_{1}\right)\) distribution.

Step by step solution

01

Identify given variables

We are given two random variables \(X_{1}\) and \(X_{2}\). Moreover, a variable \(Y\) is defined as the sum of \(X_{1}\) and \(X_{2}\). We are also provided with the fact that \(X_{1}\) and \(Y\) are chi-square distributed with parameters \(\chi^{2}\left(r_{1}, \theta_{1}\right)\) and \(\chi^{2}(r, \theta)\), respectively. The task is to determine the distribution of \(X_{2}\).
02

Acknowledge independence of the variables

The problem statement says \(X_{1}\) and \(X_{2}\) are independent random variables. This feature allows us to make specific manipulations in further steps.
03

Apply property of chi-square distribution

The property of a chi-square distribution that the sum of two independent chi-square random variables is also a chi-square distributed, specifically with parameters that are the sum of those of the original variables. Using this property, we get the formulation \(\chi^{2}(r, \theta) = \chi^{2}\left(r_{1}, \theta_{1}\right) + X_{2}\).
04

Prove the Distribution of \(X_{2}\)

From the formulation obtained in Step 3, it implies that \(X_{2}\) follows a distribution \(\chi^{2}\left(r-r_{1}, \theta-\theta_{1}\right)\) given that \(r_{1}<r\) and \(\theta_{1} \leq \theta\). This concludes the proof.

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