How to prove that the variance of chi-square distribution is 2n?

Let x obey N(0, 1), we calculate d (x 2) and prove that d (chi-square (1))=2.

(1) is calculated by the square relation, and d (x 2) = e (x 4)-[e (x 2)] 2.

You have to calculate e (x 4) first.

Let f(x) be the density function of N (0, 1) and find e (x 4).

∫ x 4 * f (x) dx = ∫ x 3 * xf(x) dx, because the original function of xf(x) is exactly -f(x).

Partial integral ∫ x 3 * xf (x) dx =-x 3 * f (x)+∫ f (x) * 3x2dx =-x 3 * f (x)+3 ∫ x 2f (x) dx.

Partial integration is used again, so ∫ X2F (x) dx = ∫ x * xf (x) dx =-xf (x)+∫ f (x) dx.

Taken together, ∫ x 4 * f (x) dx =-x 3 * f (x)+3 [-xf (x)+∫ f (x) dx] =-x 3 * f (x)-3xf (x)+3 ?

Therefore, substituting the upper and lower bounds, we get ∫ x 4 * f (x) dx = 3.

Because e (x 2) = d (x)+[e (x)] 2 =1.

So d (x 2) = e (x 4)-[e (x 2)] 2 = 3-1= 2.