Let $N$ be a zero-truncated (or positive) Poisson distribution whose pdf is $ \mathbb{P}\left\{N=n \right\} = \frac{\lambda^n}{(e^{\lambda}-1)n!}$. Then what is $$ \mathbb{E}[\log N]\quad\text{ and }\quad\mathbb{E}[(\log N)^2]? $$ If it is impossible to express in closed form formula (like infinite sum), then can we express their asymptotic formula as $\lambda\rightarrow\infty$?
This question is originated from my research: developing an Ito process for Poisson intensity $\lambda_t$. To make $\lambda_t>0$, I consider geometric Brownian motion: $d \log \lambda_t = \mu dt + \sigma dB_t $, and to estimate $\mu$ and $\sigma$, first I need $\mathbb{E}[\log N]$ and $\mathbb{E}[(\log N)^2]$. Since there is no textbook or papers related to this topic, I have no idea how to handle this.
Thanks,
Using this approach, for a smooth function $g$ and a random variable $X$ with mean and variance $\mu,\sigma^2$ we can approximate:
$$E[g(X)] \approx g(\mu) + \frac12 g''(\mu) \sigma^2 $$
$$E[g^2(X)] \approx g^2(\mu) + g''(\mu)\sigma^2 + [g'(\mu)]^2 \sigma^2 $$
$$Var(g(X)) \approx [g'(\mu)]^2\sigma^2 $$
(which might be regarded as a linear approximation of $Var(aX) = a^2 Var(X)$)
The mean and variance of a truncated Poisson are
$$ \mu= \frac{\lambda}{1-e^{-\lambda}}$$ $$ \sigma^2= \mu (1 + \lambda - \mu)$$
Then, letting $Z=\log N$, for large $\lambda$:
$$E[Z] \approx \log(\mu) - \frac{\sigma^2}{2 \mu^2}\approx \log(\lambda) - \frac{1}{2\lambda} $$
$$Var(Z) \approx \frac{\mu (1 + \lambda - \mu)}{\mu^2} \approx \frac{1-(\lambda+1)e^{-\lambda}}{\lambda}\approx \frac{1}{\lambda} $$
(this looks a bit sloppy, more careful analysis of the higher-order terms should be done, but the asymptotics seems to be correct - cf eg here)