Let the joint m.g.f (moment generating function) of $(Y,Z)$ be $$\color{blue}{M_{Y,Z} (t_1,t_2) = \frac{e^{\frac{t_1^2}{1-2t_2}}}{1-2t_2}, t_2 < 1/2}$$ (a) Find $\text{Corr}(Y,Z)$
(b) Are $Y,Z$ independent?
(c) Find the m.g.fs of $Y-Z$ and $Y+Z$.
(a) $$\text{Corr}(Y,Z) = \frac{\text{cov}(Y,Z)}{\sqrt{\text{var}(Y)}\sqrt{\text{var}(Z)}}$$ From the joint m.g.f, I can find the individual m.g.fs as $$M_Y(t_1) = M_{Y,Z}(t_1,0) \text{ and } M_Z(t_2) = M_{Y,Z}(0,t_2)$$ which gave me $E[Y] = 0, E[Z] = 2, \text{var}(Y) = 2, \text{var}(Z) = 4$. Now, $\text{cov}(Y,Z) = E[YZ] - E[Y]E[Z]$. How do I find $E[YZ]$?
(b) Is the following condition necessary and sufficient to check that $Y,Z$ are independent? $$M_{Y,Z}(t_1,t_2) = M_Y(t_1) M_Z(t_2)$$ In our case, clearly, $M_{Y,Z}(t_1,t_2) \ne M_Y(t_1) M_Z(t_2)$, so $Y,Z$ shouldn't be independent. Please let me know if there are any flaws in this reasoning.
(c) The m.g.f of $X = Y-Z$ would be $$E[e^{tX}] = E[e^{tY-tZ}]$$ Had $Y,Z$ been independent (they aren't, right?), I could've figured this out instantly. The expectation would've split and our required m.g.f would be the ratio of the m.g.fs of $Y,Z$. This doesn't seem to be the case! How do I proceed now?
Thank you very much.
a) Differentiate partially w.r.t $t_1$ and $t_2$ (in any order) and put $t_1=t_2=0$. This gives $E[YZ]$.
b) Yes. $M_{Y,Z}(t_1,t_2)=M_Y(t_1)M_Z(t_2)$ for all $t_1,t_2$ is necessary and sufficient for indepedence
c) $E[e^{tY-tZ}]$ is nothing but $M_{Y,Z}(t_1,t_2)$ when $t_1=t$ and $t_2=-t$.