# Probability density function, multivariate distribution

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

1. 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?

1. 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 $$E(X*Y)=∫xy *f(x,y)d(x,y)$$?

## 2 Answers

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

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

• This is a merely a comment (to the second question) more than an answer. – Jean Marie Feb 16 at 6:28