Dtyjlui

I am interested to understand what's the meaning of using powers in expected value.
Both mathematically and maybe intuitively, what does it even mean "the power of a random variable"?

Thanks!

Here's a simple example in the discrete case.

Suppose we have a random variable $X$ for which $$Pr[X = -1] = 1/2, \ Pr[X = 1] = 1/3, \ Pr[X = 2] = 1/6.$$ Think of the set ${-1, 1, 2}$ as the set of possible outcomes of $X$, and the probability of observing one of these outcomes is described as above.

Now, if I want to ascertain the expected value of $X$, which in the frequentist viewpoint is in a sense the "long-run average" of many observations of $X$, then $$operatorname{E}[X] = (-1)Pr[X = -1] + (1)Pr[X = 1] + (2)Pr[X = 2] = -frac{1}{2} + frac{1}{3} + frac{2}{6} = frac{1}{6}.$$ Note that the expected value need not be one of the outcomes of $X$; in fact, $Pr[X = 1/6] = 0$ as implied by the probability distribution I've provided. What $1/6$ means here is that if we were to observe many realizations of $X$, and average them together, the result would tend toward $1/6$ as the number of observations increases.

So, what does something like $operatorname{E}[X^2]$ mean? The expression $X^2$ is another random variable that can be calculated from $X$; it means that whatever outcome $X$ is, $X^2$ is its square. Since $X in {-1, 1, 2}$, then $X^2 in {1, 4}$. We get this by taking the square of each outcome, and taking all the unique values. The probability mass function of $X^2$ is $$Pr[X^2 = 1] = Pr[X = -1] + Pr[X = 1] = frac{1}{2} + frac{1}{3} = frac{5}{6}, \ Pr[X^2 = 4] = Pr[X = 2] = frac{1}{6}.$$ Then $$operatorname{E}[X^2] = (1)Pr[X^2 = 1] + (4)Pr[X^2 = 4] = frac{5}{6} + frac{4}{6} = frac{3}{2}.$$

Of course, you can similarly calculate things like $operatorname{E}[X^3]$, $operatorname{E}[X^4]$, and so forth.

The idea here is that $X^n$ corresponds to another random variable whose set of outcomes (its support) and probability distribution are determined by those of $X$ in a way that naturally extends from the familiar rules of algebra. You can do this for continuous random variables too; or random variables with more complicated definitions. Note that $operatorname{E}[X]$ may be well-defined but $operatorname{E}[X^n]$ may not for certain $n$; a trivial example is if $Pr[X = 0] > 0$ and $n < 0$, in which case you'd have division by $0$. There also exist examples where $operatorname{E}[X^2]$ fails to exist even though $operatorname{E}[X]$ does.