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

Results 1 to 6 of 6

- June 22nd 2014, 07:08 AM #1

- Joined
- Jun 2014
- From
- Little Neck, New York
- Posts
- 6

- June 22nd 2014, 02:14 PM #2

- Joined
- Oct 2009
- Posts
- 5,527
- Thanks
- 773

## 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.

- June 24th 2014, 02:02 AM #3

- Joined
- Jun 2014
- From
- Little Neck, New York
- Posts
- 6

- June 24th 2014, 02:03 AM #4

- Joined
- Jun 2014
- From
- Little Neck, New York
- Posts
- 6

- June 24th 2014, 05:43 AM #5

- June 24th 2014, 06:05 AM #6

- Joined
- Oct 2009
- Posts
- 5,527
- Thanks
- 773

## 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*}