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Chi-square distribution

A random variable X has a Chi-square distribution if it can be written as a sum of squares:[eq1]where $Y_{1}$, ..., $Y_{n}$ are mutually independent standard normal random variables. The importance of the Chi-square distribution stems from the fact that sums of this kind are encountered very often in statistics, especially in the estimation of variance and in hypothesis testing.

Definition

Chi-square random variables are characterized as follows.

Definition Let X be an absolutely continuous random variable. Let its support be the set of positive real numbers:[eq2]Let $nin U{2115} $. We say that X has a Chi-square distribution with n degrees of freedom if its probability density function is[eq3]where $c$ is a constant:[eq4]and [eq5] is the Gamma function.

The following notation is often employed to indicate that a random variable X has a Chi-square distribution with n degrees of freedom:[eq6]where the symbol $symbol{126}$ means "is distributed as".

To better understand the Chi-square distribution, you can have a look at its density plots.

Expected value

The expected value of a Chi-square random variable X is[eq7]

Proof

It can be derived as follows:[eq8]

Variance

The variance of a Chi-square random variable X is[eq9]

Proof

It can be derived thanks to the usual variance formula ([eq10]):[eq11]

Moment generating function

The moment generating function of a Chi-square random variable X is defined for any $frac{1}{2}$:[eq12]

Proof

Using the definition of moment generating function, we obtain[eq13]The integral above is well-defined and finite only when $frac{1}{2}-t>0$, i.e., when $frac{1}{2}$. Thus, the moment generating function of a Chi-square random variable exists for any $frac{1}{2}$.

Characteristic function

The characteristic function of a Chi-square random variable X is[eq14]

Proof

Using the definition of characteristic function, we obtain:[eq15]

Distribution function

The distribution function of a Chi-square random variable is[eq16]where the function[eq17]is called lower incomplete Gamma function and is usually computed by means of specialized computer algorithms.

Proof

This is proved as follows:[eq18]

Usually, it is possible to resort to computer algorithms that directly compute the values of [eq19]. For example, the MATLAB command

chi2cdf(x,n)

returns the value at the point x of the distribution function of a Chi-square random variable with n degrees of freedom.

In the past, when computers were not widely available, people used to look up the values of [eq20] in Chi-square distribution tables, where [eq21] is tabulated for several values of x and n (see the lecture entitled Chi-square distribution values).

More details

In the following subsections you can find more details about the Chi-square distribution.

The sum of independent chi-square random variables is a Chi-square random variable

Let X_1 be a Chi-square random variable with $n_{1}$ degrees of freedom and X_2 another Chi-square random variable with $n_{2}$ degrees of freedom. If X_1 and X_2 are independent, then their sum has a Chi-square distribution with $n_{1}+n_{2}$ degrees of freedom:[eq22]This can be generalized to sums of more than two Chi-square random variables, provided they are mutually independent:[eq23]

Proof

This can be easily proved using moment generating functions. The moment generating function of X_i is[eq24]Define[eq25]The moment generating function of a sum of mutually independent random variables is just the product of their moment generating functions:[eq26]where [eq27]Therefore, the moment generating function of X is the moment generating function of a Chi-square random variable with n degrees of freedom, and, as a consequence, X is a Chi-square random variable with n degrees of freedom.

The square of a standard normal random variable is a Chi-square random variable

Let Z be a standard normal random variable and let X be its square:[eq28]Then X is a Chi-square random variable with 1 degree of freedom.

Proof

For $xgeq 0$, the distribution function of X is[eq29]where [eq30] is the probability density function of a standard normal random variable:[eq31]For $x<0$, [eq32] because X, being a square, cannot be negative. Using Leibniz integral rule and the fact that the density function is the derivative of the distribution function, the probability density function of X, denoted by [eq33], is obtained as follows (for $xgeq 0$):[eq34]For $x<0$, trivially, [eq35]. As a consequence,[eq36]Therefore, [eq37] is the probability density function of a Chi-square random variable with 1 degree of freedom.

The sum of squares of independent standard normal random variables is a Chi-square random variable

Combining the two facts above, one trivially obtains that the sum of squares of n independent standard normal random variables is a Chi-square random variable with n degrees of freedom.

Density plots

This section shows the plots of the densities of some Chi-square random variables. These plots help us to understand how the shape of the Chi-square distribution changes by changing the degrees of freedom parameter.

Plot 1 - Increasing the degrees of freedom

The following plot contains the graphs of two density functions:

The thin vertical lines indicate the means of the two distributions. By increasing the number of degrees of freedom, we increase the mean of the distribution, as well as the probability density of larger values.

Chi-square density plot 1

Plot 2 - Increasing the degrees of freedom

The following plot also contains the graphs of two density functions:

As in the previous plot, the mean of the distribution increases as the degrees of freedom are increased.

Chi-square density plot 2

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let X be a chi-square random variable with $3$ degrees of freedom. Compute the following probability:[eq38]

Solution

First of all, we need to express the above probability in terms of the distribution function of X:[eq39]where the values[eq40]can be computed with a computer algorithm or found in a Chi-square distribution table (see the lecture entitled Chi-square distribution values).

Exercise 2

Let X_1 and X_2 be two independent normal random variables having mean $mu =0$ and variance $sigma ^{2}=16$. Compute the following probability:[eq41]

Solution

First of all, the two variables X_1 and X_2 can be written as[eq42]where $Z_{1}$ and $Z_{2}$ are two standard normal random variables. Thus, we can write[eq43]but the sum [eq44] has a Chi-square distribution with $2$ degrees of freedom. Therefore,[eq45]where [eq46] is the distribution function of a Chi-square random variable Y with $2$ degrees of freedom, evaluated at the point $frac{1}{2}$. With any computer package for statistics, we can find[eq47]

Exercise 3

Suppose the random variable X has a Chi-square distribution with $5$ degrees of freedom. Define the random variable Y as follows:[eq48]Compute the expected value of Y.

Solution

The expected value of Y can be easily calculated using the moment generating function of X:[eq49]Now, by exploiting the linearity of the expected value, we obtain[eq50]

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