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I have the following problem that I think I know how to solve, but I don't see why the given choices are as they are:

Suppose X and Y are jointly continuous random variables with joint 
probability density function given by

f(x, y) = 1/c, x > 0, y > 0, x^2 + y^2 ≤ 2; or 0 otherwise

where c is a normalising constant which does not depend on x and y.

What's the value of c?

What's the upper limit of the integrals? We integrate from 0 to what? 2?

Thanks!

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1 Answer

up vote 0 down vote accepted

Integrate over the part of the circle with radius $\sqrt{2}$ that lies in the first quadrant. If you really want to integrate, $y$ can travel from $0$ to $\sqrt{2-x^2}$, and then $x$ can travel from $0$ to $\sqrt{2}$. Or else we can change to polar coordinates.

There is no need to integrate, since the area is $\frac{\pi}{2}$.

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So c is essentially the area? –  Sorin Cioban Jul 23 '12 at 21:49
    
@SorinCioban: I would say exactly the area. In general, if you have a density function which is a constant $k$ over a two-dimensional region of (non-zero) area $A$, then $k=\frac{1}{A}$. Since in your case the constant is being called $\frac{1}{c}$, we have $c=A$. –  André Nicolas Jul 23 '12 at 21:57
    
Thanks a lot :) –  Sorin Cioban Jul 23 '12 at 22:03
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