Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$

Question

Let $N$ be a Poisson random variable with the mean parameter $m$. We are interested in finding the following limit \begin{align} \lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]. \end{align}

Things that I tried: First, I found an upper bound by using Jensen's inequality \begin{align} \lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le \lim_{m \to \infty} m \log\left( \frac{ E \left[ N \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}. \end{align}

For the lower bound I tried to use the CLT argument from a related question in here: \begin{align} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ] \end{align} for some $k>0$. Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that \begin{align} \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] . \end{align} Now if we use the lower bound at this point, we get \begin{align*} \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] \ge \lim_{m\to \infty}m \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty, \end{align*} for all $k \in (0,1)$.

Answer 1

This is an respectable answer, what a median in a Poisson distribution is. Your answer targets large, infinite such medians. This gives the formulae: Poisson distribution. This definition also relates the E symbol to the distribution. Expected value makes clear which operation is expected.

$$E[X]=\int_{0}^{\infty} x P(x)dx$$

$X$ is any function of the distributed parameter.

$$E[\log(\frac{N+\frac{1}{2}}{+1})]=\int_{0}^{\infty} \log(\frac{N+\frac{1}{2}}{+1}) P(N)dN$$

Now You named Poisson random variable so,

$$P(\lambda,N)=\frac{\lambda^N\exp(-\lambda)}{N!}$$

Median and $\lambda$ are : $m\approx\lambda-\frac{1}{3}-\frac{0.05}{\lambda}$. Since Your are interested in large medians only:

$$P(m,N)\approx\frac{(m+\frac{1}{3})^N\exp(-(m+\frac{1}{3}))}{N!}$$

For sufficiently high values the Poisson distribution is approximated by the standard distribution. This can only be evaluated numerically or literature.

A plot of the function over m looks like this:

$$\lim_{m->\infty} m E[\log(\frac{N+\frac{1}{2}}{+1})]=0$$

This might give some safety if the median, mean or parameter $\lambda$ has to be estimated. Since $N$ is natural a variation of $\frac{1}{2}$ is rather common as uncertainty. $\lambda$ is real and the variation in first order is estimated $1$. We approximate the entropies roughly and integration the difference between both over the "true" probability distribution with the mean as parameter. In first order the coefficient vanishes and show how good the approximation really is, if alone the Poisson distribution is selected.

This does not compare to other distribution such as the standard distribution. $\frac{1}{2}$ and $1$ might too named information uncertainty. The graph reflects teh general property of the Poisson distribution that goes over into the standard distribution. Mind this plot might have numerical errors. This is crude because it replaces the median for the parameter already $\lambda$. Use Poisson distribution for a good completion of formulae used.

Answer 2

We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.

We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$ and thus $$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+\frac12} -\frac12.$$

It's sufficient to show that $\lim_m E\ln \frac{N+ \frac12}{m+\frac12} \to -\frac{1}2$.

Condsider $m \ge 100$, $1+Y_i \sim Pois(1)$, $\xi_m \sim Pois(m - [m])$. Thus $\sum_{i=1}^{[m]} (Y_i + 1) + \xi_m = N$ in distribution.

Fix $x$. $$P(\frac{\xi_m}{\sqrt{m}} \ge x) \le \frac{E \frac{\xi_m}{\sqrt{m}} }{x} \to 0, m \to \infty$$ thus $ \frac{\xi_m}{\sqrt{m}} \to 0$ in probability. It follows from CLT that $\frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}} \to N(0,1)$. Hence $$\frac{N-m}{\sqrt{m}} = \frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}}\frac{\sqrt{[m]}}{\sqrt{m}} + \frac{\xi_m}{\sqrt{m}} \to N(0,1).$$ Put $$\eta_m = \frac{m}{m+\frac12}\cdot \frac{N-m}{\sqrt{m}} = \frac{\sqrt{m}(N-m)}{m+\frac12}$$ Then $\eta_m \to N(0,1)$ and $\eta_m \ge \frac{m}{m+\frac12}\frac{-m}{\sqrt{m}} \ge -\sqrt{m} + \frac{1}{3\sqrt{m}}$. Further, $$\lim_m E\ln \frac{N+ \frac12}{m+\frac12} = E m \ln(1 + \frac{\eta_m}{\sqrt{m}}).$$ Put $g(x) = \ln(1+x) - (x - \frac{x^2}2)$. Then $$ E m \ln(1 + \frac{\eta_m}{\sqrt{m}}) = E m( \frac{\eta_m}{\sqrt{m}} - \frac12 ( \frac{\eta_m}{\sqrt{m}})^2 + g( \frac{\eta_m}{\sqrt{m}}) ) = - \frac12 + Emg( \frac{\eta_m}{\sqrt{m}}) $$ and it's sufficient to show that $Emg( \frac{\eta_m}{\sqrt{m}}) \to 0$.

Put $A = [-\sqrt{m}+ \frac{1}{3\sqrt{m}}, - m^{\frac{3}{8}}], B= [-m^{\frac{3}{8}}, m^{\frac{3}{8}}], C = [m^{\frac{3}{8}}, \infty)$. Then

$$|Emg( \frac{\eta_m}{\sqrt{m}})| \le Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le $$ $$\le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|P(\eta_m \in A) + m \max_{x \in B} |g(\frac{x}{\sqrt{m}})|P(\eta_m \in B) + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C}$$ As $m\ge 100$, we know that $\max_{x \in A} |g(\frac{x}{\sqrt{m}})| = |\ln(\frac{1}{3m})| \le 2 \ln m $. Further, by Taylor formula $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$, $0 \le \theta \le 1$. Thus $g(x) = \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$ and $\max_{x \in B} |g(\frac{x}{\sqrt{m}})| = O(\frac{1}{ m^{\frac98 } })$.

We have $|g(x)| \le 2x^2$ for $x > 0$. Thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le E m 2 (\frac{\eta_m}{\sqrt{m}})^2 I_C = 2 E \eta_m^2 I_C$. Further there will be heuristic arguments. $2 E \eta_m^2 I_C \approx 2 E N^2(0,1) I_{(N(0,1) > m^{\frac38})} \to 0$ hence $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \to 0$.

$P(\eta_m \in A) \approx P(N(0,1) \in A) \le P(N(0,1) \le -m^{\frac38})$ and thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} \le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})| P(N(0,1) \le -m^{\frac38}) \to 0$.

It's sufficient to show that$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \to 0$.

But $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \le m \max_{x \in B} |g(\frac{x}{\sqrt{m}})| = m O(\frac{1}{ m^{\frac98 } }) = o(1)$, q.e.d.