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I'm confused about something that should be simple. Suppose I have a random uniform variable $X$ on $[0,1]$. It's fairly clear that the expected value of $X$ is 1/2. By integrating $x^2$ on $[0,1]$, I get that the expected value of $x^2$ is 1/3. I'm struggling to understand this intuitively, as I would expect it to be $1/4$ i.e. $(1/2)^2$.

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    $\begingroup$ That is the difference between $E(X^2)$ and $[E(X)]^2$, or more general, between $E[f(X)]$ and $f[E(X)]$ for non-linear function of $f$. $\endgroup$ – user158565 Dec 19 '18 at 21:40
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    $\begingroup$ $x^2$ is convex, so lookup the Jensen Inequality $\endgroup$ – kjetil b halvorsen Dec 19 '18 at 21:45
  • $\begingroup$ Because $V(X) =\frac{1}{12} = E(X^2)-[E(X)]^2 = E(X^2)-\frac 1 4,$ why don't you 'expect' $E(X^2)=\frac 1 3 ?$ $\endgroup$ – BruceET Dec 19 '18 at 21:50
  • $\begingroup$ because if I think of it as the area of a square with edge X that is standard uniform, which is completely determined by this X, I don't understand why I can't say the expected area is 1/4.. I understand how to work it out mathematically, but I don't understand intuitively why this random square isn't expected to have area 1/4 $\endgroup$ – vvv Dec 19 '18 at 21:57
  • $\begingroup$ How about this: because squaring decreases smaller numbers more than it decreases larger numbers. $\endgroup$ – The Laconic Dec 19 '18 at 22:02

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