0
$\begingroup$

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)$

$\endgroup$
3
  • 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

3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.