Independent solution

How to solve this Joint Distributions question

Answer in brief

The joint table gives E[X]=4/3, E[Y]=1/3, and E[XY]=1/3. Therefore Cov(X,Y)=1/3-(4/3)(1/3)=-1/9, so choice A is correct.

Setup

Setup

Compute the two marginal first moments from the nonzero joint masses.

E[X]=0(16)+2(12)+1(13)=43E[X]=0\left(\frac16\right)+2\left(\frac12\right)+1\left(\frac13\right)=\frac43
E[Y]=1(13)=13E[Y]=1\left(\frac13\right)=\frac13

Model

Model

Calculate the cross moment. Only the joint mass with both coordinates nonzero contributes.

E[XY]=(1)(1)(13)=13E[XY]=(1)(1)\left(\frac13\right)=\frac13

Compute

Compute

Subtract the product of the marginal means from the cross moment.

Cov⁡(X,Y)=E[XY]−E[X]E[Y]\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]
Cov⁡(X,Y)=13−43(13)=−19\operatorname{Cov}(X,Y)=\frac13-\frac43\left(\frac13\right)=-\frac19

Answer

Answer

The covariance is negative one-ninth.

−19(A)\boxed{-\frac19\quad\text{(A)}}