integrationmultivariable-calculusprobability distributions
I am supposed to show that the marginal PDF of a random variable $Y$ is:
$$g(y)= \begin{cases}
y, & 0<y \le 1\\
y^{-3}, & y>1\\
0, & \textrm{otherwise}
\end{cases}$$
given the joint PDF:
$$f(x,y)=\begin{cases}
\frac{4x^3}{y^3}, & 0<x<1, x<y\\
0, & \textrm{otherwise}
\end{cases}$$
I know one's supposed to integrate with respect to x, but I am not sure which boundries I should choose, and I can't seem to get the right answer.
Hints:
$P\left\{ X>Y\right\} =\int f_{Y}\left(y\right)P\left\{ X>Y\mid Y=y\right\} dy=\int f_{Y}\left(y\right)P\left\{ X>y\right\} dy$
Alternative:
$\int_{0}^{\infty}\int_{y}^{\infty}f_{\left(X,Y\right)}\left(x,y\right)dxdy=\int_{0}^{\infty}\int_{y}^{\infty}f_{X}\left(x\right)f_{Y}\left(y\right)dxdy=\int_{0}^{\infty}f_{Y}\left(y\right)\int_{y}^{\infty}f_{X}\left(x\right)dxdy$
Note that integrand $\int_{y}^{\infty}f_{X}\left(x\right)dx$ can be recognized as: $P\left\{ X>y\right\} $
Since $X$ and $Y$ are independent the joint PDF of $(X,Y)$ is the product of the PDFs of $X$ and $Y$.
in order to finally clarify the question, have a look at the following drawings
In the left picture, when you want to derive $f_X(x)$ you integrate $dy$, in fuction of $X$ thus your integral bounds are $y \in [0;\sqrt{x}]$
In the right picture, when you want to derivee $f_Y(y)$ you integrate in $dx$, in function of $Y$ thus your integration bouds are $x \in [y^2;1]$
Best Answer
Note that by definition, $$ g(y)=\int_{-\infty}^{\infty}f(x,y)\ dx. $$ There are three cases to consider: