A random variable $X$ takes values in $\{0, 1, 2\}$, while another random variable $Y$ takes values in $\{−1, 1\}$.
Furthermore,
$P(X = 1) = P(X = 2) = \frac{1}{3} = \frac{3}{4} \cdot P(Y = 1)$,
$P(Y = −1|X = 0) = P(Y = −1|X = 1) = P(Y = −1|X = 2).$Find the joint distribution of $(X, Y )$ and the marginal distributions.
Also, compute $E[X| Y = −1]$ and $E[Y | X = 2]$.
My solution:
From $P(X = 1) = P(X = 2) = \frac{1}{3} = \frac{3}{4} \cdot P(Y = 1)$ we obtain,
$ \begin{array}{|c|c|c|c|c|} \hline X\overset{\LARGE\setminus}{\phantom{.}}\overset{\Large Y}{\phantom{l}} & -1 & 1 & \text{Sum} \\ \hline 0 & ? & ? & ? \\ \hline 1 & ? & ? & \frac{1}{3} \\ \hline 2 & ? & ? & \frac{1}{3} \\ \hline \text{Sum} & ? & \frac{4}{9} & 1 \\ \hline \end{array} $ $\Rightarrow$ $ \begin{array}{|c|c|c|c|c|} \hline X\overset{\LARGE\setminus}{\phantom{.}}\overset{\Large Y}{\phantom{l}} & -1 & 1 & \text{Sum} \\ \hline 0 & ? & ? & \frac{1}{3} \\ \hline 1 & ? & ? & \frac{1}{3} \\ \hline 2 & ? & ? & \frac{1}{3} \\ \hline \text{Sum} & \frac{5}{9} & \frac{4}{9} & 1 \\ \hline \end{array} $
If $P(Y = −1|X = 0) = P(Y = −1|X = 1) = P(Y = −1|X = 2)$ then,
$ \begin{array}{|c|c|c|c|c|} \hline X\overset{\LARGE\setminus}{\phantom{.}}\overset{\Large Y}{\phantom{l}} & -1 & 1 & \text{Sum} \\ \hline 0 & a & ? & \frac{1}{3} \\ \hline 1 & a & ? & \frac{1}{3} \\ \hline 2 & a & ? & \frac{1}{3} \\ \hline \text{Sum} & \frac{5}{9} & \frac{4}{9} & 1 \\ \hline \end{array} $
We have $3a = \frac{5}{9} \Rightarrow a = \frac{5}{27}$.
So, we obtain,
$ \begin{array}{|c|c|c|c|c|} \hline X\overset{\LARGE\setminus}{\phantom{.}}\overset{\Large Y}{\phantom{l}} & -1 & 1 & \text{Sum} \\ \hline 0 & \frac{5}{27} & ? & \frac{1}{3} \\ \hline 1 & \frac{5}{27} & ? & \frac{1}{3} \\ \hline 2 & \frac{5}{27} & ? & \frac{1}{3} \\ \hline \text{Sum} & \frac{5}{9} & \frac{4}{9} & 1 \\ \hline \end{array} $
From subtraction with marginal probabilities, we obtain,
$ \begin{array}{|c|c|c|c|c|} \hline X\overset{\LARGE\setminus}{\phantom{.}}\overset{\Large Y}{\phantom{l}} & -1 & 1 & \text{Sum} \\ \hline 0 & \frac{5}{27} & \frac{4}{27} & \frac{1}{3} \\ \hline 1 & \frac{5}{27} & \frac{4}{27} & \frac{1}{3} \\ \hline 2 & \frac{5}{27} & \frac{4}{27} & \frac{1}{3} \\ \hline \text{Sum} & \frac{5}{9} & \frac{4}{9} & 1 \\ \hline \end{array} $
So,
$ \begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X|Y=-1) & \frac{\frac{5}{27}}{\frac{1}{3}}=\frac{5}{9} & \frac{\frac{5}{27}}{\frac{1}{3}}=\frac{5}{9} & \frac{\frac{5}{27}}{\frac{1}{3}}=\frac{5}{9} \\ \hline \end{array} $
$ \begin{array}{|c|c|c|c|c|} \hline Y & -1 & 1 \\ \hline P(Y|X=2) & \frac{\frac{5}{27}}{\frac{5}{9}}=\frac{1}{3} & \frac{\frac{4}{27}}{\frac{4}{9}}=\frac{1}{3} \\ \hline \end{array} $