Chapter 5, Problem 91SE

a.

To determine

Prove that $X_1+X_2$ has a chi-squared distribution with parameter ${\nu }_1+{\nu }_2$.

Explanation of Solution

Given info:

Reference- Example 5.22: The random variables $X_1$ and $X_2$ are independently distributed with chi-squared distribution. The parameter of $X_1$ is ${\nu }_1$ and that of $X_2$ is ${\nu }_2$.

Calculation:

The probability density function (pdf) of a random variable X having a chi-squared distribution with parameter $\nu$ is given as:

$f_X\left(x\right)=\{\begin{array}{l}\frac{1}{2^{\frac{\nu }{2}}\Gamma \left(\frac{\nu }{2}\right)}{x}^{\frac{\nu }{2}-1}{e}^{-\frac{x}{2}},x\ge 0\\ 0,\text{ otherwise}\text{.}\end{array}$

Denote $T_o=X_1+X_2$. Thus, using the above pdf and the technique in Example 5.22, the cdf of $T_o$ is obtained as follows:

$\begin{array}{c}{F}_{T_o}\left(t\right)=P\left(X_1+X_2\le t\right)\\ ={\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{t-x_1}{\int }}\left[\frac{1}{2^{\frac{{\nu }_1}{2}}\Gamma \left(\frac{{\nu }_1}{2}\right)}{x}_{{}^1}^{\frac{{\nu }_1}{2}-1}{e}^{-\frac{x_1}{2}}\right]\cdot \left[\frac{1}{2^{\frac{{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_2}{2}\right)}{x}_{{}^2}^{\frac{{\nu }_2}{2}-1}{e}^{-\frac{x_2}{2}}\right]d{x}_2}d{x}_1}\left(\text{as }{x}_1+x_2=t\right)\\ =\frac{1}{\left(2^{\frac{{\nu }_1}{2}}\Gamma \left(\frac{{\nu }_1}{2}\right)\right)\cdot \left(2^{\frac{{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_2}{2}\right)\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{t-x_1}{\int }}{x}_{{}^1}^{\frac{{\nu }_1}{2}-1}{x}_{{}^2}^{\frac{{\nu }_2}{2}-1}{e}^{-\frac{x_1+x_2}{2}}d{x}_2}d{x}_1}\\ =\frac{1}{\left(2^{\frac{{\nu }_1}{2}}\Gamma \left(\frac{{\nu }_1}{2}\right)\right)\cdot \left(2^{\frac{{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_2}{2}\right)\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{t-x_1}{\int }}{x}_{{}^1}^{\frac{{\nu }_1}{2}-1}{x}_{{}^2}^{\frac{{\nu }_2}{2}-1}{e}^{-\frac{t}{2}}d{x}_2}d{x}_1}\end{array}$

Make the transformations:

$x_1=tp$ and $x_2=t\left(1-p\right)$, so that $x_1+x_2=t$ is true and $0\le p\le 1$.

The Jacobian determinant of the transformation is:

$\begin{array}{c}\left|J\left(t,p\right)\right|=\left|\begin{array}{l}\frac{\partial x_1}{\partial t}\frac{\partial x_1}{\partial p}\\ \frac{\partial x_2}{\partial t}\frac{\partial x_2}{\partial p}\end{array}\right|\\ =\left|\begin{array}{l}pt\\ \left(1-p\right)-t\end{array}\right|\\ =\left|-tp-t+tp\right|\\ =t.\end{array}$

In that case, the above integral becomes:

$\begin{array}{c}{F}_{T_o}\left(t\right)=\frac{1}{\left(2^{\frac{{\nu }_1}{2}}\Gamma \left(\frac{{\nu }_1}{2}\right)\right)\cdot \left(2^{\frac{{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_2}{2}\right)\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(tp\right)}^{\frac{{\nu }_1}{2}-1}{\left(t\left(1-p\right)\right)}^{\frac{{\nu }_2}{2}-1}{e}^{-\frac{t}{2}}\left|J\left(t,p\right)\right|dp}dt}\\ =\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)\cdot \beta \left(\frac{{\nu }_1}{2},\frac{{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{t}^{\frac{{\nu }_1}{2}-1+\frac{{\nu }_2}{2}-1}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\frac{{\nu }_1}{2}-1}{\left(1-p\right)}^{\frac{{\nu }_2}{2}-1}tdp}dt}\\ =\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{t}^{\frac{{\nu }_1}{2}+\frac{{\nu }_2}{2}-1}\left[\frac{1}{\beta \left(\frac{{\nu }_1}{2},\frac{{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\frac{{\nu }_1}{2}-1}{\left(1-p\right)}^{\frac{{\nu }_2}{2}-1}dp}\right]dt}.\end{array}$

Now, consider the quantity:

$\frac{1}{\beta \left(\frac{{\nu }_1}{2},\frac{{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\frac{{\nu }_1}{2}-1}{\left(1-p\right)}^{\frac{{\nu }_2}{2}-1}dp}$.

