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1) So I have this joint PDF: $$ f(x,y)= \begin{cases} e^\frac{-x}{y}e^{-y} & \text{ for } x>0 , y>0\\ 0 & \text{ otherwise} \end{cases} $$

and I found marginal density of y $$ f_Y(y)= \begin{cases} ye^{-y} & \text{ for }y \leq 0\\ 0 & \text{ for }y> 0\\ \end{cases} $$

and now I have to find $P(2 < x < 3 | Y=y)$

$$ \int_2^3 f_{xy}(X|Y)/F_Y(y) dx $$

and find a value, I get like $e^{-3y}+e^{-2y}$, am I doing this question right?

2) Let (X,Y) have joint pdf $$ f(x,y)= \begin{cases} \frac{12}{5}x(2-x-y) & \text{ for } 0<x<1, 0<y<1\\ 0 & \text{ otherwise} \end{cases} $$

and I have to find $F_{x|y}(x|y)$ for $0 < y < 1$

but first I found marginal density of $Y$

$$ f_Y(y)= \begin{cases} \frac{12}{5}[\frac{2-y}{2}-\frac{1}{3}] & \text{ for } 0<y<1\\ 0 & \text{ otherwise}\\ \end{cases} $$

what do I have to do next? my professor told me I have to find marginal density in terms of $0 < x < 1$ but I don't know how to pursue that.

my professor did:

$$ f_Y(y)= \begin{cases} \frac{x(2-x-y)}{4-3y}x6 & \text{ for } 0<x<1\\ 0 & \text{ otherwise} \\ \end{cases} $$

but how did he find such that answer?

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Your marginal density for $Y$ is correct except the ranges are the wrong way around. It should be:

$$f_Y(y) = \begin{cases} ye^{-y} & \text{ for }y \gt 0\\ 0 & \text{ for }y\leq 0.\\ \end{cases} $$

\begin{align} P(2\lt X\lt 3\mid Y=y) &= \int_{x=2}^{3} f_{X|Y}(x|y)\;dx \\ &= \int_{x=2}^{3} \dfrac{f_{X|Y}(x|y)}{f_Y(y)} \;dx \\ &= \int_{x=2}^{3} \dfrac{e^{-x/y}e^{-y}}{ye^{-y}} \;dx \\ &= \int_{x=2}^{3} \dfrac{e^{-x/y}}{y} \;dx \\ &= \left[ -e^{-x/y}\right]_{x=2}^{3} \\ &= e^{-2/y}-e^{-3/y}. \end{align}

2.

Your marginal density for $Y$ is correct but you could simplify it to

$$f_Y(y) = \dfrac{2}{5}\left(4-3y\right),\quad\text{for } 0\lt y\lt 1.$$

Then, for $0\lt x\lt 1,\;0\lt y\lt 1$,

$$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} = \dfrac{\dfrac{12}{5}x\left(2-x-y\right)}{\dfrac{2}{5}\left(4-3y\right)} = \dfrac{6x\left(2-x-y\right)}{4-3y}.$$

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