integrationprobability distributions
Given that the joint probability density function associated with two continuous random
variables X and Y is given by f(x,y): \begin{cases}
Pe^{-y}, & \text{− y < x < y and 0 < y < ∞ } \\
0, & \text{otherwise}
\end{cases}
I need to find the value of the constant P.
I know that to do this I need to find the double integral which will equal 1. However I'm confused as to what the limits for the double integral will be?
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 $Y$.
$$ \Bbb P (X+Y>3)=\Bbb P (Y>3-X)=1-\Bbb P(Y<3-X) \\=1-\int_0^{3}\int_0^{3-y}e^{-x-y}dx~dy=4e^{-3} $$
We are looking for the region to the right of the triangle enclosed between the x axis, y axis and the line, which should be the same as the converse of the region to the left, im not sure why your answer is like that, unless there is some other piece of information that is missing from the question, or maybe I've made a mistake.
Best Answer
So, as you said
$$1= \int_{\Bbb R^2} f_{X,Y}(x,y)dxdy$$
Now, the limits of $y$ are $0$ and $\infty$ and of $x$ are $-y $ and $y$ so you get
$$1 = \int_0^{\infty} \int_{-y}^{y} Pe^{-y}dx dy $$