Independent solution

How to solve this Joint Distributions question

Setup

Setup

Obtain the marginal law of X by summing the joint probability function over every possible value of the second variable.

pX(x)=∑y=022x+y+136p_X(x)=\sum_{y=0}^{2}\frac{2x+y+1}{36}

Model

Model

Collect the three terms in the marginal sum and simplify.

pX(x)=3(2x+1)+(0+1+2)36=x+16p_X(x)=\frac{3(2x+1)+(0+1+2)}{36}=\frac{x+1}{6}
pX(0)=16,pX(1)=26,pX(2)=36p_X(0)=\frac16,\quad p_X(1)=\frac26,\quad p_X(2)=\frac36

Compute

Compute

Compute the first two marginal moments and subtract the squared mean.

E⁡[X]=0(1)+1(2)+2(3)6=43\operatorname{E}[X]=\frac{0(1)+1(2)+2(3)}{6}=\frac43
E⁡[X2]=02(1)+12(2)+22(3)6=73\operatorname{E}[X^2]=\frac{0^2(1)+1^2(2)+2^2(3)}{6}=\frac73
Var⁡(X)=73−(43)2=59=0.555555…\operatorname{Var}(X)=\frac73-\left(\frac43\right)^2=\frac59=0.555555\ldots

Answer

Answer

The marginal variance rounds to 0.56.

0.56(A)\boxed{0.56\quad\text{(A)}}