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Show that the following function satisfies the proper-
ties of a joint probability mass function.

$\begin{array}{|c|c|c|}\hline x & {y} & {f_{X Y}(x, y)} \\ \hline 1 & {1} & {1 / 4} \\ \hline 1.5 & {2} & {1 / 8} \\ {1.5} & {3} & {1 / 4} \\ {2.5} & {4} & {1 / 4} \\ {3} & {5} & {1 / 8} \\ \hline\end{array}$

Determine the following probabilities:

$\begin{array}{ll}{\text { (a) } P(X<2.5, Y<3)} & {\text { (b) } P(X<2.5)} \\ {\text { (c) } P(Y<3)} & {\text { (d) } P(X>1.8, Y>4.7)}\end{array}$

$f_{x,y}(x, y)>0$

$\sum_{x} \sum_{y} f(x, y)=\frac{1}{4}+\frac{1}{8}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}=1$

(a) $P(x<2.5, y<3)=F(1,1)+F(1.5,2)$

$=1 / 4 +\frac{1}{8}=\frac{3}{8}$

(b) $P(x<2.5)=F(1,1)+F(1.5,2)+F(1.5,3)$

$=1 / 4+\frac{1}{8}+\frac{1}{4}=\frac{5}{8}$

(c) $P(y<3)=F(1,1)+F(1.5,2)=\frac{1}{4}+\frac{1}{8}=3 / 8$

(d) $P(x>1.8, y>4.7)=F(3,5)=\frac{1}{8}$

$\begin{array}{|c|c|c|}\hline x & {y} & {f_{X Y}(x, y)} \\ \hline 1 & {1} & {1 / 4} \\ \hline 1.5 & {2} & {1 / 8} \\ \hline 1.5 & {3} & {1 / 4} \\ \hline 2.5 & {4} & {1 / 4} \\ \hline 3 & {5} & {1 / 8} \\ \hline\end{array}$

Determine:
(a) $\mathrm{E}(\mathrm{X}), \mathrm{E}(\mathrm{Y})$ .
(b) The marginal probability distribution of the random
variable $X .$
(c) The conditional probability distribution of $Y$ given that
$X=1.5 .$

(d) The conditional probability distribution of $X$ given that
$Y=2 .$
(e) $E(Y | X=1.5)$ .
(f) Are $X$ and $Y$ independent?

(a) $E(x)=1 \times \frac{1}{4}+1.5 \times \frac{1}{8}+1.5 \times \frac{1}{4}+2.5 \times \frac{1}{4}+3 \times \frac{1}{8}$

$=1.8125$

$E(y)=1 \times \frac{1}{4}+2 \times \frac{1}{8}+3 \times \frac{1}{4}+4 \times \frac{1}{4}+5 \times \frac{1}{8}$

$=2.875$

(b) $\begin{array}{|c|c|}\hline x & {f(x)} \\ \hline 1 & {1 / 4} \\ {1.5} & {3 / 8} \\ {2.5} & {1 / 4} \\ {3} & {1/8} \\ \hline\end{array}$

(c) $F_{y | 1.5}(y)=\frac{F_{x y}(1.5, y)}{F_{x}(1.5)}$

$\begin{array}{|c|c|}\hline y & {F_{y|1.5}(y)} \\ \hline 2 & {1 / 3} \\ {3} & {2 / 3} \\ \hline\end{array}$

(d) $F_{x | 2}=\frac{F_{x y}(x , 2)}{F_{y}(2)}$

$=\frac{1 / 8}{1 / 8}=1$

(e) $E\left(y | x=1.5\right)=2 \times \frac{1}{3}+3 \times \frac{2}{3}=\frac{8}{3}$

(f) $F_{y | 1.5} \neq f_{y}(y) \rightarrow x, y$ Not independant