What Is E(X^3) When X~(μ, σ^2)?

  • Thread starter Thread starter jsndacruz
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating Cov(X, Y) where X is a normally distributed random variable and Y is defined as X^2. The key formula used is Cov(X, Y) = E(X^3) - E(X)E(X^2), but the user is uncertain about how to compute E(X^3). It is clarified that if X is standard normal, E(X^3) is zero, leading to Cov(X, Y) also being zero. The conversation also touches on using moment generating functions or integration techniques to find E(X^3) for non-standard normal distributions. The user has submitted their problem set and is awaiting grading for confirmation of their solution.
jsndacruz
Messages
3
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
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.
 
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.
 
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!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top