Question

W1 and W2 are discrete random variables (and are independent) with probability functions:

p1(w1) = 1/6 for w1 = −2 ,−1 , 0

p2(w2) = 1/4 for w2 = 1, 6

Let Y = W1 + W2

Find the distribution and probability mass function of Y (Hint: First find MGF of W1 and W2 and then find MGF of Y)

Answer 1

$p_1(w_1)~ and~p_2(w_2)~are~not~correct~since\\ p_1(-2)+p_1(-1)+p_1(0)=3/6=1/2~and\\ p_2(1)+P_2(6)=1/2\\ However~we~define~p_1(w_1)~and~p_2(w_2)~as~follows~such~that~both~are~pmfs:\\ p_1(w_1)=1/3;~w_1=-2,-1,0\\ =0~otherwise\\ p_2(w_2)=1/2;~w_2=1,6\\ =0~otherwise\\ P_Y(Y=W_1+W_2=-1)=p_1(-2)p_2(1)=1/6\\ P_Y(Y=W_1+W_2=4)=p_1(-2)p_2(6)=1/6\\ P_Y(Y=W_1+W_2=0)=p_1(-1)p_2(1)=1/6\\ P_Y(Y=W_1+W_2=5)=p_1(-1)p_2(6)=1/6\\ P_Y(Y=W_1+W_2=1)=p_1(0)p_2(1)=1/6\\ P_Y(Y=W_1+W_2=6)=p_1(0)p_2(6)=1/6\\ Since~W_1~and~W_2~are~independent~so\\ ~P(W_1=w_1,W_2=w_2)=P(W_1=w_1)P(W_2=w_2)=p_1(w_1)p_2(w_2)~\forall~w_1,w_2\\$

$Hence~the~distribution~of~Y~is\\ p_Y(y)=\frac{1}{6}~if~y=-1,0,1,4,5,6\\ =0~otherwise\\ MGF~of~W_1~is\\ M_{W_1}(t)=E(e^{tW_1})=\frac{e^{-2t}+e^{-t}+1}{3}\\ MGF~of~W_2~is\\ M_{W_2}(t)=E(e^{tW_2})=\frac{e^{t}+e^{6t}}{2}\\ MGF~of~K~is\\ M_Y(t)=E(e^{W_1t+W_2t})=E(e^{W_1t})E(e^{W_2t}),~since~W_1~and~W_2~are~independent\\ =\frac{(e^{-2t}+e^{-t}+1)(e^{t}+e^{6t})}{6}=\frac{e^{-t}+1+e^{t}+e^{4t}+e^{5t}+e^{6t}}{6}$