Question

3. A fair die is rolled twice. Let the first outcome be X and the second outcome be Y (a) (5 points) Calculate Cov(XY, X -Y) and simplify. (hint: What is Cov(aX+bY) in terms of Cov(X, Y)?) (b) (5 points) Are X Y and X -Y independent? Explain. (c) (5 points) Calculate the moment generating function Mx+y(t) of X+Y (the answer should be a function of t and can contain unsimplified sums)

Answer 1

$(a)~The~joint~distribution~of~X,Y~is\\ P(X=x,Y=y)=\frac{1}{36},~x,y=1(1)6,~(since~the~die~is~fair)\\ =0~otherwise\\ Again~P(X=x,Y=y)=P(X=x)P()Y=y)~\forall~x,y=1(1)6\\~X~and~Y~are~independent~so~Cov(X,Y)=0.\\ E(X)=\sum_{x=1}^6\frac{x}{6}=\frac{7}{2},~E(X^2)=\sum_{x=1}^6\frac{x^2}{6}=\frac{91}{6},\\ ~Var(X)=E(X^2)-E^2(X)=\frac{35}{12}\\ Similarly,~Var(Y)=\frac{35}{12}\\ Hence~Cov(X+Y,X-Y)=Var(X)-Var(Y)=0\\ (b)~P(X+Y=2,X-Y=0)=P(\{(1,1)\})\\ =\frac{1}{36}\neq P(X+Y=2)P(X-Y=0)\\ since~P(X+Y=2)=P(\{(1,1)\})=\frac{1}{36},~\\ P(X-Y=0)=P(\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}) =\frac{6}{36}=\frac{1}{6}\\ Hence~X+Y~and~X-Y~are~not~independent.\\ (c)~M_{X+Y}(t)=E(e^{t(X+Y)})=E(e^{tX}e^{tY})\\ =E(e^{tX})E(e^{tY}),~since~X~and~Y~are~independent\\ =(E(e^{tX}))^2,~since~X~and~Y~are~identical\\ =\left (\sum_{x=1}^6e^{tx}\times \frac{1}{6} \right )^2=\frac{1}{36}\times \left ( \frac{e^t(e^{6t}-1)}{e^{t}-1} \right )^2=\frac{e^{2t}}{36}\left ( \frac{e^{6t}-1}{e^{t}-1} \right )^2$