Exchange paradox: PDF over the positive reals such that f(2x) = f(x/2)?

Question

I was reading about the exchange paradox (two envelopes, one with $m$ dollars and another with $2m$ dollars, you get one at random, would you switch before looking?). The paradox is that if you get $X$, the expected value of switching seems to be $(1/2)*2X + (1/2)*X/2 > X$, so switching seems optimal. A resolution I read was that if you do this, you're assuming that for all $x \in\mathbb{R}^+$, that $x/2$ and $2x$ have the same probability, and that no such pdf exists because it does not integrate to 1. This statement was not supported. Is there anyway to prove or reason through this?

Answer 1

It's done with integrals and $u$-substitution. If $\displaystyle\int_0^\infty f(2x)\,dx=1$, then $$\int_0^\infty f\left(x\over 2\right)\,dx = \int_0^\infty f(2u)\cdot4\, du = 4 \cdot 1 = 4.$$ (Here, I let $u=\displaystyle{x\over4}$.)