$\displaystyle k \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (6-x-y)dx dy = 8k$
Can someone describe why this is true? I have no idea what steps to take to get this answer.
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?
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.
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:
$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx \, dy ~=~ \int_2^4 \int_0^2 k(6-x-y) \, dx \, dy$
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?
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.