What Is E(X^3) When X~(μ, σ^2)?

[jsndacruz]

Got stuck on an intermediate step in a larger problem. We are given X~(μ, σ^2), i.e. X is a random variable with standard normal distribution, and that Y=X^2. The question then asks to compute Cov(X,Y):

Cov(X,Y) = Cov(X,X^2) = E(X^3) - E(X)E(X^2) = E(X^3) - (μ)(μ^2 + σ^2)

I can't go any further however, because I don't know what E(X^3) is! I computed the last term using the variance equation and re-arranging, but I can't use that same trick.

 

[Stephen Tashi]

What techniques are you permitted to use in this problem? Moment generating functions?

I suppose you could write down the integral the defines E(X^3) and use integration by parts to reduce the X^3 to an X^2.

 

[alan2]

You say that X is standard normal which means N(0,1) but then you say it is normal. If it is standard normal then the integral is trivially zero as is Cov(X,Y). If it is not then the moment generating function is surely the least painful method.

 

[jsndacruz]

Thank, alan2 and Stephen. Alan2, I did mean normal distribution - not standard normal. I haven't visited Probability/Stats in a while so I had to look over moment generating functions before I could solve. I handed in the problem set yesterday, so I can't report back the final answer until it's graded. Thanks for your help!

 

[blue_raver22]

To compute E(X^3), we can use the moment generating function (MGF) of X, which is defined as M(t) = E(e^(tX)). Using this, we can find the third moment of X by taking the third derivative of the MGF at t=0.

So, E(X^3) = M'''(0) = (μ^3 + 3μσ^2)e^(tμ + t^2σ^2)|_(t=0) = μ^3 + 3μσ^2.

Substituting this into the equation for Cov(X,Y), we get:

Cov(X,Y) = E(X^3) - (μ)(μ^2 + σ^2) = (μ^3 + 3μσ^2) - (μ)(μ^2 + σ^2) = 2μσ^2

Therefore, Cov(X,Y) = 2μσ^2.