Expand Expectation with Absolute Value $^3$

Question

How could I expand the next expression? The Value $E(X) = p$ and $Var(X) = p(1-p)$

$$E(|X - E(X)| ^3 ) $$

I have the answer which is $p(1-p)^3 + p^3(1-p)$

It might be something quite easy but I just don't see how they got rid of the absolute value.

Answer 1

The random variable $X$ has not been fully described. Presumably $X$ is intended to have Bernoulli distribution with "probability of success" (probability that $X=1$) equal to $p$.

Then $X$ has mean $p$, and we want $E(|X-p|^3)$. Note that $X-p=1-p$ with probability $p$, and $X-p=-p$ with probability $1-p$.

It follows that $|X-p|^3=(1-p)^3$ with probability $p$, and $|X-p|^3=p^3$ with probability $1-p$.

Thus the required expectation is $$(1-p)^3 p +p^3(1-p).$$