Solutions - 5240-07

(a)

$$\begin{array}{c}E[X]=\frac{3}{2}{\int }_0^1{\int }_0^{1}x(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1{x}^3+\frac{x}{3}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^4}{4}+\frac{x^2}{6}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{4}+\frac{1}{6}]\gg \gg \frac{5}{8}\end{array}$$

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(b)

By symmetry, $E[Y]=E[X]$:

$$E[Y]=\frac{5}{8}$$

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(c)

$$\begin{array}{c}E[XY]=\frac{3}{2}{\int }_0^1{\int }_0^{1}xy(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1\frac{x^3}{2}+\frac{x}{4}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^4}{8}+\frac{x^2}{8}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{8}+\frac{1}{8}]\gg \gg \frac{3}{8}\end{array}$$

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(d)

(1) Compute $E\left[X^2\right]$:

$$\begin{array}{c}E\left[X^2\right]=\frac{3}{2}{\int }_0^1{\int }_0^1{x}^2(x^2+y^2)dydx\gg \gg \frac{3}{2}{\int }_0^1{x}^4+\frac{x^2}{3}dx\\ \\ \gg \gg \frac{3}{2}{[\frac{x^5}{5}+\frac{x^3}{9}]}_0^1\gg \gg \frac{3}{2}[\frac{1}{5}+\frac{1}{9}]\gg \gg \frac{7}{15}\end{array}$$

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(2) Compute $\text{Var}[X]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2=\frac{7}{15}-{\left(\frac{5}{8}\right)}^2\gg \gg \frac{73}{960}\end{array}$$

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(e)

(1) Compute $E\left[Y^2\right]$:

By symmetry, $E\left[Y^2\right]=E\left[X^2\right]=\frac{7}{15}$.

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(2) Compute $\text{Var}[Y]$:

$$\begin{array}{c}\text{Var}[Y]=\text{Var}[X]\gg \gg \frac{73}{960}\end{array}$$

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(f)

$$\begin{array}{c}\text{Cov}[X,Y]=E[XY]-E[X]E[Y]=\frac{3}{8}-\frac{25}{64}\gg \gg -\frac{1}{64}\end{array}$$

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(g)

$$\begin{array}{c}\rho [X,Y]=\frac{\text{Cov}[X,Y]}{\sqrt{\text{Var}[X]\text{Var}[Y]}}=\frac{-\frac{1}{64}}{\sqrt{{\left(\frac{73}{960}\right)}^2}}\gg \gg -\frac{15}{73}\end{array}$$

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(h)

Since $\text{Cov}[X,Y]\ne 0$, we can conclude that $X$ and $Y$ are not independent.