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I am now being introduced to rigorous statistics and doing some self-learning. A question recently came to mind:

Suppose I have a continuous random variable $X$ and I know $E(X)$ (expected value) and $V(X)$ (variance). Given an arbitrary function $g(X)$, can I find $E(g(X))$ and $V(g(X))$? If not, does there exist a non-constant function $g_0(X)$ for which I can find these values with no additional information?

My intuitive answer is no, since the computation of these involves an integral.

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As the expectation value is linear, it follows that $E(aX + b)=aE(X) + b$. So for $g(x)=ax+b$ you can compute $E(g(X))$ exactly from only knowing $E(X)$.

In other cases, you can try a Taylor expansion of $g(x)$ around $x=\mu:=E(X)$, but this is obviously only a (possibly crude) approximation: \begin{eqnarray*} g(x) & \approx & g(\mu) + g'(\mu)\cdot (x-\mu) + \frac{g''(\mu)}{2} (x-\mu)^2\\ \Rightarrow E(g(X)) & \approx & g(\mu) + 0 + \frac{g''(\mu)}{2} Var(X) \end{eqnarray*}

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If $g$ is convex you can bound the value using Jensen's inequality, but with that little information I think that's probably the best you can do (assuming you don't know the pdf of $X$).

If $g$ is affine and/or $X$ is constant almost surely then you get equality of $E[g(X)]$ and $g(EX)$. Since you also know $VarX$ you can also compute $Var[g(X)]$ with some algebra in that case.

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