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

Question

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.

Answer 1

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.

Answer 2

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

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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 $$ \mathbb{E} \psi(X) - \psi(\mathbb{E} X) = G(\kappa, \theta) - \psi(\kappa \theta) = - \Delta(\kappa,\theta). $$

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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 $k$.

Answer 3

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.