Protective index
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The following are proofs of several characteristics related to the chi-squared distribution.
Derivations of the pdf
Derivation of the pdf for one degree of freedom
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.
Derivation of the pdf for two degrees of freedom
To derive the chi-squared distribution with 2 degrees of freedom, there could be several methods. Here presented is one of them which is 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:
and
Simply, we can derive the joint distribution of and :
where is replaced by . Further, let and , we can get that:
and
or, inversely
and
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:
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
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: