Proofs related to chi-squared distribution

From Wikipedia, the free encyclopedia
Jump to: navigation, search

The following are proofs of several characteristics related to the chi-squared distribution.

Derivations of the pdf[edit]

Derivation of the pdf for one degree of freedom[edit]

Let random variable Y be defined as Y = X2 where X has normal distribution with mean 0 and variance 1 (that is X ~ N(0,1)).


Where and are the cdf and pdf of the corresponding random variables.


Alternative proof directly using the change of variable formula[edit]

The change of variable formula (implicitly derived above), for a monotonic transformation , is:

In this case the change is not monotonic, because every value of has two corresponding values of (one positive and negative). However, because of symmetry, both halves will transform identically, i.e.

In this case, the transformation is: , and its derivative is

So here:

And one gets the chi-squared distribution, noting the property of the gamma function: .

Derivation of the pdf for two degrees of freedom[edit]

There are several methods to derive chi-squared distribution with 2 degrees of freedom. Here is one based on the distribution with 1 degree of freedom.

Suppose that and are two independent variables satisfying and , so that the probability density functions of and are respectively:


Simply, we can derive the joint distribution of and :

where is replaced by . Further, let and , we can get that:


or, inversely


Since the two variable change policies are symmetric, we take the upper one and multiply the result by 2. The Jacobian determinant can be calculated as:

Now we can change to :

where the leading constant 2 is to take both the two variable change policies into account. Finally, we integrate out to get the distribution of , i.e. :

Let , the equation can be changed to:

So the result is:

Derivation of the pdf for k degrees of freedom[edit]

Consider the k samples to represent a single point in a k-dimensional space. The chi square distribution for k degrees of freedom will then be given by:

where is the standard normal distribution and is that elemental shell volume at Q(x), which is proportional to the (k − 1)-dimensional surface in k-space for which

It can be seen that this surface is the surface of a k-dimensional ball or, alternatively, an n-sphere where n = k - 1 with radius , and that the term in the exponent is simply expressed in terms of Q. Since it is a constant, it may be removed from inside the integral.

The integral is now simply the surface area A of the (k − 1)-sphere times the infinitesimal thickness of the sphere which is

The area of a (k − 1)-sphere is:

Substituting, realizing that , and cancelling terms yields: