Question

Problem 2. (3 marks; 3, 2) Let Yi and Y, be two independeet discrete Let K-Y+Y a) Find the moment generating function of Y.Ys and K. For bistre ee.t) Le 64 24 Page 6 of

  

Answer 1

Problem 2(a)

$p_1(y_1)~ and~p_2(y_2)~are~not~correct~since\\ p_1(-2)+p_1(-1)+p_1(0)=3/6=1/2~and\\ p_2(1)+P_2(6)=1/2\\ However~we~define~p_1(y)~and~p_2(y)~as~follows~such~that~both~are~pmfs:\\ p_1(y_1)=1/3;~y=-2,-1,0\\ =0~otherwise\\ p_2(y_2)=1/2;~y=1,6\\ =0~otherwise\\ P_K(K=Y_1+Y_2=-1)=p_1(-2)p_2(1)=1/6$

$P_K(K=Y_1+Y_2=4)=p_1(-2)p_2(6)=1/6\\ P_K(K=Y_1+Y_2=0)=p_1(-1)p_2(1)=1/6\\ P_K(K=Y_1+Y_2=5)=p_1(-1)p_2(6)=1/6\\ P_K(K=Y_1+Y_2=1)=p_1(0)p_2(1)=1/6\\ P_K(K=Y_1+Y_2=6)=p_1(0)p_2(6)=1/6\\ Since~Y_1~and~Y_2~are~independent.$

$Hence~the~distribution~of~K~is\\ p_K(k)=\frac{1}{6}~if~k=-1,0,1,4,5,6\\ =0~otherwise\\ MGF~of~Y_1~is\\ M_{Y_1}(t)=E(e^{ty_1})=\frac{e^{-2t}+e^{-t}+1}{3}\\ MGF~of~Y_2~is\\ M_{Y_2}(t)=E(e^{ty_2})=\frac{e^{t}+e^{6t}}{2}\\ MGF~of~K~is\\ M_K(t)=E(e^{Y_1t+Y_2t})=E(e^{Y_1t})E(e^{Y_2t}),~since~Y_1~and~Y_2~are~independent\\ =\frac{(e^{-2t}+e^{-t}+1)(e^{t}+e^{6t})}{6}=\frac{e^{-t}+1+e^{t}+e^{4t}+e^{5t}+e^{6t}}{6}$