This is the integration of the probability density function (pdf) of the beta distribution (first kind) taken over the whole range $\left(0,1\right)$. Hence,

$\frac{1}{\beta \left(\frac{{\nu }_1}{2},\frac{{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\frac{{\nu }_1}{2}-1}{\left(1-p\right)}^{\frac{{\nu }_2}{2}-1}dp}=1$.

Thus, the cdf reduces to:

$\begin{array}{c}{F}_{T_o}\left(t\right)=\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{t}^{\frac{{\nu }_1}{2}+\frac{{\nu }_2}{2}-1}\left[\frac{1}{\beta \left(\frac{{\nu }_1}{2},\frac{{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\frac{{\nu }_1}{2}-1}{\left(1-p\right)}^{\frac{{\nu }_2}{2}-1}dp}\right]dt}\\ =\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}{t}^{\frac{{\nu }_1}{2}+\frac{{\nu }_2}{2}-1}dt}\\ \underset{\_}{F_{T_o}\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)}{t}^{\frac{{\nu }_1}{2}+\frac{{\nu }_2}{2}-1}dt}}.\end{array}$

Compare this with the pdf of the chi-squared distribution. It is clearly observed that the quantity within the integral, $\frac{e^{-\frac{t}{2}}}{2^{\frac{{\nu }_1+{\nu }_2}{2}}\Gamma \left(\frac{{\nu }_1+{\nu }_2}{2}\right)}{t}^{\frac{{\nu }_1}{2}+\frac{{\nu }_2}{2}-1}$, is the pdf of a chi-squared distribution where the degrees of freedom is $\nu =\frac{{\nu }_1+{\nu }_2}{2}$.

It is known that the cdf of a distribution is unique. Thus, using the uniqueness property of the cdf of a random variable, it is proved that $X_1+X_2$ has a chi-squared distribution with parameter $\underset{\_}{{\nu }_1+{\nu }_2}$.

b.

To determine

Find the distribution of $Z_1^2+Z_2^2+\mathrm{...}+Z_n^2$.

Answer to Problem 91SE

The distribution of $Z_1^2+Z_2^2+\mathrm{...}+Z_n^2$ is a chi-squared distribution with parameter $\underset{\_}{\nu =n}$.

Explanation of Solution

Given info:

Reference- Exercise 71: The random variables $Z_1,Z_2,\mathrm{...},Z_n$ are independently distributed standard normal variables.

Calculation:

The result in Exercise 71 shows that, for a standard normal random variable, Z, the variable $Z^2$ has a chi-squared distribution with $\nu =1$.

The result in part a shows that the addition of two variables having chi-squared distributions with parameters ${\nu }_1$ and ${\nu }_2$ produces another variable, which also a chi-squared distribution with parameter $\nu ={\nu }_1+{\nu }_2$. This result can be generalized for the addition of n such variables with chi-squared distributions.

For the random variables $Z_1,Z_2,\mathrm{...},Z_n$, each $Z_i^2$ has a chi-squared distribution with ${\nu }_i=1$, for $i=1,2,\mathrm{...},n$.

Thus, by generalization of the result in part a and the result in Exercise 71, the random variable $Z_1^2+Z_2^2+\mathrm{...}+Z_n^2$ will have a chi-squared distribution. The parameter of this distribution is:

$\begin{array}{c}1+1+\mathrm{...}+1\left(\text{upto }n\text{ terms}\right)={\displaystyle \sum _{i=1}^n\left(1\right)}\\ =n.\end{array}$

Hence, the distribution of $Z_1^2+Z_2^2+\mathrm{...}+Z_n^2$ is a chi-squared distribution with parameter $\underset{\_}{\nu =n}$.

c.

To determine

Find the distribution of the sum $Y={\displaystyle \sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2}$.

Answer to Problem 91SE

The distribution of the sum $Y={\displaystyle \sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2}$ is chi-squared distribution with parameter $\underset{\_}{\nu =n}$.

Explanation of Solution

Given info:

The random variables $X_1,X_2,\mathrm{...},X_n$ form a random sample taken from a normal distribution with mean $\mu$ and variance ${\sigma }^2$.

Calculation:

For a normally distributed random variable, X, with mean $\mu$ and variance ${\sigma }^2$, the standard normal variable, Z, obtained is:

$Z=\frac{X-\mu }{\sigma }$.

Thus, for the normally distributed random variables $X_i$ each having n $\mu$ and variance ${\sigma }^2$, the corresponding standard normal variable is

$Z_i=\frac{X_i-\mu }{\sigma }$, for $i=1,2,\mathrm{...},n$.

Now,

$\begin{array}{c}Y={\displaystyle \sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2}\\ ={\displaystyle \sum _{i=1}^n{Z}_i^2}\\ =Z_1^2+Z_2^2+\mathrm{...}+Z_n^2.\end{array}$

Thus, Y is the sum of the squares of n standard normal variables.

The result in part b shows that the distribution of $Z_1^2+Z_2^2+\mathrm{...}+Z_n^2$ is a chi-squared distribution with parameter $\nu =n$.

Hence, the distribution of the sum $Y={\displaystyle \sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2}$ is chi-squared distribution with parameter $\underset{\_}{\nu =n}$.