Answer to Question #99283 in Statistics and Probability for Fatima1

Question

Problem 1 : The joint probability distribution of two random variables X and Y is given in the following table

Y
X 0 1 2 3
f(x)
2 1/12 1/12 1/12 1/12
3 1/12 1/6 1/12 0
4 1/12 1/12 0 1/6
f(y)
a) Find the marginal density of X and the marginal density of Y. (add them to the above table)
b) Are X and Y independent?
c) Compute the P{Y>1| X>2}
d) Compute the expected value of X.
e) Compute the probability that X is greater or equal to Y+1.
f) Compute the conditional probability distribution of X given Y=3.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X=0 & X=1 & X=2 & X=3 & \\ \hline Y=2 & 1/12 & 1/12 & 1/12 & 1/12 & \bold{1/3} \\ \hdashline Y=3 & 1/12 & 1/6 & 1/12 & 0 & \bold{1/3} \\ \hdashline Y=4 & 1/12 & 1/12 & 0 & 1/6 & \bold{1/3} \\ \hdashline & \bold{1/4} & \bold{1/3} & \bold{1/6} & \bold{1/4} & \bold{1} \end{array}$$

a) Find the marginal density of $X$ and the marginal density of $Y.$ (add them to the above table) 

$$P(X=0)={1 \over 12}+{1 \over 12}+{1 \over 12}={1 \over 4}$$

$$P(X=1)={1 \over 12}+{1 \over 6}+{1 \over 12}={1 \over 3}$$

$$P(X=2)={1 \over 12}+{1 \over 12}+0={1 \over 6}$$

$$P(X=3)={1 \over 12}+0+{1 \over 6}={1 \over 4}$$

$$P(Y=2)={1 \over 12}+{1 \over 12}+{1 \over 12}+{1 \over 12}={1 \over 3}$$

$$P(Y=3)={1 \over 12}+{1 \over 6}+{1 \over12}+0={1 \over 3}$$

$$P(Y=4)={1 \over 12}+{1 \over 12}+0+{1 \over6}={1 \over 3}$$

b) Are $X$ and $Y$ independent? 

To check whether $X$ and $Y$ are independent, we need to check that $P(X=x_i, Y=y_j)=P(X=x_i)P(Y=y_j)$

$i=1,2,3,4; j=1,2,3$

$$P(X=2,Y=2)={1 \over 12}\not={1 \over 6}\cdot{1 \over 3}=P(X=2)P(Y=2)$$

Thus, we conclude that $X$ and $Y$ are not independent.

c) Compute the P{Y>1| X>2} 

$$P(Y>1|X>2)={P(X>2, Y>1) \over P(X>2)}$$

$$P(Y>1|X>2)={{1 \over 12}+0+{1 \over 6}\over {1 \over 4}}=1$$

d) Compute the expected value of X. 

$$E(X)=0({1 \over 4})+1({1 \over 3})+2({1 \over 6})+3({1 \over 4})={17 \over 12}$$

e) Compute the probability that $X$ is greater or equal to $Y+1.$

$$P(X=3,Y=2)={1 \over12}$$

f) Compute the conditional probability distribution of $X$ given $Y=3.$

$$P(X=x_i|Y=3)={P(X=x_i,Y=3)\over P(Y=3)}$$

$$P(X=0|Y=3)={P(X=0,Y=3)\over P(Y=3)}={1/12\over 1/3}={1\over4}$$

$$P(X=1|Y=3)={P(X=1,Y=3)\over P(Y=3)}={1/6\over 1/3}={1\over2}$$

$$P(X=2|Y=3)={P(X=2,Y=3)\over P(Y=3)}={1/12\over 1/3}={1\over4}$$

$$P(X=3|Y=3)={P(X=3,Y=3)\over P(Y=3)}={0\over 1/3}=0$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline f_{X|Y=3}(x) & 1/4 & 1/2 & 1/4 & 0 \end{array}$$