Expected value and standard deviation of $X^2$ in terms of moments of $X$

Question

If $X$ is a discrete random variable with expected value $\mu$ and standard deviation $\sigma$, and $Y = X^2$, how can we describe the expected value and standard deviation of $Y$?

Would expected value of $Y = \mu^2$?

Answer 1

For the mean, the answer is simple: $$E[Y]=E[X^2]=\operatorname{var}(X)+E[X]^2=\sigma^2+\mu^2$$

But, you can't find the variance of $Y$, without the fourth moment of $X$ because: $$\operatorname{var}(Y)=E[Y^2]-E[Y]^2=E[X^4]-(\sigma^2+\mu^2)^2$$

Answer 2

The variance is given in this related question. The mean of $X^2$ depends on the first two moments of $X$ and the variance of $X^2$ depends on the first four moments of $X$. These are:

$$\begin{align} \mathbb{E}(X^2) &= \mu^2 + \sigma^2, \\[6pt] \mathbb{V}(X^2) &= 4 \mu^2 \sigma^2 + 4 \mu \gamma \sigma^3 + (\kappa-1) \sigma^4, \\[6pt] \end{align}$$

where $\gamma$ and $\kappa$ are the skewness and kurtosis of $X$. (To get the standard deviation you just take the square-root of the variance.)