Suppose $X$ and $Y$ have joint density $f(x,y)=c(x+y)$ for $0<x$, $y<1$. Now my question is, what is $c$? I tried to solve, which is reasonable in my opinion. But it don't seem to work: \begin{align} \int\limits_0^\infty \int\limits_0^1 f(x,y)\ dx\ dy = \int\limits_0^\infty \int\limits_0^1 c(x+y) \ dy\ dx = \int\limits_0^\infty \big[cxy + \frac{1}{2}cy^2 \big]^{y=1}_{y=0}\ dx = \int\limits_0^\infty cx + \frac{1}{2}c \ dx. \end{align} What to do, are the restrictions on $x$ related to $y$?
Asked
Active
Viewed 56 times
0
-
The joint density function "lives" on the square $S$ with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$, so you need the double integral $\iint_S c(x+y),dy,dx$. Express as an iterated integral. If you draw the square, the limits will be clear. – André Nicolas Jun 03 '15 at 15:53
1 Answers
0
I quess you should take the first integral between $0$ and $1$, by the condition $0 <x,y<1$. Then you would get: $$1= \int_0^1 \int_0^1f(x,y) dx dy = \int_0^1 cx +\frac{1}{2}c dx = [cx^2+\frac{1}{2}cx]_0^1=\frac{3}{2}c$$.
This implies $c= \frac{2}{3}$.
N.B. I would have posted a comment, but my reputation doesn't allow it.
Greetings
Tim Huijgens
- 342