Answer to Question #242557 in Statistics and Probability for PRS

Question

Suppose X₁ and X₂ are random variables of the discrete type which have the joint pmf p(x_1,x₂)=(x₁+2x₂)/18, (x_1,x₂)=(1,1),(1,2),(2,1),(2,2), zero elsewhere. Determine the conditional mean and variance of X₂,given X₁=x₁, for x₁=1 or 2. Also ,compute E(3X₁-2X₂)

Answer 1

Let $x_1=1$

Conditional distribution $f(x_2|x_1=1)$ is:

$\def\arraystretch{1.5} \begin{array}{c:c:c} x_2 & 1 & 2 \\ \hline f(x_2|x_1=1) & \frac{1/6}{8/18}=\frac{3}{8} &\frac{ 5}{8} \\ \end{array}$

Then conditional mean

$M(x_2|x_1=1)=1\cdot \frac{3}{8}+2\cdot \frac{5}{8}=\frac{13}{8};$

$M(x_2^2|x_1=1)=1\cdot \frac{3}{8}+4\cdot \frac{5}{8}=\frac{23}{8}$

Conditional variance

$Var(x_2|x_1=1)=M(x_2^2|x_1=1)-\left( M(x_2|x_1=1) \right)^2=\\ \frac{23}{8}-\left(\frac{13}{8}\right)^2=\frac{15}{64}$

Now let be $x_1=2$

In this case Conditional distribution $f(x_2|x_1=2)$ is:

$\def\arraystretch{1.5} \begin{array}{c:c:c} x_2 & 1 & 2 \\ \hline f(x_2|x_1=2) & \frac{\frac{4}{18}}{\frac{10}{18}}=\frac{2}{5} & \frac{3}{5} \\ \end{array}$

And further we have:

$M(x_2|x_1=2)=1\cdot \frac{2}{5}+2\cdot \frac{3}{5}=\frac{8}{5};$

$M(x_2^2|x_1=2)=1\cdot \frac{2}{5}+4\cdot \frac{3}{5}=\frac{14}{5}$

$Var(x_2|x_1=2)=\frac{14}{5}-(\frac{8}{5})^2=\frac{6}{25}$

For computing $E(3X_1-2X_2)$ we use the table of $(x_1,x_2)$ distribution:

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} (x_1,x_2) & (1,1) & (1,2)&(2,1)&(2,2) \\ \hline f(x_1,x_2) & \frac{3}{18} & \frac{5}{18}&\frac{4}{18}&\frac{6}{18}\\ \hdashline 3\cdot x_1-2\cdot x_2 & 1 & -1&4&2 \end{array}$

Therefore

$E(3X_1-2X_2)=\frac{3\cdot 1+5\cdot(-1)+4\cdot 4+6\cdot 2}{18}=\frac{26}{18}=\frac{13}{9}$