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)).
Then,


Where
and
are the cdf and pdf of the corresponding random variables.
Then 
Derivation of the pdf for two degrees of freedom
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:

and of course
. Then, we can derive the joint distribution of
:

where
. Further[clarification needed], 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[clarification needed]:

Now we can change
to
[clarification needed]:

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

Substituting
gives:

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:

Substituting, realizing that
, and cancelling terms yields:
