The approximated function of $\mathbb{E}\left\{ \ln\left(1+X\right)\right\}$, where $X\sim\operatorname{Gamma}\left(\kappa\ge1,\theta>0\right)$

Given $$X \sim \operatorname{Gamma}(\kappa, \theta)$$ with CDF $$F_X(\kappa, \theta)$$, where $$\kappa \geq 1$$ and $$\theta > 0$$, the expected value of $$\mathbb{E} \left\{ \ln(1+X) \right\}$$ is calculated by $$\int_{x} \ln(1+x) \, dF_X(\kappa, \theta)$$.

For calculation simplification, we want to get its approximated expression $$f(\kappa, \theta)$$ with $$\mathbb{E} \left\{ \ln(1+X) \right\} \geq f(\kappa, \theta)$$.

One example is to set $$f(\kappa, \theta) = \ln \left(1 + \alpha \mathbb{E} \left\{ X \right\} \right)$$, and an approximated solution is to set $$\alpha = \left( e^{\psi(\kappa) + \ln(\theta)} - 1 \right) / \kappa \theta$$, such that

$$f\left(\kappa,\theta\right)=\ln\left(1+\alpha\mathbb{E}\left\{ X\right\} \right)=\psi\left(\kappa\right)+\ln\left(\theta\right)=\mathbb{E}\left\{ \ln\left(X\right)\right\} \le\mathbb{E}\left\{ \ln\left(1+X\right)\right\}.$$

However, this approximation is too loose when $$\theta$$ is small.

In summary, we want to get a tighter approximation, for example, finding the function of $$\alpha$$ over $$\theta$$ and $$\kappa$$, as shown in the figure.

• MathJax note: Confusingly, although it works fine in preview, something about the order of parsing means that unescaped MathJax in alt text for images escapes from the alt text and leaks out as partial raw HTML. (See below the image at revision 4.) I edited to escape the alt text. Commented Jul 7 at 10:46
• Sorry, I didn't get your point. Should I revise somewhere? Commented Jul 7 at 11:11
• Re, no, here I did it. It is just something that you might want to keep in mind for future posts, especially since it does not show up in preview. Commented Jul 7 at 12:01
• THX! I will keep it in mind. Commented Jul 7 at 12:25

I don't know if the following is sharp enough to answer your question, but it is an improvement at least.

One can pretty easily get the bound $$\mathbb{E}[\ln(1+X)]\ge\psi(\kappa)+\ln(\theta)+\ln(1+\frac{1}{\theta\kappa}).$$ This improves on your bound somewhat for small $$\theta\kappa$$, and follows a simple argument.

\begin{align} \mathbb{E}[\ln(1+X)] &= \mathbb{E}[\ln(1+X)-\ln(X)+\ln(X)] \\ &= \mathbb{E}[\ln(X)] + \mathbb{E}[\ln(1+\frac{1}{X})]\\ &= \psi(\kappa)+\ln(\theta)+\mathbb{E}[f(X)],\qquad f(x) := \ln(1+\frac{1}{x})\\ &\geq \psi(\kappa)+\ln(\theta)+f(\mathbb{E}[X])\\ &=\psi(\kappa)+\ln(\theta)+\ln(1+\frac{1}{\kappa\theta}). \end{align} Essentially, in the above I rewrite $$\ln(1+X)$$ to be the sum of your current bound, plus the remainder error, and use Jenson's inequality to get a non-trivial lower bound on the remainder error.

The above is (clearly) an identity, except for the loss in Jenson's inequality. One might wonder how large this is. Examining a sharper lower bound on $$\mathbb{E}[f(X)]-f(\mathbb{E}[X])$$, I get in your case that

$$\mathbb{E}[f(X)]-f(\mathbb{E}[X]) \geq \inf_{x\in(0,\infty)}\frac{f(x)-f(\kappa\theta)}{(x-\kappa\theta)^2}+\frac{f'(\kappa\theta)}{x-\kappa\theta} = \inf_{x\in(0,\infty)}\frac{\ln(1+\frac{1}{x})-\ln(1+\frac{1}{\kappa\theta})}{(x-\kappa\theta)^2}+\frac{1}{\kappa^2\theta^2+\kappa\theta}\frac{1}{x-\kappa\theta}.$$ Numerically, this does not appear to be too large, but if the above bound is not good enough, it might make sense to get a better bound on this.

