If $\left< e^{-x}\right>=1$ then $\left<x\right>\ge 0$.

Question

I am wanting to prove that: $$\left< e^{-x}\right>=1\Rightarrow \left<x\right>\ge 0\tag{a}$$ which comes up in nonequilbrium physics with $x$ being the difference in entropy. Intuitively it seems correct as we have: $$ \int dx\; e^{-x} P(x)=1\tag{b}$$ thus $P(x)$ will need to be larger for the $x \gt 0$ terms then for the $x\lt 0$ due to the $e^{-x}$ factor weighing down $x\gt 0$ terms. But how can I rigorously prove (a)?

Answer 1

Jensen's inequality $\langle e^{-x} \rangle \geq e^{-\langle x \rangle}$ holds due to the convexity of $\exp(-x)$. From that we obtain $$1= \langle e^{-x} \rangle \geq e^{-\langle x \rangle}$$ and after taking $\log$ $$ 0\geq - \langle x \rangle \Rightarrow 0 \leq \langle x \rangle.$$

Answer 2

Assuming $\left <e^{-x}\right>=1$ we can write $$\left <x\right >=\int (e^{-x}-1+x)P(x)dx$$ This follows, since $\left <x\right >=\left <e^{-x}\right >-1+\left < x\right >$. Now since $P(x)\geq 0$, we only have to show that $e^{-x}-1+x\geq 0$.

Let $f(x)=e^{-x}-1+x$, and consider $$f'(x)=1-\frac{1}{e^x}$$ We observe that:

• $f'(x)<0$ when $x<0$
• $f'(x)=0$ when $x=0$
• $f'(x)>0$ when $x>0$

Thus, $f(x)$ has a minimum at $x=0$, in which case $f(0)=0$. So $f(x)\geq 0$.