If $Y=X^n$, with $n$ and the expectation and variance of $X$ known, what is the expectation and variance of $Y$?
Asked
Active
Viewed 1.8k times
5
-
7Knowing the expectation and variance of $X$ is not enough information to know the moments of all orders. – Stefan Hansen Nov 01 '13 at 06:47
-
I have to agree with @StefanHansen. Since expectation is just the integral of the function, it does not really give us information about the integral of a function raised to the $n$th power, in general. Although, there may be a nice form in certain cases. – Suugaku Nov 01 '13 at 06:52
-
What if we assume X is distributed normal? – frank Nov 01 '13 at 06:57
-
Have a look at this question. – Stefan Hansen Nov 01 '13 at 08:46
2 Answers
3
To give this an answer, as Stefan Hansen comments:
Knowing the expectation and variance of X is not enough information to know the moments of all orders.
As an explicit example, let $X$ be normally distributed with mean zero and variance 1, and let $W$ be +1 or -1 with probability 1/2 (the Rademacher distribution). Then $EX = EW = 0$ and $E X^2 = E W^2 = 1$, but $E X^4 = 3$ while $E W^4 = 1$.
In the special case of the normal distribution, the distribution of a normal random variable are completely determined by its mean and variance, hence so are its higher moments. You can find various formulas for those moments on Wikipedia.
Nate Eldredge
- 101,664
-5
Here's a partial answer: If $X$ is a Bernoulli variable, and $n\gt0$, then $E[X^n]=E[X]$ and $Var(X^n)=Var(X)$.
bof
- 6,376