# Thread: f(x,y)= 2e^-2e^-2y

1. ## f(x,y)= 2e^-2e^-2y

Does anyone know the solution to this type of problem?

f(x,y)= 2e^-2e^-2y } for 0<x<inf

P(x<a) for a constant a

P(x<1, y<1)

P(x<y)

I have no idea how to approach this.
Thanks!!!
Toni

2. ## Re: f(x,y)= 2e^-2e^-2y

The joint density function has the property that for any set $D$ of points on the plane
$\operatorname{Prob}((X,Y)\in D)=\int_D f(x,y)\,dxdy$
So for each probability that you have to find determine the appropriate $D$ and calculate the required integral.

How?

4. ## Re: f(x,y)= 2e^-2e^-2y

Can you be more specific and provide one of the solutions? Not sure how to plug in the probabilities here.

5. ## Re: f(x,y)= 2e^-2e^-2y

Whoops! Wrong D.

-Dan

6. ## Re: f(x,y)= 2e^-2e^-2y

For example, for (c) you need to find a set $D$ such that $\operatorname{Prop}(X<Y)=\operatorname{Prop}((X,Y )\in D)$. Then you can calculate the right-hand side using the formula in post #2. Obviously, $D=\{(x,y)\in\Bbb R^2\mid x<y\}$.

Then
\begin{align*}
\int_D f(x,y)\,ds&= \int_0^\infty\int_x^\infty 2e^{-x-2y}\,dy\,dx\\
&=\int_0^\infty e^{-x}\int_x^\infty 2e^{-2y}\,dy\,dx\\
&=\int_0^\infty e^{-x}\left(\left.-e^{-2y}\right|_x^\infty\right)\,dx\\
&=\int_0^\infty e^{-3x}\,dx\\
&=\frac{1}{3}
\end{align*}