Question

Let X and Y have Joint probability function given by f(x,y) = xy/66 x = 2,3,4 ; y = 1,2,3 Form the p.d table and find marginal probabilities function of X and Y? Also a) Find P (X=4 | Y=2)? b) Find P (X + Y < 3)? c) Are X and Y independent?

Answer 1

The probability that $X$ takes value $x$ and $Y$ takes value $y$ is: $p_{x,y}=f(x,y)=\frac{xy}{66}$. The probability distribution table has the form:

$\begin{array}{ c c c } (x,y) & p_{x,y}\\ (2,1) & \frac{2}{66} \\ (2,2) & \frac{4}{66} \\ (2,3) & \frac{6}{66} \\ (3,1) & \frac{3}{66} \\ (3,2) & \frac{6}{66} \\ (3,3) & \frac{9}{66} \\ (4,1) & \frac{4}{66} \\ (4,2) & \frac{8}{66} \\ (4,3) & \frac{12}{66} \\ \end{array}$

As we can see from the table, $\frac{1}{66}(2+4+6+3+6+9+4+8+12)=\frac{54}{66}$ . Since the sum is less than 1, $X$, $Y$ with the function $f(x,y)$ is not a proper distribution. For correct distribution the formulae have the form:

a). $P(X=4|Y=2)=\frac{P((X=4)\cap(Y=2))}{P(Y=2)}=\frac{f(4,2)}{P((X=2)\cap(Y=2)) +P((X=3)\cap(Y=2))+P((X=4)\cap(Y=2))}=$

$=\frac{f(4,2)}{f(2,2) +f(3,2)+f(4,2)}$

b). $P(X+Y<3)=P(X+Y=2)=0$ It follows from fact that the minimum value of $X$ is 2.

c). By definition, two events $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$. It is enough to check that $P(X=x,Y=y)=f(x,y)=P(X=x)P(Y=y)=(\sum_{l}f(x,l))(\sum_{k}f(k,y))$