Independent solution

How to solve this Joint Distributions question

Setup

Setup

Sum across each row of the joint table to obtain the marginal probabilities for Y.

Pr⁡(Y=0,1,2,3)=128(6,8,8,6)\Pr(Y=0,1,2,3)=\frac{1}{28}(6,8,8,6)
E[Y]=4228=32,E[Y2]=9428=4714\mathbb{E}[Y]=\frac{42}{28}=\frac{3}{2},\qquad \mathbb{E}[Y^2]=\frac{94}{28}=\frac{47}{14}

Model

Model

Use the X=0 column and divide by its column total to form the required conditional distribution.

Pr⁡(Y=0,1,2,3∣X=0)=16(3,2,1,0)\Pr(Y=0,1,2,3\mid X=0)=\frac{1}{6}(3,2,1,0)
E[Y∣X=0]=46=23,E[Y2∣X=0]=66=1\mathbb{E}[Y\mid X=0]=\frac{4}{6}=\frac{2}{3},\qquad \mathbb{E}[Y^2\mid X=0]=\frac{6}{6}=1

Compute

Compute

Convert each pair of raw moments into a variance and divide in the requested order.

Var⁡(Y)=4714−(32)2=3128\operatorname{Var}(Y)=\frac{47}{14}-\left(\frac{3}{2}\right)^2=\frac{31}{28}
Var⁡(Y∣X=0)=1−(23)2=59\operatorname{Var}(Y\mid X=0)=1-\left(\frac{2}{3}\right)^2=\frac{5}{9}
Var⁡(Y)Var⁡(Y∣X=0)=31/285/9=279140=1.992857…\frac{\operatorname{Var}(Y)}{\operatorname{Var}(Y\mid X=0)}=\frac{31/28}{5/9}=\frac{279}{140}=1.992857\ldots

Answer

Answer

The ratio rounds to the value listed in choice D.

1.99(D)\boxed{1.99\quad\text{(D)}}