# Expectation of XY

1. Apr 29, 2005

### gimmytang

X~N(0,1), Y=X^2~$$\chi^2$$(1), find E(XY).

My thoughts are in the following:
To calculate E(XY), I need to know f(x,y), since $$E(XY)=\int{xyf(x,y)dxdy}$$. To calculate f(x,y), I need to know F(x,y), since f(x,y)=d(F(x,y)/dxdy.

$$F(x,y)=P(X\leq x, Y\leq y) \\ =P(X\leq x, X^{2} \leq y)\\ =P(X\leq x, -\sqrt{y} \leq X \leq \sqrt{y})$$
Thus,
$$F(x,y) =P(-\sqrt{y} \leq X \leq x)P(x<\sqrt{y})+P(-\sqrt{y} \leq X \leq \sqrt{y})P(x > \sqrt{y})$$
Then I don't know how to calculate the four components of probabilities accordingly. Anyone gives a hand?

Thanks!
gim

Last edited: Apr 29, 2005
2. Apr 29, 2005

### juvenal

I think you're making this too hard for yourself. x and y have 100% correlation. I think you essentially want to calculate the third moment of a normal distribution, since x*y = x^3. So find the expected value of x^3 = the third moment.

3. May 1, 2005

### gimmytang

Thank you for your useful hint! The result following your method is E(XY)=0, then cor(X,Y)=0. In this sense X and Y are uncorrelated, but they are fully associated.
gim

4. May 2, 2005

### gimmytang

I am still wondering the joint distribution of X and Y. There must be a solution to that. If it is not too difficult, please give me some hints.
Thanks!
gim

5. May 2, 2005

### juvenal

f(x,y) = f(x) * f(y|x)

So, you need f(y|x). However, once you know x, you know y exactly, so
f(y|x) = delta function(y - x^2).

So f(x,y) = f(x) * delta function(y-x^2).

I'm not sure if I've seen delta functions outside of physics, actually. Here's a writeup I found:

http://www.tutorfusion.com/eTutor/physics/e&m/1/5/1_5_dirac_delta_function.htm

If you don't want to use delta functions, I guess you could just say:

f(x,y) = f(x) when y= x^2
= 0 otherwise

Last edited: May 2, 2005