Is there a probability distribution defined over $\mathbb{R}^{+}$ by the pdf $f$ such that, $$\forall x > 0, f(x)=\frac{1}{x}f\left(\frac{1}{x}\right)$$ and $$\int_0^{\infty} x~\mathrm{d}f = 1 $$

The first condition is easily satisfied by picking functions of the form

$$\left(\int_{-\infty}^{\infty} v(t)(x^t+x^{1-t})~\mathrm{d}t\right)^{-1}$$

However, in all cases, the first moment is always greater than one. Only by pushing the standard deviation to zero can it approach one.

I realize the moment can be written as $$\int_0^1 \left(x+\frac{1}{x^2}\right)f(x)~\mathrm{d}x$$

and we have the constraint

$$\int_0^1 \left(1+\frac{1}{x}\right)f(x)~\mathrm{d}x=1$$

but I'm not sure if somehow that implies it has to be greater than one...

  • $\begingroup$ Calculus of variations with a lagrange multipler? $\endgroup$
    – abnry
    Commented May 5, 2015 at 17:52

2 Answers 2


Since you already rewrote the moment and the density constraints, let's start from that.

We have $$1 = \int_0^1 \frac{x^3 + 1}{x^2}f(x)dx$$ and $$1 = \int_0^1 \frac{x^2 +x}{x^2}f(x)dx$$.

This gives $$0 = \int_0^1 (x^3 + 1 - x^2 - x) \frac{f(x)}{x^2}$$

But it is easy to see that $\forall x \in [0,1], x^3 + 1 - x^2 - x \geq 0$, hence we have a null integral with a non-negative integrand. This implies that $f(x) = 0$ almost everywhere, a contradiction.


The penultimate expression equals 1, which is correct. That implies that

f(x) = (x+1/x^2)^-1,


  • $\begingroup$ It does not verify $f(x) = \frac{1}{x} f(\frac{1}{x})$, and it doesn't even have a finite integrale over $\mathbb{R}^+$ ?? $\endgroup$
    – Arthur B.
    Commented May 5, 2015 at 19:28

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