# Calculate E(XY) when X~N(0,1), Y=X^2~\chi^2(1)

• gimmytang
In summary, To find E(XY), we need to know the joint distribution of X and Y. By using the definition of joint distribution and the independence of X and Y, we can calculate the joint distribution as f(x,y) = f(x) * delta function(y-x^2). Alternatively, we can also represent the joint distribution as f(x,y) = f(x) when y=x^2 and 0 otherwise. This information can then be used to calculate the expected value of XY, which is equivalent to finding the third moment of a normal distribution.
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:
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.

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

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

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 by a moderator:

## 1. What is the formula for calculating E(XY) when X~N(0,1) and Y=X^2~\chi^2(1)?

The formula for calculating E(XY) in this scenario is E(XY) = E(X) * E(Y) + Cov(X,Y), where E(X) is the expected value of X, E(Y) is the expected value of Y, and Cov(X,Y) is the covariance between X and Y.

## 2. How do you find the expected value of a normal distribution with mean 0 and standard deviation 1?

The expected value for a normal distribution with mean 0 and standard deviation 1 is simply 0, as the mean of a normal distribution represents the center or average value of the data.

## 3. What is the expected value of a chi-square distribution with 1 degree of freedom?

The expected value of a chi-square distribution with 1 degree of freedom is 1, as the expected value for a chi-square distribution is equal to its degrees of freedom.

## 4. How do you calculate the covariance between two variables?

The covariance between two variables, X and Y, can be calculated using the formula Cov(X,Y) = E[(X-E(X))(Y-E(Y))]. This involves finding the expected value of the product of the difference between each variable and its expected value.

## 5. What is the relationship between a normal distribution and a chi-square distribution?

A chi-square distribution with 1 degree of freedom is the square of a standard normal distribution. This means that if X~N(0,1), then Y=X^2~\chi^2(1). In general, a chi-square distribution with k degrees of freedom is the sum of the squares of k independent standard normal random variables.

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