How could I show that the chi-squared distribution with $k$ degrees of freedom has mean $k$ and variance $2k$?

Question

I see that I could possibly use the probability density function to find the mean and variance, but is there an easier way?

Thanks in advance.

Answer 1

If $U_1,\dots,U_k$ are iid with standard normal distribution then:$$X:=U_1^2+\cdots+U_k^2$$ has chi-squared distribution with $k$ degrees of freedom.

With linearity of expectation and symmetry you find:$$\mathbb EX=k\mathbb EU_1^2$$

If two random variable are independent then the variance of their sum is the sum of their variances. Applying that here on the $U_i$ we find:

$$\text{Var}X=k\text{Var}U_1^2$$ So it remains to find $\mathbb EU_1^2$ and $\text{Var}U_1^2$ which I will leave up to you.

Answer 2

To complement drhab's anwer, we want $\mathbb{E}U_1^2,\,\mathbb{E}U_1^4$. But $$\mathbb{E}U_1^{2n}=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}x^{2n} e^{-ax^2/2}$$ with $a=1$. Start from $$a^{-1/2}=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-ax^2/2}.$$Differentiating under the integral sign with respect to $-a/2$ and then setting $a=1$ gives $$\mathbb{E}U_1^2=1,$$which is what you'd expect since $U_1$ has mean $0$ and variance $1$. If we instead differentiate twice before setting $a=1$, we get$$\mathbb{E}U_1^4=3,\,\text{Var}U_1^2=3-1^2=2.$$To make the calculation more explicit, we note that if we instead differentiate with respect to $a$ twice we get $-\frac{1}{2}a^{-3/2}$ and then $\frac{3}{4}a^{-5/2}$. The derivatives with respect to $-a/2$ require us to multiply by $\left( -2\right)^n$, giving $a^{-3/2},\,3a^{-5/2}$.

Answer 3

You could also go the gamma distribution route. The chi-square distribution is a special case of the gamma, that is to say, it is a gamma distribution with shape parameter $k/2$ and scale parameter $2$: $$X \sim \operatorname{ChiSquare}(k) \sim \operatorname{Gamma}(k/2,2).$$ Since the gamma (for positive integer shape) is the sum of iid exponentials, we then have by linearity of expectation $$\operatorname{E}[X] = \frac{k}{2} \cdot 2 = k,$$ and by linearity of variance in the iid case, $$\operatorname{Var}[X] = \frac{k}{2} \cdot 2^2 = 2k.$$ Of course, you would need to justify this reasoning when $k$ is not a positive even integer.