Independent solution

How to solve this Conditional Distributions question

Setup

Setup

Condition on the stated value of the first variable. The corresponding five joint masses have total probability 0.33.

Pr⁡(S=2)=0.10+0.03+0.07+0.05+0.08=0.33\Pr(S=2)=0.10+0.03+0.07+0.05+0.08=0.33
Pr⁡(B=b∣S=2)=p2,b0.33,b=1,…,5\Pr(B=b\mid S=2)=\frac{p_{2,b}}{0.33},\qquad b=1,\ldots,5

Model

Model

Compute the first two conditional raw moments and then use the variance identity.

Var⁡(B∣S=2)=E⁡[B2∣S=2]−E⁡[B∣S=2]2\operatorname{Var}(B\mid S=2)=\operatorname{E}[B^2\mid S=2]-\operatorname{E}[B\mid S=2]^2

Compute

Compute

Weight the five possible values by the normalized row probabilities.

E⁡[B∣S=2]=1(0.07)+2(0.10)+3(0.08)+4(0.05)+5(0.03)0.33=8633\operatorname{E}[B\mid S=2]=\frac{1(0.07)+2(0.10)+3(0.08)+4(0.05)+5(0.03)}{0.33}=\frac{86}{33}
E⁡[B2∣S=2]=12(0.07)+22(0.10)+32(0.08)+42(0.05)+52(0.03)0.33=27433\operatorname{E}[B^2\mid S=2]=\frac{1^2(0.07)+2^2(0.10)+3^2(0.08)+4^2(0.05)+5^2(0.03)}{0.33}=\frac{274}{33}
Var⁡(B∣S=2)=27433−(8633)2=16461089=1.5114784206…\operatorname{Var}(B\mid S=2)=\frac{274}{33}-\left(\frac{86}{33}\right)^2=\frac{1646}{1089}=1.5114784206\ldots

Answer

Answer

The conditional variance rounds to 1.51.

1.51(B)\boxed{1.51\quad\text{(B)}}