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Let $f: [a, b] \to \mathbb [0, \infty)$ be an integrable function. By the Cauchy-Schwarz inequality:

$$ \left(\int_a^b f(x) dx\right)^2 \leq (b-a) \int_a^b f(x)^2 dx$$

with equality iff $f$ is constant.

If we notate $\mu(f) := \frac{1}{b-a} \int_a^b f$, dividing with $\frac{1}{(b-a)^2}$ we get

$$\mu(f)^2 \leq \mu(f^2)$$

i.e. the square of the mean of a function is no larger than the mean of the square of a function, and the equality occurs iff $f$ is constant.

Is there an elegant intuitive way to see this geometrically? The best I can do is to use a discrete analogue while approximating $f$ and $f^2$ with rectangles (the definition of the integral). Is there a more visual approach?

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Not a geometric statement, but a rather obvious algebraic meaning: Note that you can rewrite this to $\mu\left[(f-\mu(f))^2\right]\geq 0$. In other words, the mean of the squared deviations from the mean can't be negative. – Semiclassical Aug 22 '14 at 15:45
up vote 2 down vote accepted

Combining JiK's (and implicitly john mangual's) answer, and Semiclassical's comments: If the region enclosed by the $x$-axis and the graph $y = f(x) - \mu(f)$ (subject to $a \leq x \leq b$) is revolved about the $x$-axis, the (non-negative) volume swept out (multiplied by $1/\pi$) is given by \begin{align*} \int_{a}^{b} \bigl[f(x) - \mu(f)\bigr]^{2}\, dx &= \int_{a}^{b} \bigl[f(x)^{2} - 2f(x)\mu(f) + \mu(f)^{2}\bigr]\, dx \\ &= \left[\int_{a}^{b} f(x)^{2}\, dx\right] - \mu(f)^{2}. \end{align*} Region being revolved

The same volume is swept out by revolving the graph $y = \bigl|f(x) - \mu(f)\bigr|$, if you want to avoid revolving a region that dips below the $x$-axis.

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Here's one way to look at it. I don't know how intuitive it is, because I can't find a good way to express the geometric reasoning, but maybe it says something to someone.

If the curve $f$ between $a$ and $b$ rotates around the $x$-axis, the resulting object $A$ has volume $$ V_A = \pi \int_a^b f(x)^2\,\mathrm{d}x = \pi (b-a) \mu(f^2). $$

Let $g=\mu(f)$, the average of $f$. If the curge $g$ rotates around the $x$ axis, the resulting object $B$ is a cylinder with volume $$ V_B = \pi (b-a) \mu(f)^2. $$

Now compare $A\setminus B$, that is, the part of $A$ which is outside $B$, with $B\setminus A$, the part of $B$ which is outside $A$. Your result is equivalent to the fact that the volume of $A\setminus B$ is larger than the volume of $B \setminus A$.

Unfortunately I can't think of a good way to explain this geometrically, but it feels intuitive, if you remember that the radius of $B$ is the average of the radius of $A$. My intuitive idea is similar to comparing three circles of radii $r-r'$, $r$, and $r+r'$: clearly the difference between the areas of the two smallest circles is smaller than the difference between the areas of the two largest circles.

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Wouldn't it be simpler just to say that $A$ has more volume than $B$? Tossing out the common volume of $A\cap B$ just makes things look odder than they really are. – Semiclassical Aug 22 '14 at 16:27
Actually, wouldn't the intuition just be this? If $f(x)>\mu(f)$, then $f(x)$ is tracing out a larger circle upon rotation around the $x$-axis than $\mu(f)$ is. The opposite is true if $f(x)<\mu(f)$, and so the contributions from $f(x)>\mu(f)$ are weighted more than those of $f(x)<\mu(f)$. But that means that $A$ must certainly have more volume than $B$. (Case in point: A cone of height 1 and radius 2 has a volume of $4\pi/3$ whereas a cylinder of height 1 and radius 1 has volume of $\pi$.) – Semiclassical Aug 22 '14 at 16:35
@Semiclassical It would indeed be simpler to say that. But to me it's most intuitive to think that when "mass" is moved from $(0,\mu(f))$ to $(\mu(f),\infty)$ to transform $g$ into $f$, the volume increases, so what matters is where and how much the radius differs from $\mu(f)$. That's why I thought about the differences first. – JiK Aug 22 '14 at 17:33

In physics this is known as the parallel axis theorem.

The Moment of intertia of a body $A \in \mathbb{R}^2$ is the average distance squared relative to the center of Mass:

$$ I_{CM} = \mathbb{E}\big[ || r - \overline{r}||^2\big] = \int_A \big[(x-x_0)^2 + (y-y_0)^2\big] \;dA$$

How do you compute moments relative to other axes? Then we have formula:

$$ I = I_{CM} + m d^2 $$

$m$ is the mass of your object and $d$ is the distance between the center of your two axes. In probability theory we could say:

$$ \mathbb{E}[ (X - a)^2] = \mathbb{E}[ X^2] - 2a \mathbb{E}[ X] + a^2 \geq 0 $$

So this is certainly minimized when $a = \mathbb{E}[X]$

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Your last sentence is incorrect: you want to arrive at $a = E[X]$, not at $a = E[X^2]$. More simply, $$\begin{align}E[(X-1)^2]&=E[((X-\mu) + (\mu-a))^2]\\&= E[(X-\mu)^2] + (\mu-a)^2 + 2(\mu-a)E[X-\mu]\\&= E[(X-\mu)^2] + (\mu-a)^2\\&=\sigma^2 + (\mu-a)^2\\&\geq \sigma^2\end{align}$$ with equality when $a = \mu = E[X]$. – Dilip Sarwate Aug 22 '14 at 16:37

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