Need Help?

Subscribe to Probability

Subscribe
  • Notes
  • Comments & Questions

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

No comments yet

Join the conversation

Join Notatee Today!