Question

Let $X_1$ and $X_2$ be two independent random variables. Let $X_1$ and $Y=$ $X_1+X_2$ be ${\chi }^2(r_1,{\theta }_1)$ and ${\chi }^2(r,\theta )$, respectively. Here \(r_{1}

Short answer

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 ${\chi }^2(r-r_1,\theta -{\theta }_1)$ distribution.

Step-by-step solution

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 ${\chi }^2(r_1,{\theta }_1)$ and ${\chi }^2(r,\theta )$, respectively. The task is to determine the distribution of $X_2$.

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.

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 ${\chi }^2(r,\theta )={\chi }^2(r_1,{\theta }_1)+X_2$.

Prove the Distribution of $X_2$

From the formulation obtained in Step 3, it implies that $X_2$ follows a distribution ${\chi }^2(r-r_1,\theta -{\theta }_1)$ given that $r_1<r$ and ${\theta }_1\le \theta$. This concludes the proof.