Joint Density function Question

**padawan** · January 10th 2011, 01:49 PM

Hi all,

I have been going though a Q and have no idea how to even start it:

Q:

Suppose that $X,Y$ have the joint distribution:

$f(x,y) = c(1-x^2-y^2)^{1/2}$ , $x^2 + y^2 \le 1$

Find c..?

Now my initial thoughts are that the bounded unit cirrcle region gives the limits of integration such that:

$P(X,Y \subset A)= _A\int \int f(x,y) dydx$ , in our case A is bound by the unit circle, giving:

$= c \displaystyle \int_{-(1-x^2)^{1/2}}^{(1-x^2)^{1/2}} \int_{-(1-y^2)^{1/2}}^{(1-y^2)^{1/2}} 1-x^2-y^2 dx dy$

does this look like the right method to employ? I am not convinced..... the answer in the text book (no method though) $3/2 \pi$ ... can anyone offer any guidance on this one??

Many thanks for reading!

**snowtea** · January 10th 2011, 01:59 PM

Try to integrate in polar coordiantes.

$r^2 = x^2 + y^2$

**matheagle** · January 10th 2011, 04:08 PM

yes you need polar, but your bounds and integrand are wrong in rectangular as well

**matheagle** · January 10th 2011, 04:10 PM

It would be

$1= c \displaystyle \int_{-1}^1\int_{-(1-y^2)^{1/2}}^{(1-y^2)^{1/2}} \sqrt{1-x^2-y^2} dx dy$

which becomes, in polar....

$1= c \displaystyle \int_0^{2\pi}\int_0^1 r\sqrt{1-r^2}dr d\theta$

**padawan** · January 11th 2011, 11:05 AM

thats great, thanks a lot guys.

just seen my mistake... dont worry!

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