We now give some additional technical details about the chi-square distribution and provide proofs for some of the key propositions. Except for the proof of Corollary 2 knowledge of calculus is required.

**Observation**: For any positive real number *k*, per Definition 1, the chi-square distribution with *k* degrees of freedom, abbreviated χ^{2}(*k*), has probability density function

Using the notation of Gamma Function Advanced, the cumulative distribution function for *x *≥ 0 is

**Property A**: The moment generating function for a random variable with χ^{2}(*k*) distribution is

Proof:

Set y = *x*/2 (2θ–1). Then *x* = 2y/(1–2*θ*) and d*x*=2 dy/(1–2*θ*). Thus

The result follows from the fact that by Definition 1 of Gamma Function

**Property 1**: The χ^{2}(*k*) distribution has mean* k* and variance *2k*

Proof: By Property A, the moment generating function of χ^{2}(*k*) is

it follows that

**Theorem 1**: Suppose *x* has standard normal distribution *N*(0,1) and let *x _{1},…, x_{k} *be

*k*independent sample values of

*x*, then the random variable

has a chi-square distribution χ^{2}(*k*).

Proof: Since the *x _{i}* are independent, so are the . From Property 4 of General Properties of Distributions, it follows that

But since all the *x _{i}* are samples from the same population, it follows that

Since *x* is a standard normal random variable, we have

Set y = *x *. Then *x* = y/ and d*x* = dy/, and so

is the pdf for *N*(0, 1), it follows that

Combining the pieces, we have

But by Property A this is the same moment generating function as for χ^{2}(*k*). Thus, by Theorem 1 of General Properties of Distributions, it follows that *w* has distribution χ^{2}(*k*).

**Property 2**: If *x* and y are independent and *x* has distribution χ^{2}(*m*) and y has distribution χ^{2}(*n*), then *x* + y has distribution χ^{2}(*m + n*)

Proof: Since *x* and y are independent, by Property A and Property 4 of General Properties of Distributions

But this is the moment generating function for χ^{2}(*m + n*), and so the result follows from the fact that a distribution is completely determined by its moment generating function (Theorem 1 of General Properties of Distributions).

**Theorem 2**: If *x* is drawn from a normally distributed population *N*(*μ, σ*) then for samples of size *n* the sample variance *s*^{2} has distribution

Proof: Since (*x̄* – *µ*)^{2} is a constant and

Dividing both sides by *σ*^{2} and rearranging terms we get

Now we use the fact that *x* ~ *N*(*μ, σ*) with each *x _{i} *independently sampled from this distribution. Thus the left side of the equation is the sum of

*n*variables, each the square of a z-score, and so by Theorem 1, the left side of the equation has chi-square distribution with

*n*degrees of freedom:

Next consider the last term of the equation above. Since the sampling distribution of the mean for data sampled from a normal distribution *N*(*μ, σ*) is normal with mean *μ* and standard deviation *σ/*, it follows that the last term is the square of a z-score, and so by Theorem 1, it has chi-square distribution with 1 degree of freedom

By Property 2, it follows that the remaining term in the equation is also chi-square with *n* – 1 degrees of freedom.

Dividing by *n* – 1, it follows that

It follows that

which completes the proof.

**Corollary 2**: *s*^{2} is an unbiased, consistent estimator of the population variance

Proof (unbiased): From the theorem and Property 1 it follows that:

Proof (consistent): From Property 1 and Property 3b of Expectation:

Excellent, helps me explain a lot of things!

Thanks for your analysis I take it for my assignment

I loved your explanation, I’ve been looking in many books for this.

Thank you for this easy to follow proofs.

This is refreshing and easy to follow for begginers. All too often many qualified/advanced staticians skip basic explanations and say ” …it is trivial to prove that… ” ; yet for the begginner, she/he would be expecting those very basics in order to come to a level where they will see the triviality! Thanks a lot and I will keep coming back for more as I am just starting statistics especially the mathematical statistics.

Sydeny,

Thanks for your comment. I try hard to give some of the theory and basic explanations, without getting too theoretical, for just the reasons you stated.

Charles