Difference in Means Between Two Chi Square Distribution Continuous Outcome
A random variable has a Chi-square distribution if it can be written as a sum of squares of independent standard normal variables.
Sums of this kind are encountered very often in statistics, especially in the estimation of variance and in hypothesis testing.
In this lecture, we derive the formulae for the mean, the variance and other characteristics of the chi-square distribution.
Table of contents
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Degrees of freedom
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Definition
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Symbol
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Expected value
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Variance
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Moment generating function
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Characteristic function
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Distribution function
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More details
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The sum of independent chi-square random variables is a Chi-square random variable
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The square of a standard normal random variable is a Chi-square random variable
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The sum of squares of independent standard normal random variables is a Chi-square random variable
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Density plots
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Plot 1 - Increasing the degrees of freedom
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Plot 2 - Increasing the degrees of freedom
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Solved exercises
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Exercise 1
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Exercise 2
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Exercise 3
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We will prove below that a random variable has a Chi-square distribution if it can be written as
where
, ...,
are mutually independent standard normal random variables.
The number of variables is the only parameter of the distribution, called the degrees of freedom parameter. It determines both the mean (equal to
) and the variance (equal to
).
Chi-square random variables are characterized as follows.
To better understand the Chi-square distribution, you can have a look at its density plots.
The following notation is often employed to indicate that a random variable has a Chi-square distribution with
degrees of freedom:
where the symbol
means "is distributed as".
The expected value of a Chi-square random variable is
Proof
It can be derived as follows:
The proof above uses the probability density function of the distribution. An alternative, simpler proof exploits the representation (demonstrated below) of as a sum of squared normal variables.
Proof
The variance of a Chi-square random variable is
Proof
Again, there is also a simpler proof based on the representation (demonstrated below) of as a sum of squared normal variables.
Proof
The moment generating function of a Chi-square random variable is defined for any
:
Proof
The characteristic function of a Chi-square random variable is
Proof
Using the definition of characteristic function, we obtain:
The distribution function of a Chi-square random variable is where the function
is called lower incomplete Gamma function and is usually computed by means of specialized computer algorithms.
Proof
This is proved as follows:
Usually, it is possible to resort to computer algorithms that directly compute the values of . 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 in Chi-square distribution tables, where
is tabulated for several values of
and
(see the lecture entitled Chi-square distribution values).
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 be a Chi-square random variable with
degrees of freedom and
another Chi-square random variable with
degrees of freedom. If
and
are independent, then their sum has a Chi-square distribution with
degrees of freedom:
This can be generalized to sums of more than two Chi-square random variables, provided they are mutually independent:
Proof
The square of a standard normal random variable is a Chi-square random variable
Let be a standard normal random variable and let
be its square:
Then
is a Chi-square random variable with 1 degree of freedom.
Proof
For , the distribution function of
is
where is the probability density function of a standard normal random variable:
For
,
because
, 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
, denoted by
, is obtained as follows (for
):
For , trivially,
. As a consequence,
Therefore,
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 independent standard normal random variables is a Chi-square random variable with
degrees of freedom.
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 first graph (red line) is the probability density function of a Chi-square random variable with
degrees of freedom;
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the second graph (blue line) is the probability density function of a Chi-square random variable with
degrees of freedom.
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.
Plot 2 - Increasing the degrees of freedom
The following plot also contains the graphs of two density functions:
-
the first graph (red line) is the probability density function of a Chi-square random variable with
degrees of freedom;
-
the second graph (blue line) is the probability density function of a Chi-square random variable with
degrees of freedom.
As in the previous plot, the mean of the distribution increases as the degrees of freedom are increased.
Below you can find some exercises with explained solutions.
Exercise 1
Let be a chi-square random variable with
degrees of freedom.
Compute the following probability:
Solution
First of all, we need to express the above probability in terms of the distribution function of :
where the values 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 and
be two independent normal random variables having mean
and variance
.
Compute the following probability:
Solution
First of all, the two variables and
can be written as
where
and
are two standard normal random variables. Thus, we can write
but the sum
has a Chi-square distribution with
degrees of freedom. Therefore,
where
is the distribution function of a Chi-square random variable
with
degrees of freedom, evaluated at the point
. With any computer package for statistics, we can find
Exercise 3
Suppose that the random variable has a Chi-square distribution with
degrees of freedom.
Define the random variable as follows:
Compute the expected value of .
Solution
The expected value of can be easily calculated using the moment generating function of
:
Now, by exploiting the linearity of the expected value, we obtain
Please cite as:
Taboga, Marco (2021). "Chi-square distribution", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/probability-distributions/chi-square-distribution.
Source: https://www.statlect.com/probability-distributions/chi-square-distribution
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