$\displaystyle k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k$

Can someone describe why this is true? I have no idea what steps to take to get this answer.

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- Mar 11th 2010, 04:49 PMgarymarkhovDouble integration problem
$\displaystyle k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k$

Can someone describe why this is true? I have no idea what steps to take to get this answer. - Mar 11th 2010, 05:01 PMTKHunny
It's a joke, right? Maybe the 8 was bent over in the original statement, "$\displaystyle \infty$"?

That region is very much unbounded. You will be hard-pressed to drag a finite solution from it. - Mar 11th 2010, 05:20 PMgarymarkhov
Whew, so I'm not crazy.

No, it's not a joke. I've got it in my probability notes.

The question is: find $\displaystyle k$ such that

$\displaystyle f_{X,Y}(x,y)dx dy = \left\{ \begin{array}{lr}(6-x-y), &0<x<2, 2<y<4\\ 0, &otherwise \end{array} \right.$

is a joint density. Find $\displaystyle P(X+Y<3)$ and the marginal density of $\displaystyle X$.

The first step in the solution was

$\displaystyle 1= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f_{X,Y}(x,y)dx dy = k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k$

which makes no sense to me. Is there something I'm missing? - Mar 11th 2010, 07:56 PMTKHunny
This shows us the value of presenting the actual problem statement. The limits make all the difference.

$\displaystyle (-\infty,\infty)$ means $\displaystyle (0,2)$ in the x-direction. It is no longer unbounded.

Please evaluate the iterated integral with these very useful finite limits.

For P(X + Y < 3), notice that this is nicely related to P(X < 3 - Y) and you should be almost done. - Mar 11th 2010, 09:09 PMdrumist
To clarify, your notation is bad.

Consider for a moment a simpler function:

$\displaystyle f(x)= \begin{cases} 1/2, & \mbox{if } 3<x<5 \\ 0, & \mbox{elsewhere} \end{cases}$

If you are asked to evaluate this integral:

$\displaystyle \int_{-\infty}^{\infty} f(x) \, dx$

then you would have to split it into separate regions because it's a piecewise function, i.e.,

$\displaystyle \int_{-\infty}^{\infty} f(x) \, dx ~=~ \int_{-\infty}^3 f(x) \, dx + \int_3^5 f(x) \, dx + \int_5^{\infty} f(x) \, dx$

$\displaystyle =~ \int_{-\infty}^3 0 \, dx + \int_3^5 (1/2) \, dx + \int_5^{\infty} 0 \, dx$

Naturally, the integrals on the sides equal zero, so we just write it as

$\displaystyle \int_{-\infty}^{\infty} f(x) \, dx ~=~ \int_3^5 (1/2) \, dx$

Extending this logic into two variables with your problem shows that you should be evaluating this:

$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx \, dy ~=~ \int_2^4 \int_0^2 k(6-x-y) \, dx \, dy$ - Mar 12th 2010, 05:08 AMgarymarkhov
- Mar 12th 2010, 01:37 PMTKHunny
It might be easier to worry about P(Y < 3 - X). You have new limits. Your Y-limits are no longer all of (2,4). You might get (2,3-X)? Since x runs in (0,2), that seems quite reasonable.

- Mar 14th 2010, 05:46 PMgarymarkhov
- Mar 14th 2010, 10:14 PMdrumist
Close, but you made a mistake with the limits of integration.

Graph the region that the density is defined on. Then plot the line x+y=3. You will then be able to identify the region corresponding with x+y<3. You should integrate the joint density over this region only.

Can you identify your error? - Mar 16th 2010, 12:48 PMgarymarkhov
- Mar 17th 2010, 10:24 PMdrumist
Sorry for taking so long to respond.

The region you displayed is correct! However, your endpoints are still not correct. :\

This is the correct setup:

$\displaystyle \int_2^3 \int_0^{3-y} k(6-x-y) \, dx \, dy$

The error you originally made was that you took y from 2 to 4, but the region goes from 2 to 3.