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Chi-Square Distribution

Chi-Square Distribution

If \(Z_{1}, Z_{2}, \ldots, Z_{n}\) are \(n\) independent standard normal variables, then the random variable \(X\)

\[ \begin{aligned} X &= Z_{1}^{2} + Z_{2}^{2} + \cdots + Z_{n}^{2}\newline \text{then,} \quad X &\sim \chi_{n}^{2} \end{aligned} \]

i.e., \(X\) follows the chi-square distribution with \(n\) degrees of freedom.

If we add two chi-square distributed variables with degrees of freedom \(n_{1}\) and \(n_{2}\), then the resultant variable itself is chi-square distributed with \(n_{1} + n_{2}\) degrees of freedom. This simply follows from the fact that the sum of the two random variables is nothing but sum of \(n_{1} + n_{2}\) standard normal squared variables which is nothing but a chi-square variable with \(n_{1} + n_{2}\) degrees of freedom.

If \(X \sim \chi_{n}^{2}\), then \(\chi_{\alpha, n}^{2}\) is

\[ \begin{aligned} P(X \geq \chi_{\alpha, n}^{2}) = \alpha \end{aligned} \]

This quantity is usually listed in mathematical tables since they are heavily used in hypothesis testing.

Relation between Chi-Square and Gamma Distribution

Consider the moment generating function for a chi-square random variable with \(n=1\) degrees of freedom

\[ \begin{aligned} E[e^{tX}] &= E[e^{tZ^{2}}] \quad\text{$Z \sim \mathcal{N}(0, 1)$}\newline &= \int_{-\infty}^{\infty} e^{tx^{2}} f_{Z}(x) dx \quad\text{since $E[g(x)] = \int_{x} g(x)p(x)$}\newline &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{tx^{2}} e^{-x^{2}/2}\newline &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-x^{2}(1/2 - t)}\newline \text{Using}\quad \int_{-\infty}^{\infty} e^{-a(x+b)^{2}} &= \sqrt{\frac{\pi}{a}} \quad \text{by rearranging the integral to get a normal distribution expression}\newline E[e^{tX}] &= \frac{1}{\sqrt{2\pi}} \sqrt{\frac{\pi}{1/2 - t}}\newline &= \frac{1}{\sqrt{1 - 2t}} \end{aligned} \]

Extending this idea to the case of \(n\) degrees of freedom,

\[ \begin{aligned} E[e^{tX}] &= E[e^{t(Z_{1}^{2} + Z_{2}^{2} + \cdots + Z_{n}^{2})}]\newline &= E[\prod_{i=1}^{n} e^{t Z_{i}^{2}}]\newline &= \prod_{i=1}^{n} E[e^{t Z_{i}^{2}}] \quad\text{since $Z_{i}$ are independent}\newline &= (1 - 2t)^{-n/2} \quad\text{from the derivation above} \end{aligned} \]

But, the quantity just derived is nothing but the moment generating function of the Gamma distribution with parameters \((n/2, 1/2)\). Hence, by the uniqueness of the moment generating function, we are forced to conclude that the probability density function of a chi-square variable with n degrees is same as that of a Gamma distribution with parameters (n/2, 1/2).

Thus,

\[ \begin{aligned} f_{X}(x) = \frac{\frac{1}{2} e^{-x/2} (\frac{x}{2})^{(n/2) - 1}}{\Gamma(\frac{n}{2})} \quad\text{$x > 0$} \end{aligned} \]

Sum of Exponentially Distributed Random Variables to Chi-Square Distribution

We say that a Gamma distributed random variable with \(\lambda = 1/2\) and \(\alpha\) can be considered equivalent to a \(\chi^{2}\_{2\alpha}\) variable. Here, \(\lambda\) is constrained to be \(1/2\). By transforming the variables appropriately, we can extend the idea to a sum of exponentially distributed random variables.

Consider \(n\) independent and identically exponentially distributed random variables \(X_{i}\) with parameter \(\lambda\). Consider for any of those random variables \(X_{i}\),

\[ \begin{aligned} Y &= 2\lambda X\newline F_{Y}(y) &= P(Y \leq y) = P(X \leq \frac{y}{2\lambda})\newline &= F_{X}(\frac{y}{2\lambda}) = 1 - \exp \bigg( -\frac{y}{2} \bigg)\newline f_{Y}(y) &= \frac{d}{dy} 1 - \exp \bigg( -\frac{y}{2} \bigg)\newline &= \frac{1}{2}\exp \bigg( -\frac{y}{2} \bigg)\newline &= Exp(\frac{1}{2}) \end{aligned} \]

i.e., \(2\lambda Exp(\lambda) \sim Exp(\frac{1}{2})\). Now, consider the sum of these transformed random variables

\[ \begin{aligned} 2\lambda\bigg( X_{1} + \cdots X_{n} \bigg) &\sim Gamma(n, 1/2) \sim \chi_{2n}^{2} \end{aligned} \]

Hence, we can convert the sum of \(n\) exponentially distributed random variables with parameter \(\lambda\), to a \(\chi^{2}\_{n}\) variable by multiplying the individual variables by \(2\lambda\).

Mean and Variance

Since the distribution of a chi-square variable is identical to a \(Gamma(n/2, 1/2)\) distribution,

\[ \begin{aligned} E[X] &= n\newline Var(x) &= 2n \end{aligned} \]

Sum of \(\chi^{2}\) Variables

Sum of \(n\) independent \(\chi^{2}\) variables where \(X_{i} \sim \chi^{2}\_{r_{i}}\) has a \(\chi^{2}\) distribution as well.

\[ \begin{aligned} r &= \sum_{i=1}^{n} r_{i}\newline \sum_{i=1}^{n}\chi^{2}\_{r_{i}} &= \chi^{2}\_{r} \end{aligned} \]

which follows from the fact that each \(\chi^{2}\) variable is a sum of independent standard normals, and sum of \(\chi^{2}\) variables will itself be a sum of even more independent standard normals.