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Joint probability- mean and variance? Pl refer image attached

Dec 26, 2017

The answer is $\text{(d) } \frac{3}{4} {x}_{2} \mathmr{and} \frac{3}{80} {x}_{2}^{2}$.

Explanation:

Part 1: Conditional Mean
By definition, $\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$,

where

$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$,

where

$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{A} f \left({x}_{1} , {x}_{2}\right) {\mathrm{dx}}_{1}$.

In this case, since ${x}_{1}$ is bounded by $0 < {x}_{1} < {x}_{2}$, our integration interval is $A = \left(0 , {x}_{2}\right) .$ Thus:

$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{0}^{{x}_{2}} 21 \text{ "x_1^2" "x_2^3" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ "x_2^3int_0^(x_2)x_1^2" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ } {x}_{2}^{3} {\left[\frac{1}{3} {x}_{1}^{3}\right]}_{0}^{{x}_{2}}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 7 \text{ } {x}_{2}^{6}$

Thus,

$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$

$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \left(21 \text{ "x_1^2" "x_2^3)/(7" } {x}_{2}^{6}\right)$

$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{3 \text{ } {x}_{1}^{2}}{{x}_{2}^{3}}$.

Finally,

$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = {\int}_{A} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$

Since we are once again integrating with respect to ${x}_{1}$, the integration interval is the same as before:

$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{0}^{{x}_{2}} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$

color(white)("E"(X_1|X_2))=3/x_2^3 int_0^(x_2)x_1^3" "dx_1

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{x} _ {2}^{3} {\left[{x}_{1}^{4} / 4\right]}_{0}^{{x}_{2}}$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4 {x}_{2}^{3}} \left[{x}_{2}^{4}\right]$

$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4} {x}_{2}$

Part 2: Conditional Variance

The conditional variance is

"Var"(X_1|X_2)="E"(X_1^2|X_2)-["E"(X_1|X_2)]^2

I'll leave the calculation of $\text{E} \left({X}_{1}^{2} | {X}_{2}\right)$ as an exercise. (Hint: just replace ${x}_{1}$ with ${x}_{1}^{2}$ in the formula for $\text{E} \left({X}_{1} | {X}_{2}\right)$.)

The result is:

$\text{Var} \left({X}_{1} | {X}_{2}\right) = \frac{3 {x}_{2}^{2}}{5} - {\left[\frac{3 {x}_{2}}{4}\right]}^{2}$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3 {x}_{2}^{2}}{5} - \frac{9 {x}_{2}^{2}}{16}$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{1}{5} - \frac{3}{16}\right]$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{16 - 15}{80}\right]$

$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{80} {x}_{2}^{2}$.