Answer to Question #242527 in Statistics and Probability for PRS

Question

Given that the joint probability distribution of the discrete random variables X and Y is given by

f(x,y)= { k(4x+3y) x=1,2,3 y=1,2

{ 0 otherwise

Find

i) The value of the constant k

ii) The marginal probability distribution function X

iii)The conditional probability distribution Y given x=2

iv) The conditional mean given X=2

Answer 1

i)

$$X$$ $$Y\ \def\arraystretch{1.5} \begin{array}{c:c:c:c} & 1 & 2 & 3 \\ \hline 1 & 7k & 11k & 15k \\ \hdashline 2 & 10k & 14k & 18k \end{array}$$

$$7k+11k+15k+10k+14k+18k=1$$

$$k=\dfrac{1}{75}$$

ii)

$$Y\ \def\arraystretch{1.5} \begin{array}{c:c:c:c} & 1 & 2 & 3 \\ \hline 1 & 7/75 & 11/75 & 15/75\\ \hdashline 2 & 10/75 & 14/75 & 18/75 \end{array}$$

$X=1:$

$$P(X=1|Y=1)+P(X=1|Y=2)$$

$$=7/75+10/75=17/75$$

$X=2:$

$$P(X=2|Y=1)+P(X=2|Y=2)$$

$$=11/75+14/75=25/75=1/3$$

$X=3:$

$$P(X=3|Y=1)+P(X=3|Y=2)$$

$$=15/75+18/75=33/75=11/25$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} x & 1 & 2 & 3 \\ \hline f_X(x) & 17/75 & 1/3 & 11/25 \end{array}$$

iii)

$$P(Y=1|X=2)=11/25$$

$$P(Y=2|X=2)=14/25$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} y & 1 & 2 \\ \hline f_{Y|X=2}(y) & 11/25 & 14/25 \end{array}$$

iv)

$$E(Y|X=2)=1(11/25)+2(14/25)=39/25$$