Independent random variables and their PDF

**courteous** · September 15th 2010, 11:15 AM

Let $f_{X,Y}(x,y)=C(x^2+y^2+1)^{-2}$ . Set $C$ so that the two vector components will be independent.

$f_{X,Y}(x,y)=f_X(x)*f_Y(y)$

I can get $f_X(x)$ (and similarly $f_Y(y)$ ) as $\int_{-\infty}^\infty f_{X,Y}(x,y)dy$ , but I can't integrate it. I suspect this isn't the right way to proceed.

The solution given is: "There is no such $C$ ."

**mr fantastic** · September 15th 2010, 01:57 PM

Originally Posted by courteous

$f_{X,Y}(x,y)=f_X(x)*f_Y(y)$

I can get $f_X(x)$ (and similarly $f_Y(y)$ ) as $\int_{-\infty}^\infty f_{X,Y}(x,y)dy$ , but I can't integrate it. I suspect this isn't the right way to proceed.

The solution given is: "There is no such $C$ ."

In each case the integration to get the marginal pdf's will involve the inverse tan function ....

**Moo** · September 15th 2010, 02:10 PM

Hello,

There is a unique C such that f_{X,Y} is a pdf (it will be the inverse of the integral of the function). We don't care about the value, we can just keep it as C. With C being different from 0, would it be possible to write the joint pdf as a product in the form g(x)*h(y) ? (which would mean that X and Y are independent)
The answer is obviously no...

Hence the answer "there is no such C"

**courteous** · September 25th 2010, 08:16 AM

Originally Posted by Moo

There is a unique C such that f_{X,Y} is a pdf (it will be the inverse of the integral of the function).

I don't understand what you mean by "inverse of the integral" ... isn't the "function" already p.d.f.?

Originally Posted by Moo

With C being different from 0, would it be possible to write the joint pdf as a product in the form g(x)*h(y) ? (which would mean that X and Y are independent)
The answer is obviously no...

Not obvious to me.

Why is $C\neq 0 \Rightarrow X, Y \text{ are independent }$ not feasible?

All I know is that $f_{X,Y}(x,y)=f_X(x)*f_Y(y)$ and that $\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)dxdy=1$ ... which is not a lot.

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