Independent solution

How to solve this Conditional Distributions question

Setup

Setup

Let A denote the conditioning event and express the monetary variable in thousands so that its possible values are 0, 25, and 50.

A={X≤25,000},Yk=Y/1000A=\{X\le 25{,}000\},\qquad Y_{\mathrm{k}}=Y/1000
Pr⁡(A)=0.80\Pr(A)=0.80

Model

Model

Aggregate the joint masses inside A for each possible value of the target variable, then normalize by the probability of A.

Pr⁡(Yk=0∣A)=0.100.80=0.125\Pr(Y_{\mathrm{k}}=0\mid A)=\frac{0.10}{0.80}=0.125
Pr⁡(Yk=25∣A)=0.500.80=0.625\Pr(Y_{\mathrm{k}}=25\mid A)=\frac{0.50}{0.80}=0.625
Pr⁡(Yk=50∣A)=0.200.80=0.250\Pr(Y_{\mathrm{k}}=50\mid A)=\frac{0.20}{0.80}=0.250

Compute

Compute

Calculate the first two conditional moments and subtract the square of the conditional mean.

E[Yk∣A]=25(0.625)+50(0.250)=28.125\mathbb{E}[Y_{\mathrm{k}}\mid A]=25(0.625)+50(0.250)=28.125
E[Yk2∣A]=252(0.625)+502(0.250)=1015.625\mathbb{E}[Y_{\mathrm{k}}^2\mid A]=25^2(0.625)+50^2(0.250)=1015.625
Var⁡(Yk∣A)=1015.625−28.1252=224.609375\operatorname{Var}(Y_{\mathrm{k}}\mid A)=1015.625-28.125^2=224.609375

Answer

Answer

Because one squared thousand equals one million in squared base monetary units, the requested variance rounds to 225 million.

Var⁡(Y∣A)≈225 million(A)\boxed{\operatorname{Var}(Y\mid A)\approx 225\text{ million}\quad\text{(A)}}