If X Xp2 use the MGF of a chisquare random variable and MGF

If X ~ X_p^2, use the MGF of a chi-square random variable, and MGF differentiation results for finding moments, to prove directly that E(X) = p and Var(X) = 2p.

Solution

Mgf of chi square distrinbution is M_X(t)= ( 1 - 2t)^-p/2 where, p < 1/2

so E(X) = d/dt(M_X(t)) at t=0 and E(X²) = d²/dt²((M_X(t)) at t=0

therefore,E(X) = d/dt(M_X(t)) = p/(1-2t)^(p+2)/2 =p/1 = p

E(X²)= (p(p+2))/((1-2t)^((p+4)/2)) at t=0 = p²+2p

Var(X) = E(X²) - (E(X))² =  p²+2p - p² = 2p

hence proved