Question regarding functions of 1 random variable & their expectation

Question

Say you have random variables $X$ and $Y = g(X)$.

Consider this statement: $$ E[Y] = g(E[X]) $$ A friend of mine & I have seen this statement be true a few times, but most of the time the statement is false. We don't understand why this is so, and feel like there should be some sort of intuition behind the statement.

Any clarification would be appreciated!

Answer 1

If you want a formal definition :

$\mathbb{E}[X] = \int_\Omega X(\omega)d\mathbb{P}(\omega)$

then $g(\mathbb{E}[X])=g(\int_\Omega X(\omega)d\mathbb{P}(\omega))$

and is not $\int_\Omega g(X(\omega))d\mathbb{P}(\omega)=\mathbb{E}[Y]$

Informally you can apply the same reasoning to any discrete random variable and see that in most cases $g(\sum p_k x_k) \neq \sum p_k g(x_k) $.

Answer 2

For a discrete r.v. $X$

\begin{align} g(E[X])&= g\left(\sum_x{xp_X(x)}\right)\\ E[g(X)]&=\sum_x{g(x)p_X(x)} \end{align}

There is no reason for the above expressions to coincide except the special case of linear $g(x)=Ax+B$