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)

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