# Math Help - Double integration problem

1. ## Double integration problem

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

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

2. It's a joke, right? Maybe the 8 was bent over in the original statement, " $\infty$"?

That region is very much unbounded. You will be hard-pressed to drag a finite solution from it.

3. Originally Posted by TKHunny
It's a joke, right? Maybe the 8 was bent over in the original statement, " $\infty$"?

That region is very much unbounded. You will be hard-pressed to drag a finite solution from it.
Whew, so I'm not crazy.

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

The question is: find $k$ such that

$f_{X,Y}(x,y)dx dy = \left\{ \begin{array}{lr}(6-x-y), &0

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

The first step in the solution was

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

which makes no sense to me. Is there something I'm missing?

4. This shows us the value of presenting the actual problem statement. The limits make all the difference.

$(-\infty,\infty)$ means $(0,2)$ 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.

Consider for a moment a simpler function:

$f(x)= \begin{cases} 1/2, & \mbox{if } 3

If you are asked to evaluate this integral:

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

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

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

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

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

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

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

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx \, dy ~=~ \int_2^4 \int_0^2 k(6-x-y) \, dx \, dy$

6. Originally Posted by TKHunny
This shows us the value of presenting the actual problem statement. The limits make all the difference.

$(-\infty,\infty)$ means $(0,2)$ 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.
Sorry, I should have included the whole problem statement.

Can you expand on the P(X < 3 - Y) bit?

7. 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.

8. Originally Posted by TKHunny
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.
Sorry, I'm still not sure where to go with the solution. Do you mean I should evaluate $\int^4_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy$?

9. Originally Posted by garymarkhov
Sorry, I'm still not sure where to go with the solution. Do you mean I should evaluate $\int^4_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy$?
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.

10. Originally Posted by drumist
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.

Is it that I need to restrict both integrals (see attachment)? So maybe $\int^{3-x}_2 \int^{3-y}_0 k(6-x-y) \, dx \, dy$
$\int_2^3 \int_0^{3-y} k(6-x-y) \, dx \, dy$