Upper bound for $E(\log(1+e^X))$

Question

I'm looking for an upper bound for the $E(\log(1+\exp(X)))$, where $X$ is a normal random variable. I tried Jensen's inequality for the $\log$ function, but Im searching for a more tight bound, or maybe an upper bound for the function itself.

Answer 1

Suggestion:

For a standard normal distribution distribution, a rough upper bound -just to conclude that $E[\log(1+e^X)]$ is finite- can be obtained by breaking the integral in two pieces:

\begin{align} g(x)=\log(1+e^x)&=\log(1+e^{x})\mathbb{1}(x\leq0)+(x+\log(1+e^{-x}))\mathbb{1}(x>0)\\ &=x\mathbb{1}(x>0) + \log(1+e^{-|x|})\\ &\leq x\mathbb{1}(x>0) + \log2 \end{align}

Then $$E[g(X)]\leq E[\log(2)+\mathbb{1}(X>0)X]=\log(2)+\frac{1}{\sqrt{2\pi}}$$

Once the integrability of $g(X)$ has been established, the ultimate sharpest bound is the value of $E[g(X)]$ itself.

Better upper bounds can be obtained by further exploiting the identity \begin{align} g(x)&=x\mathbb{1}(x>0) + \log(1+e^{-|x|}) \end{align}

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Edit:

1. Numerical integration gives $E[g(X)]\approx 0.8060592$,
2. simple Monte Carlo integration ($10^6$ samples) gives $E[g(X)]\approx 0.8066063$.
3. The upper bound that I presented is $\log(2)+\frac{1}{\sqrt{2\pi}}\approx 1.092089$

Answer 2

We may split the integral for $\mathbb{E}[\ln(1+e^X)]$ like so:

$$ \begin{array}{ll} \displaystyle \int_{-\infty}^{\infty} \ln(1+e^x)\,\mathrm{d}F & \displaystyle = \int_{-\infty}^0 \ln(1+e^x)\,\mathrm{d}F+\int_0^\infty \ln(1+e^x) \,\mathrm{d}F \\ & =\displaystyle \int_0^\infty \ln(1+e^{-x})\,\mathrm{d}F+\int_0^\infty \big[x+\ln(1+e^{-x})\big]\,\mathrm{d}F \\ & \displaystyle =2\int_0^\infty \ln(1+e^{-x})\,\mathrm{d}F+\int_0^\infty x\,\mathrm{d}F. \end{array} $$

We can explicitly evaluate

$$ \int_0^\infty x\,\mathrm{d}F=\int_0^\infty xe^{-x^2/2}\frac{\,\mathrm{d}x}{\sqrt{2\pi}}=\frac{1}{\sqrt{2\pi}}\left[-e^{-x^2/2}\right]_0^\infty =\frac{1}{\sqrt{2\pi}}. $$

We may also use $\ln(1+u)\le u$ for $0\le u\le1$, where $u=e^{-x}$, to get

$$ \int_0^\infty \ln(1+e^{-x})\,\mathrm{d}F \le \int_0^\infty e^{-x}e^{-x^2/2}\,\frac{\mathrm{d}x}{\sqrt{2\pi}} = \int_0^\infty e^{-(x+1)^2/2+1/2}\,\frac{\mathrm{d}x}{\sqrt{2\pi}} $$

$$ = \sqrt{e}\int_1^\infty e^{-x^2/2} \,\frac{\mathrm{d}x}{\sqrt{2\pi}}=\sqrt{e}\left(\frac{1}{2}-\frac{1}{\sqrt{2\pi}}\int_0^1 e^{-x^2/2}\,\mathrm{d}x\right). $$

This gives

$$ \mathbb{E}[\ln(1+e^X)]\le \frac{1}{\sqrt{2\pi}}+2\sqrt{e}\left(\frac{1}{2}-\frac{1}{\sqrt{2\pi}}\int_0^1 e^{-x^2/2}\,\mathrm{d}x\right)\approx 0.9221. $$

We can do better by bounding $\ln(1+u)$ by an odd number of terms from its Mercator expansion (I chose only the very first term), but then there are more finite integrals to approximate.