• Thanks for your answers. I checked by numerical; it is a good bound. Commented Jul 9 at 14:28
• To improve the bound for small $\theta \kappa$, I did the analysis on $\mathbb{E}[f(X)]-f(\mathbb{E}[X]))$. When $x\to \infty$, $\mathbb{E}[f(X)]-f(\mathbb{E}[X]))\ge \lim_{x\to\infty}\frac{0-\ln\left(1+1/\kappa\theta\right)}{\left(x-\kappa\theta\right)^{2}}+\frac{1}{\kappa^{2}\theta^{2}+\kappa\theta}\cdot0=0$. Commented Jul 9 at 14:38
• We can get this bound using a different method, as shown in the next answer. Commented Jul 9 at 15:40
• Thanks for your answer again. After reading the paper "Sharpening Jensen’s Inequality," I think I can try to find a sharper bound as the example in section 4. Commented Jul 9 at 17:07

To set notation define $$G(\kappa, \theta) = \mathbb{E}_{X \sim \mathsf{Gamma}(\kappa, \theta)}\Big[\psi(X)\Big], \quad \psi(z) = \log(1 + z).$$

Explicit formula for the Jensen gap:

By Taylor's theorem with integral remainder, we have \begin{align*} G(\kappa, \theta) &= \log(1 + \kappa \theta)-\int_0^1 (1-t) \mathbb{E}_{X \sim \mathsf{Gamma}(\kappa, \theta)} \Big[\Big( \frac{X- \kappa \theta}{1 + \kappa \theta + t(X - \kappa \theta)}\Big)^2\Big] \, \mathrm{dt} \\ &= \log(1 + \kappa \theta)-\frac{1}{(1+\tau)}\int_{1+\tau}^\infty \frac{s-(1+\tau)}{s} \,\mathbb{E}_{Y \sim \mathsf{Gamma}(\kappa, 1/\kappa)} \Big[\Big( \frac{Y-1}{s + Y - 1}\Big)^2\Big] \, \mathrm{ds}. \end{align*} Above, $$\tau = 1/(\kappa \theta)$$, and we made the substitution $$t = (1+\tau)/s$$.

Hence, if we define $$\Delta(\kappa, \theta) = \int_{1+\tau}^\infty \frac{s-(1+\tau)}{s(1+\tau)} \,\mathbb{E}_{Y \sim \mathsf{Gamma}(\kappa, 1/\kappa)} \Big[\Big( \frac{Y-1}{s + Y - 1}\Big)^2\Big] \, \mathrm{ds}, \quad \mbox{where} \quad \tau = \frac{1}{\kappa \theta},$$ then the Jensen gap satisfies $$\mathbb{E} \psi(X) - \psi(\mathbb{E} X) = G(\kappa, \theta) - \psi(\kappa \theta) = - \Delta(\kappa,\theta).$$

Limit relations for the gap:

Note that since $$\mathbb{E}[(Y-1)^2] = 1/\kappa$$, we have $$\Delta(\kappa, \theta) \leq \frac{1}{\kappa(1+\tau)}\int_{1+\tau}^\infty \frac{s - (1+\tau)}{s(s-1)^2} \, \mathrm{ds} = \frac{1}{\kappa} \Big(\log(1 + \kappa \theta) - \frac{\kappa \theta}{1 + \kappa \theta}\Big).$$ Thus, the gap vanishes in the limits $$k \to \infty, k \to 0^+$$ for each fixed $$\theta > 0$$ and similarly in the limit $$\theta \to 0$$ for each fixed $$k$$.

• Thanks for your answer. But I can't follow your result of "Taylor's theorem with integral remainder". Could you please provide more hints? Commented Jul 10 at 2:53
• For a twice differentiable function $f$, you have $f(1) = f(0) + f'(0) + \int_0^1 f''(t) (1-t) dt$. Apply this to the function $f(t) = \psi( \mathbb{E} X + t (X - \mathbb{E} X))$ and take expectations. The first order term cancels (this is essentially the content of Jensen's inequality when $f$ is differentiable.) Then $f(1) - f(0)$ is the Jensen gap, and the final step is to use Fubini to swap the order of integration. Commented Jul 10 at 2:56
• I did a numerical result on $\kappa=\theta=1$, and calculate $\Delta(\kappa,\theta)=69.3647$. This bound is not right. Is my understand right? Commented Jul 10 at 2:56
• Well--it is exact. It is exactly the gap between your function and the result of Jensen's inequality. Commented Jul 10 at 2:57
• In numerical experiments with $\kappa=\theta=1$. $\mathbb{E}\left[\ln\left(1+X\right)\right]-\ln\left(1+\mathbb{E}\left[X\right]\right)=0.403$. But here $\Delta(\kappa,\theta)=69.3647$ Commented Jul 10 at 3:02

We can get the bound as $$\mathbb{E}[\ln(1+X)]\le\ln(1+\mathbb{E}[X]) + \ln\left(\kappa\right)-\psi\left(\kappa\right).$$ This is because
$$\mathbb{E}\left[\ln\left(1+X\right)\right]-\ln\left(1+\mathbb{E}\left[X\right]\right) \le\mathbb{E}\left[\ln\left(X\right)\right]-\ln\left(\mathbb{E}\left[X\right]\right) =\psi\left(\kappa\right)+\ln\left(\theta\right)-\ln\left(\kappa\theta\right) =\psi\left(\kappa\right)-\ln\left(\kappa\right)$$ The inequality holds because the function $$\mathbb{E}\left[a+\ln\left(X\right)\right]-\ln\left(a+\mathbb{E}\left[X\right]\right)$$ is decreasing over $$a$$.

This result is the same as the answer from Mark Schultz-Wu.