Question

Prove that if random variable X follows a standard normal distribution (with mean u= 0 and standard deviation o = 1), then Y = X2 follows a chi-square distribution with 1 degree of freedom. In particular, show that My(t) = Mx2(t) = E[etX?), which equals the moment generating function of a chi-square distribution with 1 degree of freedom.

Answer 1

which is the mgf of chi square random variable with 1 df

Given X ~ N(8,1). We need to prove that y=x² follows X? we know the paf of x, f(x2= 1/ 3, XER Yz xa , dyex {70, 470 il-) If aco, y is a decreasing functions of x and if xxo, y is an increasing function of x. we now find the cumulative distribution a yett lyzo function (df) of Y, Hly)=P(Asy) =P(x²2y)=P(-ry sxs ug) • Plys vy) - P(xs - sý) - Ty (sy) - En C-vy) Now differentiating Hey to obtain the pdf of y, se, pelf of Y, beg) - fx Crg ) x + - fxl-rg) x Lara - facog) webatfel-ng) - 2x som en samlejlar Beggo at met gol, yoo (ima i which is the pdf of a chisquare distorbution with 1 of. Now the Moment generating function (Mgf) of to My Ct). Myect) - hit Ecete) • Setaforo dx = 'g était en de - Ide*. It išti. 4) dan 36.237 den de folhush the maig de 2. l te the par in NC 012-25').