The expectation of the $K^{th}$ raw moment of $X$ given $X$: $E(X^k|X)$

Question

Intuitively I believe $E(X^k|X)=X^k$, $k$ is a non-negative integer. One obvious special case is when $k=1$. Anyone has an idea how to prove it?

Answer 1

So long as $k$ is not itself a random variable (in which case the expectation would have to be taken over the conditional distribution of $k$), then that result is correct. How to prove it really depends on the level of rigour you want to use. Formally, the conditional expectation with respect to a random variable is any random variable that satisfies a required integral equation with respect to the distribution of the conditioning variable. Translating the formal definition for your problem means that the conditional expectation of interest is any random variable $g$ that satisfies:

$$\int \limits_\mathscr{X} g(x) dF_X(x) = \int \limits_\mathscr{X} x^k dF_X(x).$$

It is trivial to see that the result $g(x) = x^k$ satisfies this equation, and hence $\mathbb{E}(X^k|X=x) = x^k$ is a valid conditional expectation. (All valid conditional expectations are almost surely equal to this function, but formally, the conditional expectation is defined only up to almost-sure equivalence.)

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It is also possible to give a more intuitive heuristic "proof" that eschews the formal definition of conditional expectation and proceeds using the standard density machinery. We first establish that the conditional density for $X$ given $X=x$ is a point-mass on its conditional value $x$:

$$p(X=x'|X=x) = \frac{p(X=x',X=x)}{p(X=x)} = \frac{p(X=x) \cdot \mathbb{I}(x=x')}{p(X=x)} = \mathbb{I}(x=x').$$

Now, assuming $k$ is a constant, we then have:

$$\mathbb{E}(X^k|X=x) = \int \limits_\mathscr{X} x'^k dP(X=x'|X=x) = x^k.$$

There are some aspects of this "proof" that are non-rigorous, but it gives a heuristic demonstration of the result, which is a bit more intuitive than the formal proof.