# Find distribution that minimises a function of its moments

Imagine a probability density function $$f(x)$$, defined for positive $$x$$, and let's note its $$n$$th non-centred moment $$x_{n}$$. The mean $$x_{1}$$ is fixed (and positive).

How can I find $$f(x)$$ that minimises some given function of its moments? In my case, $$\frac{ x_{3}+x_{1}^{3}-2x_{1}x_{2} }{ (x_{2}-x_{1}^{2})^{2} }$$

I tried to take the Gateaux derivative of that expression in the direction of a test function $$h(x)$$, and setting the result to be zero for any $$h(x)$$. In the end, I find a relation involving a few moments of $$f(x)$$ and the variable $$x$$, which makes no sense. Would you have any idea of the correct approach here?

Many thanks!

Let $$X$$ be a positive random variable (r.v.) with probability density function $$f$$. The exact lower bound on $$r(X):=\frac{x_3+x_1^3-2x_1x_2}{(x_2-x_1^2)^2}$$ is $$0$$, and it is not attained at any $$f$$.

Indeed, by the Cauchi--Schwarz inequality, $$x_2\le x_3^{1/2}x_1^{1/2}$$, and $$x_2=x_3^{1/2}x_1^{1/2}$$ 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, $$x_2 and hence $$x_3+x_1^3-2x_1x_2>x_3+x_1^3-2x_1x_3^{1/2}x_1^{1/2}=(x_3^{1/2}-x_1^{3/2})^2\ge0.$$ So, $$r(X)>0.$$

Note also that, for any real $$t$$ and any natural $$k$$, if we replace $$X$$ by $$tX$$, then $$x_k$$ gets replaced by $$t^k x_k$$. So, $$r(tX)=\frac{t^3}{t^4}\,r(X)=\frac{r(X)}t\to0$$ as $$t\to\infty$$. Therefore and because $$r(X)>0$$, we see that indeed the exact lower bound on $$r(X)$$ is $$0$$, and it is not attained at any $$f$$.

• I believe $x_1$ is fixed in the question, while here $x_1$ goes to infinity. Apr 5, 2020 at 10:00

This is to complete Mateusz Kwaśnicki's answer by proving that $$EY^2(1+Y)\ge(EY^2)^2\tag{1}$$ if $$Y\ge-1$$ and $$EY=0$$.

Since $$Y\ge-1$$, for any real $$v$$ we have \begin{align} Y^3=(Y+1)(Y-v)^2&+(2v-1)Y^2+(2v-v^2)Y-v^2 \\ &\ge (2v-1)Y^2+(2v-v^2)Y-v^2. \end{align} So, choosing now $$v=EY^2$$, we have $$EY^3 \ge (2v-1)EY^2+(2v-v^2)EY-v^2 =(2v-1)v+(2v-v^2)0-v^2=v^2-v,$$ so that $$EY^3 \ge v^2-v$$, which is equivalent to (1).

(This is not an answer, rather an extended comment.)

If $$X = a \frac{n}{n-1}$$ with probability $$\frac{n-1}{n}$$ and $$X = 0$$ otherwise, then $$x_1 = \mathbb{E}X = a$$, $$x_2 = \mathbb{E}X^2 = a^2 \frac{n}{n-1}$$ and $$x_3 = \mathbb{E}X^3 = a^3 (\frac{n}{n-1})^2$$, so that $$\frac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2} = \frac{1}{a} .$$ Of course, one can smooth out $$X$$ a little bit to get an absolutely continuous distribution with the above ratio arbitrarily close to $$\frac{1}{a}$$.

My guess would be that $$\dfrac{1}{a}$$ is the lower bound for $$\dfrac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2}$$ if $$x_1$$ is required to be equal to $$a$$.

Let $$X$$ have density function $$f(x)$$, and let $$Y = X/a - 1$$, so that $$Y \geqslant -1$$ and $$\mathbb{E} Y = 0$$ (recall that we assume that $$x_1 = a$$). Observe that $$x_3 + x_1^3 - 2 x_1 x_2 = \mathbb{E}(X^3 + a^3 - 2 a X^2) = a^3 \mathbb{E}(Y^2 + Y^3)$$ and $$x_2 - x_1^2 = \mathbb{E}(X^2 - a^2) = a^2 \mathbb{E} Y^2 .$$ Thus, $$\frac{x_3 + x_1^3 - 2 x_1 x_2}{(x_2 - x_1^2)^2} - \frac{1}{a} = \frac{1}{a} \, \frac{\mathbb{E}Y^2 + \mathbb{E}Y^3}{(\mathbb{E}Y^2)^2} - \frac{1}{a} = \frac{1}{a} \, \frac{\mathbb{E}(Y^2 (1 + Y)) - (\mathbb{E}Y^2)^2}{(\mathbb{E}Y^2)^2} .$$ My guess is therefore equivalent to $$\mathbb{E}(Y^2 (1 + Y)) \geqslant (\mathbb{E}Y^2)^2$$ whenever $$\mathbb{E} Y = 0$$ and $$Y \geqslant -1$$.

I do not an immediate proof of the above inequality, nor do I see a counter-example. I though I would share it anyway, perhaps someone else can help. Edit: the proof is completed in Iosif Pinesis's answer.