# Variational problem: how to minimise the second moment?

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!

• I think it does not satisfy the conditions for applying the E-L equation. May 6, 2020 at 2:43

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 $$x_2$$.
• Thanks! So $f(x)$ should be a Dirac? Apr 5, 2020 at 15:50
• For the lower bound to be attained, the distribution of $X$ should indeed be Dirac at the point $x_1$. Apr 5, 2020 at 16:06