Let $U \perp V$ two random variables such that $\mathbb{P}(U=1)=\mathbb{P}(V=1)=\frac{1}{4}$ and $\mathbb{P}(U=-1)=\mathbb{P}(V=-1)=\frac{3}{4}$. Let $X=\frac{U}{V}$ and $Y=U+V$. Find:
a) the joint law of $(X,Y)$.
b) $\operatorname{Cov}(X,Y)$.
c) If we have $50$ i.i.d. with the same law of $U$ ($U_{1},\ldots,U_{50}$), and $Z=U_{1}+\cdots+U_{50}$, what is the distribution of $Z$? And the value of $\mathbb{P}(Z\leq-10)$?
For b) we have $\operatorname{Cov}(X,Y)=E[XY]+\frac{1}{4}$. Now, how I find $E[XY]=E[E[XY|X=i]]=?$
For c)… Pitch dark.
Any hints?
Thanks in advance!
Since $U$ and $V$ are independent, there are four possibilities for the values of $(U,V)$ with probabilities $$ \mathbb P(U=-1, V=-1) = \frac34\cdot\frac34=\frac9{16} $$ $$ \mathbb P(U=-1, V=1) = \frac34\cdot\frac14=\frac3{16} $$ $$ \mathbb P(U=1, V=-1) = \frac14\cdot\frac34=\frac3{16} $$ $$ \mathbb P(U=1, V=1) = \frac14\cdot\frac14=\frac1{16} $$
Note that $$ (U=-1, V=-1)\quad \Longrightarrow \quad (X=1, Y=-2) $$ $$ (U=-1, V=1)\quad \Longrightarrow \quad (X=-1, Y=0) $$ $$ (U=1, V=-1)\quad \Longrightarrow \quad (X=-1, Y=0) $$ $$ (U=1, V=1)\quad \Longrightarrow \quad (X=1, Y=2) $$ As you can see, there are three values for the pair $(X,Y)$ only: $$ \mathbb P(X=1, Y=-2) = \frac9{16}, \quad \mathbb P(X=-1, Y=0) = \frac6{16}, \quad \mathbb P(X=1, Y=2) = \frac1{16}. $$ Return to (b): $$ \mathbb E[XY] = \sum_{i,j}x_iy_i\mathbb P(X=x_i,Y=y_j)= 1\cdot(-2) \cdot \frac9{16}+(-1)\cdot 0 \cdot \frac6{16} + 1\cdot 2 \cdot \frac1{16} $$
For (c). You can transform every $U_i$ to a Bernoulli random variable with success probability $\frac14$ by $B_i=\frac{U_i+1}{2}$ or $U_i=2B_i-1$. And then $$ Z=U_1+\ldots+U_{50}=2(B_1+\ldots+B_{50})-50. $$ The distribution of the sum of independent Bernoulli random variables $B_1+\ldots+B_{50}$ is binomial $B(50,\frac14)$. So every time when $B_1+\ldots+B_{50}=k$, $Z=2k-50$ with the same probability.