## Expectation of a Uniform Distribution

Let \(X\) be a random variable with uniform distribution over the integers from \(-n\) to \(n\). Let \(Y := X^2\).

Which of the following are true?

- Since \(PDF_X(x)\) is symmetric around 0, we know the mean has to be 0.
- All values of \(Y\) are nonnegative and most of them are actually positive with non-zero probability. There is no way for the mean to be 0. It has to be some positive value.
- Obvious, since \(E[X]=0\) and \(E[Y]>0\).
- True by linearity of expectation, no matter what \(X\) and \(Y\) are.
- This is tricky. The equation does not hold in general for non-independent \(X\) and \(Y\). However, in this particular case, it happens to hold. To see this, note that the right hand side is 0, since \(E[X]=0\). At the same time, the random variable \(XY=X^3\). Its PDF is symmetric around 0, so its mean must be 0 as well.
- \(X\) and \(Y\) are obviously not independent.
- \(E[Y]=\sum_{i=0}^n 2i^2\frac{1}{2n+1} < \sum_{i=0}^n 2i^4\frac{1}{2n+1} = E[Y^2]\).