Solutions - 5200-02

(1) Compute the marginal distribution of $X$ by integrating $f_1$ with respect to $y$:

$$\begin{array}{c}{f}_X(x)={\int }_0^{\infty }6e^{-2x}{e}^{-3y}dy\\ \\ \gg \gg e^{-2x}\underset{b\to \infty }{\lim }{[-2e^{-3y}]}_0^b\gg \gg 2e^{-2x},x>0\end{array}$$

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(2) Compute the marginal distribution of $Y$ by integrating $f_1$ with respect to $x$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{\infty }6e^{-2x}{e}^{-3y}dx\gg \gg 3e^{-3y},y>0\end{array}$$

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(3) Determine independence by multiplying the marginal PDFs:

$$f_X(x)f_Y(y)=6e^{-2x}{e}^{-3y}=f_1$$

Since the product of the marginal PDFs equals the joint PDF, $X$ and $Y$ are independent.

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(4) Compute the marginal distribution of $X$ by integrating $f_2$ with respect to $y$:

$$\begin{array}{c}{f}_X(x)={\int }_0^{1-x}24xydy\\ \\ \gg \gg 24x{\left[\frac{y^2}{2}\right]|}_0^{1-x}\gg \gg 12x{(1-x)}^2\end{array}$$

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(5) Compute the marginal distribution of $Y$ by integrating $f_2$ with respect to $x$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{1-y}24xydx\\ \\ \gg \gg 24y{\left[\frac{x^2}{2}\right]|}_0^{1-y}\gg \gg 12y{(1-y)}^2\end{array}$$

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(6) Determine independence by multiplying the marginal PDFs:

$$f_X(x)f_Y(y)=144xy{(1-x)}^2{(1-y)}^2\ne 24xy=f_2$$

Therefore, $X$ and $Y$ are not independent.