Question

(a) Let x, have a chi-squared distribution with parameter y, and let x, be independent of x, and have a chi-squared distribution with parameter vz Show that x2 + x, has a chi-squared distribution with parameter v, + vz Let y = x2 + x2. Identify the correct expression for Fyn). "1/2 - 2x212/2-12-(x1 + x2)/2 dx1 OFyly) = ' -X1/2 dx1 + -x212 dxz e 109) = 6 {1*(***) ** (*)*" + x2)120mg +6 (277(3)) -<*****@*)*** .(-)70%as C (7-(1) (3+2). *27.n.az Fuly = 6 {019 5)(27+() -x2)/2 din Fy(y) = OFyly) -X1/2 + 1/2 22/2-16-(*1 + x2)/2 2012 dx1 Simplifying Fyl results in which of the following? Fyly) = (' 2"z * »W21 (v1 + v2)/2-1-1/2 1 ()) - 42 ou Jo OF) -u/2 2e du NO 1/2-u/(v1 + v2) OFyly) = " +6+(3*3*3*+ () X +x2)du O FHD) = 6” (2641 = 6°(26+ v212; (**) (-172 * v2)du OFylv) = 5" (2603 +1272; (***)(*2 + 2 1/ 2-1/(v1 + v2) ("1 + v2)/2 - 1 -1/2 euzau Thus we see the integrand is the density chi-squared distribution with parameter v+ Vy

Answer 1

then 2-) e Sol". If an chi-squared distribution fal = 1 6-x/2 (%) with parametern n's M/3) Given: x, and X are two independent chi-squared random variables with parameter Vi and Va respectively. so f(x,) - fol. food 1/2-1 Valet brothy/ ve f(x, y) M e vill to whom 2 Let Y=Xitxa fly) = P(Yay) = P(x + X2 Ey) px * fm, sal day day y=0 Vila- Valz - p (@, +xz)/2 dx X2 das fadin so fy())= V= Nz 4(1-1) option fifth is correct. Let us transform to u and v defined as U=Mithy so that H,=uv Xitky As a, and by both range from o tooo uranges from oto oo and from o to l. Jacobian of transformation J dhika) aurus -u ox, du axy DV anz V TV -u 볼 5-6/ ** [-vsaj d Fluvs - (uvo Il dudu Sit Wella rete de 2 2012 e Vitveyal du dv yht (1-vje, = (VI+V2)/2 2 GN XV) -1 du (1 + 2)/2 (Vitva uvitull du e- » df(u) = (1, +WJ12 » Fy (Y) (Vitvall vitvy so option first is correct.