Independent solution

How to solve this Joint Distributions question

Setup

Setup

Evaluate the supplied joint mass formula separately for the two possible values of Y. When Y=1, its numerator simplifies to one for every allowed X.

p(x,1)=x−2−x+318=118,x∈{1,2,3}p(x,1)=\frac{x-2-x+3}{18}=\frac1{18},\qquad x\in\{1,2,3\}

Model

Model

Apply the expectation-of-a-function formula. Every state with Y=0 contributes zero because its function value Y/X is zero.

E⁡ ⁣[YX]=∑x=13∑y=01yxp(x,y)\operatorname{E}\!\left[\frac{Y}{X}\right]=\sum_{x=1}^{3}\sum_{y=0}^{1}\frac{y}{x}p(x,y)
E⁡ ⁣[YX]=∑x=131x118\operatorname{E}\!\left[\frac{Y}{X}\right]=\sum_{x=1}^{3}\frac{1}{x}\frac1{18}

Compute

Compute

Add the three nonzero terms using a common denominator.

E⁡ ⁣[YX]=118(1+12+13)\operatorname{E}\!\left[\frac{Y}{X}\right]=\frac1{18}\left(1+\frac12+\frac13\right)
E⁡ ⁣[YX]=118116=11108\operatorname{E}\!\left[\frac{Y}{X}\right]=\frac1{18}\frac{11}{6}=\frac{11}{108}

Answer

Answer

The expected ratio is 11/108.

11108(B)\boxed{\frac{11}{108}\quad\text{(B)}}