Variational problem: how to minimise the second moment?

Question

This is a neater version of a question I posted here, on which I'm also stuck.

The problem: Say I have a probability density function $f(x)$, defined for positive $x$, and let's note its $n$th non-centred moment $x_{n}$. Now, if the mean $x_{1}$ is fixed, what $f(x)$ minimises the second moment $x_{2}$? My intuition says it should be an exponential distribution (or similar shape).

My attempt: I want to minimise the functional

$J(y)=\int x^{2}f(x)dx$

with the constraints of mean and unit integral:

$\int xf(x)dx=x_{1}$

$\int f(x)dx=1$

So I need to minimise the modified functional:

$J^{\ast}(y)=\int x^{2}f(x)dx+\alpha \int xf(x)dx + \beta \int f(x)dx$

$J^{\ast}(y)=\int (x^{2}f(x) +\alpha xf(x) + \beta f(x))dx$

$J^{\ast}(y)=\int (x^{2} +\alpha x + \beta )f(x)dx$

Applying Euler-Lagrange gives $x^{2}+\alpha x +\beta =0$, where $f(x)$ doesn't appear!

So I don't know how to do, especially for finding the values of $\alpha$ and $\beta$. I also tried adding a constraint of positivity on $f(x)$, by substituting it with $u(x)^{2}$ (always positive by construction), but it leads to the same Euler-Lagrange equation.

Could anyone help me on that? Many thanks!

Answer 1

Let $X$ be a positive random variable (r.v.) with probability density function $f$. By the Cauchi--Schwarz inequality, $x_1^2=(EX)^2$ is a lower bound on $x_2=EX^2$, and this lower bound is attained if and only if the r.v. $X$ is a constant. Since $X$ has a pdf $f$, it is not discrete and hence not a constant. So, the lower bound $x_1^2$ on $EX^2$ is not attained here. However, by considering e.g. a r.v. $X$ uniformly distributed on the interval $[x_1-h,x_1+h]$ with $h\in(0,x_1)$, we see that here $x_2=x_1^2+h^2/3$, which converges to $x_1^2$ as $h\to0$.

We conclude that $x_1^2$ is the exact lower bound on $x_2$.