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 χ2(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 χ2(k) distribution is
Set y = x/2 (2θ–1). Then x = 2y/(1–2θ) and dx=2 dy/(1–2θ). Thus
The result follows from the fact that by Definition 1 of Gamma Function
Property 1: The χ2(k) distribution has mean k and variance 2k
Proof: By Property A, the moment generating function of χ2(k) is
it follows that
Theorem 1: Suppose x has standard normal distribution N(0, 1) and let x1, …, xk be k independent sample values of x, then the random variable
has a chi-square distribution χ2(k).
Proof: Since the xi are independent, so are the . From Property 4 of General Properties of Distributions, it follows that
But since all the xi 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 dx = 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 χ2(k). Thus, by Theorem 1 of General Properties of Distributions, it follows that w has distribution χ2(k).
Property 2: If x and y are independent and x has distribution χ2(m) and y has distribution χ2(n), then x + y has distribution χ2(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 χ2(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 s2 has distribution
Proof: Since (x̄ – µ)2 is a constant and
Dividing both sides by σ2 and rearranging terms we get
Now we use the fact that x ~ N(μ, σ) with each xi 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: s2 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: