# exponential of expected value

If for a random variable $$X$$ we have that $$e^{\mathbb{E}[X]} = \mathbb{E}[{e^{X}}],$$

how can I show that $$X= c$$ almost surely, where $$c$$ is constant?

Proof: Suppose that $$\exp\left(\int_X f(x)d\mu(x)\right) = \int_X e^{f(x)}d\mu(x). \qquad (1)$$ We want to show that if equality holds then $$f$$ is constant.

Let us denote $$c=\int_X f(x) \,\mathrm{d}\mu(x)$$. The above inequality can now be rewritten as $$e^c = \int_X e^{f(x)} \,\mathrm{d}\mu(x)$$.

Suppose that $$f$$ is not constant, i.e. $$f(x)\ne c$$ on a set of positive measure. Then both sets $$A=\{x; f(x)>c\}$$ and $$B=\{x; f(x) must have positive measure. (Otherwise we would get a contradiction with the definition of $$c$$ as the mean value of $$f$$.)

For any $$t\ne c$$ we have $$e^t>e^c+e^c(t-c)$$, since the graph of $$e^x$$ lies above the tangent line at the point $$c$$. (The factor $$e^c$$ is the slope at the point $$c$$.) This means $$e^t-e^c>e^c(t-c).$$

Thus for any $$x\in A\cup B$$ we have strict inequality $$e^{f(x)}-e^c>(f(x)-c)e^{c}$$. For any $$x$$ we have $$e^{f(x)}-e^c \ge (f(x)-c)e^c$$. Integrating gives $$\int_X e^{f(x)} \,\mathrm{d}\mu(x) - e^c > e^c \left(\int_X f(x) \,\mathrm{d}\mu(x) -c \right) = 0,$$ contradicting the equality (1).

Am I getting it right?

• Apply Jensen's Inequality (or the AM-GM inequality).
– whuber
Feb 4, 2021 at 18:20
• $$e^{\sup_{x} \Phi(x)} = \sup_{x} e^{\Phi(x)}$$ for every real-valued family $\{\Phi(x)\}_x$. This holds more generally for any increasing (and continuous) function $f\colon \mathbb{R}\to\mathbb{R}$: $$f(\sup_{x} \Phi(x)) = \sup_{x} f(\Phi(x))$$
– user310375
Feb 4, 2021 at 18:30
• @whuber I edited my answer
– user310375
Feb 4, 2021 at 18:40

Without invoking any special results, a direct demonstration might go like this.

Let (for notational convenience) $$\mu=E[X],$$ which I suppose is finite. The line with equation $$e^\mu(x-\mu) + e^\mu$$ is a support line for the graph of the exponential function: it is tangent to the graph at the point $$(\mu, e^\mu)$$ and otherwise lies strictly below the graph. Thus, for all $$x,$$

$$e^x \ge e^\mu(x-\mu) + e^\mu$$

with equality if and only if $$x=\mu.$$ Applying this to the random variable $$X$$ and taking expectations gives

$$E\left[e^X\right] \ge E\left[e^\mu(X-\mu)+e^\mu\right] = e^\mu E[X]+e^\mu(1-\mu)=e^\mu=e^{E[X]},$$

with equality if and only if $$X$$ is almost surely equal to $$\mu,$$ QED.

### Post script

You posted essentially the same argument as an edit while I was posting this answer, so all I have contributed (apart from affirmation of your approach) is an indication of how statistical language (of expectations) might streamline the presentation.

• thanks @whuber for the support and the confirmation
– user310375
Feb 4, 2021 at 18:48