Variance of a distribution's product with itself

Question

I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) > Var(X)$ where $i > 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.

Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?

Answer 1

$Var(X^i)$ = $\mathbb{E}[(X^i - \mathbb{E}[X^i])^2]$ = $\mathbb{E}[X^{2i}] - (\mathbb{E}[X^i])^2$ by definition. This expresses the variance of $X^i$ in terms of moments of $X$.

The generalization is false, because $\mathbb{E}[(\lambda X)^{2i}]$ = $|\lambda|^{2i}\mathbb{E}[X^{2i}]$ implies that the scale parameter of $(\lambda X)^i$ will be smaller than the scale parameter of $\lambda X$ when $\lambda$ is sufficiently close to zero and $i \gt 1$.

Answer 2

Quick note, you may find useful discussion of why the formula for estimating the standard deviation for a sample uses $n-1$ as opposed to $n$.

• Paul Savory Why divide by (n-1) for sample standard deviation
• Graphpad Why use n-1 when calculating a standard deviation?
• Andrew Hardwick Why there is a Minus One in Standard Deviations