If $a_{n} = E(X-c)^n$ then $P(X=c) = 1$ iff $a_{n } = 0$ for how many values of $n$?

Question

Suppose $X$ is a random variable, $c$ is a constant and $a_{n} = E(X-c)^n$ is finite $\forall n \geq 1$.

Then $P(X = c) = 1$ iff $a_{n} = 0$ for

1) atleast one $n \geq 1$

2) atleast one odd $n$

3) atleast one even $n$

4) atleast two values of $n$

I could do the forward direction as for $P(X =c) =1$ then $x= c$ with probability 1,so $a_{n} = E(X-c)^n = E(0) = 0$ so this is true for any $n \geq 1$, but what happens to the converse case that is if $a_{n} = 0$ or $E(x-c)^n = 0$ then can $P(X= c) =1$ then how can we conclude that it occurs for some finite values of $n$ or for all values of $n$ ?

Atleast i think this question is not in MSE as i tried searching in Approach0 here.

Answer 1

Some hints:

First, try to rule out some of those options by finding a counterexample. Consider the case where $X$ is a standard normal random variable, and $c = 0$. What is $a_n = E[X^n]$ for any odd integer $n \ge 1$? What is $P[X = 0]$? This should help you rule out most of those options.

Next, if $n$ is an even integer, then $(X-c)^n$ is a non-negative random variable. So, theorems like this one apply.