# intuitive explanation for expected value of the square of a uniform variable

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. $$(1/2)^2$$.

• 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 $f$. – user158565 Dec 19 '18 at 21:40
• $x^2$ is convex, so lookup the Jensen Inequality – kjetil b halvorsen Dec 19 '18 at 21:45
• Because $V(X) =\frac{1}{12} = E(X^2)-[E(X)]^2 = E(X^2)-\frac 1 4,$ why don't you 'expect' $E(X^2)=\frac 1 3 ?$ – BruceET Dec 19 '18 at 21:50
• 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 – vvv Dec 19 '18 at 21:57
• How about this: because squaring decreases smaller numbers more than it decreases larger numbers. – The Laconic Dec 19 '18 at 22:02