Why $z=f(x)$ does not imply $E[z]=f(E[x])$ when f is not linear?

I can give an example, but I couldn't derive the general form.

Let $z = {x^2}$

Let $g(x)$ be the pdf of $x$. Then:

$ E\left[ z \right] = E\left[ {{x^2}} \right] = \int\limits_{ - \infty }^\infty {\left( {{x^2}} \right)g\left( x \right)dx} $

$f\left( {E\left[ x \right]} \right) = {\left( {\int\limits_{ - \infty }^\infty {\left( x \right)g\left( x \right)dx} } \right)^2}$

$ \Rightarrow E\left[ z \right] \ne f\left( {E\left[ x \right]} \right)$

  • 1
    $\begingroup$ It is slightly difficult to understand the question, but consider $X$ uniformly distributed on $[-1,1]$ and $f(x)=x^3$. You have $(E[X])^3 = 0 = E[X^3].$ $\endgroup$
    – Henry
    Jun 19, 2011 at 17:55
  • 1
    $\begingroup$ Jensen's inequality is basically what you are looking for. $\endgroup$
    – Tim
    Jun 19, 2011 at 20:40
  • $\begingroup$ @Tim Why not posting this as an answer? Please? $\endgroup$
    – user88
    Jun 19, 2011 at 22:50

1 Answer 1


You are basically looking for the Jensen's inequality. Proofs are in the article.


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