Probability density function, multivariate distribution

Question

I have no idea what i have to do.

I have pdf

$$f(x,y)=a│xy│\exp(-(x^2+y^2))$$

• 1. Find constant a. With

$$∫∫f(x,y)d(x,y)=1$$

It's ok, not sure but $a=0.25$ counted in the head...

(here $\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(x,y)d(x,y)=4a \exp(-(x^2+y^2))=1$.)

• 2. Find pdf of X and pdf of Y. here I have really no idea.

Or should I find F(x,y) and derive on x for pdf of X and the same for Y?

• 3. Find E(X),E(X(X-Y))

so for $X(X-Y)$, can i find $E(X(X-Y))=E(X*X)-E(X*Y)$ like $E(X*X)=∫x^2 *f(x,y)d(x,y)$ and $E(X*Y)=∫xy *f(x,y)d(x,y)$?

Answer 1

It is obvious that $X$ and $Y$ are independent since

$$f(x,y)=a*\left(|x| e^{-x^2}\right)\left(|y| e^{-y^2}\right)=a*g(x)*h(y)$$

so $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) dx dy =a* \left(\int_{-\infty}^{\infty} |x| e^{-x^2} dx\right)\left(\int_{-\infty}^{\infty} |y| e^{-y^2}dy\right)$$

It is easy to check that $$\int_{-\infty}^{\infty} |x| e^{-x^2}dx=2\int_{0}^{\infty} x e^{-x^2}dx=-e^{-x^2}|_{0}^{+\infty}=1$$

For question 2 since two variables are independents, you already calculated it!!

$f(x)=|x| e^{-x^2}$ and $f(y)=|y| e^{-y^2}$

for question 3

$$E(X)=\int_{-\infty}^{\infty} x|x| e^{-x^2} dx\overset{?}{=}0$$

$$E(X(Y-X))=E(XY)-E(X^2)=E(X)E(Y)-E(X^2)=0-E(X^2)$$

$$E(X^2)=\int_{-\infty}^{\infty} x^2|x| e^{-x^2} dx$$

$$=2\int_{0}^{\infty} x^2|x| e^{-x^2} dx=2\int_{0}^{\infty} x^3 e^{-x^2} dx$$

in last integrate you can use $t=x^2$ and use gamma distribution

Answer 2

The probability density functions of $X$ and $Y$ can be obtained as follows: $$f_X(x) = \int_{-\infty}^\infty f(x,y)dy$$ and $$f_Y(y) = \int_{-\infty}^\infty f(x,y)dx.$$