Let $X \sim \mathrm{Pois}(\lambda)$ and $x_1, \cdots , x_n$ observations following this distribution. I want to derive the analytical solution of the following series: $$l(\lambda):=\lim_{ n \to \infty}\frac{1}{n}\sum_{i=1}^{n} \log P(X = x_i).$$

After a few trials, I found a good numerical approximation of the solution: $$l(\lambda)=-\log(\sqrt{17.08\cdot \lambda}).$$

See the graph below, where the dots represent an approximation of the solution by simulating poisson distributions, while the blue line represents the approximated numerical solution.

y=sapply(x,function(x) mean(log(dpois(rpois(100000,x),x))))

<span class=$l(\lambda)$ according to $\lambda$">

  • $\begingroup$ I don't know how you found this numerical fit, but it is amazingly close to the analytical approximation by Gaussian for $\lambda$ that is "large enough" (say, $\lambda > 10$) for Central Limit Theorem to start kicking in practically. The expression is $l(\lambda) \approx \frac{-1}2(1 + \log(2\pi \lambda))$ $\endgroup$ – Lee David Chung Lin Feb 2 '19 at 14:30
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    $\begingroup$ See the added graph. I just tried different transformations. Can you elaborate a bit more about the approximation of $l(\lambda)$ for a Gaussian distribution? $\endgroup$ – Anthony Hauser Feb 2 '19 at 14:45
  • $\begingroup$ @LeeDavidChungLin I'm following your basic idea but I'm not seeing how you get that exact result. $\endgroup$ – Ian Feb 2 '19 at 14:59
  • $\begingroup$ I'm impressed by how you found this transformation by trial-and-error. Anyway, another quick observation that most likely leads to nowhere is that $\exp[ l(\lambda) ] = (\Pi P(X = x_i))^{1/n}$, the geometric mean of the observations. $\endgroup$ – Lee David Chung Lin Feb 3 '19 at 6:02

$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$By the Law of Large Numbers the sample mean converges to the expectation

$$l_n(\lambda)\equiv \frac{1}{n}\sum_{i=1}^{n} g(x_i)~, \quad \text{then}\quad \lim_{ n \to \infty}l_n(\lambda) \to E[g(X)]$$

where the expression of interest here is $g(x_i) = \log P(X = x_i)$.

As @lan mentioned, it boils down to finding $E[\log(X!)]$ which seem challenging. I personally am not aware of any justification that a closed form even exists.

Below is the (cheap) asymptotic analysis I mentioned in the comment, where $\lambda$ is "large enough".

The sum of independent Poisson distributions again follows a Poisson distribution. Namely, consider independent $X_i \sim \mathrm{Pois}(\lambda_i)$ where $\lambda_i$ do NOT have to be identical, then $X \equiv \sum X_i \sim \mathrm{Pois}(\lambda)$ where $\lambda = \sum \lambda_i$.

As soon as one sees the sum of independent random variables, one knows there's a variation of Central Limit Theorem (CLT) that applies.

Let's just keep things simple and consider the iid case where all $\lambda_i = \lambda_0$ share a common and FINITE value.

\begin{align} &\text{for}~i = 1 \sim m & X_i &\sim \mathrm{Pois}(\lambda_0) & E[X_i] &= V[X_i] = \lambda_0 & X_i &\perp X_j ~\text{for}~ i \neq j && \\ && X &\equiv \sum_{i = 1}^m X_i & E[X] &= V[X_i] = m\lambda_0 \equiv \lambda \end{align}

Here the CLT is equivalent to a limit-statement about a "runaway" distribution that moves-and-stretches to infinity ($\lambda \to \infty$ as $m \to \infty$).

\begin{align} \frac{ \displaystyle\frac1m \sum_{i = 1}^m X_i - \lambda_0}{ \lambda_0 } &\xrightarrow{~~d~~} \mathcal{N}(0,1) & \begin{aligned}[t] \implies& & \frac{X - \lambda }{ \lambda } &\xrightarrow{~~d~~} \mathcal{N}(0,1) \\ \implies& & X &\overset{d}{ \approx } \mathcal{N}(\lambda, \sqrt{\lambda}) \end{aligned} \end{align}

This is essentially the same procedure when people treat Binomial distribution as roughly Gaussian when $np$ is large enough. Here the approximation is good when $\mathbb{\lambda}$ is large "enough". The same practical Normal approximation can be applied to various distributions that are "sums", like Negative Binomial or Gamma distribution.

Our expression of interest now becomes $$\lim_{ n \to \infty}l_n(\lambda) \approx E[g(X)] = \int_{-\infty}^{\infty} f(x) \log\bigl( f(x) \bigr)\dd{x} \tag*{Eq.(1.a)}$$ where $f$ is the Gaussian density with mean $\lambda$ and variance $\lambda$ $$f(x) = \frac1{\sqrt{2\pi \lambda}} e^{ \frac{-(x - \lambda)^2}{ 2 \lambda }} \quad \text{so that} \quad \log\bigl( f(x) \bigr) = \frac{-(x - \lambda)^2}{ 2 \lambda } - \frac12 \log(2 \pi \lambda)$$ The integration Eq.(1.a) is not trivial but manageable. Allow me to omit typing up the calculation and just state that the result is

$$E[g(X)] = \frac{-1}2 \bigl( 1 + \log(2\pi \lambda) \bigr) \tag*{Eq.(1.b)}$$

One can be pedantic about the lower integration limit, arguing that the actual density is Poisson and non-negative. For that matter, one can consider \begin{align} E_{+}[g(X)] &= \int_{\mathbf{0}}^{\infty} f(x) \log\bigl( f(x) \bigr)\dd{x} \qquad\text{, skipping to result} \\ &= \frac12 \sqrt{ \frac{\lambda}{ 2\pi} } e^{-\lambda/2} -\frac14 \bigl( 1 + \log(2\pi \lambda) \bigr) \Bigl( 1 + \mathrm{erf}\bigl( \sqrt{ \frac{\lambda}2 } \bigr) \Bigr) \tag*{, or} \\ &= \frac12 \left( \sqrt{ \frac{\lambda}{2\pi} } e^{-\lambda/2} - \bigl( 1 + \log(2\pi \lambda) \bigr)\Phi(\sqrt{\lambda} )\right) \tag*{Eq.(2)} \end{align} where erf is the error function and $\Phi$ is the CDF of standard Normal. There's practically no difference between Eq.(1.b) and the supposedly more sensible Eq.(2) in the applicable region where $\lambda$ is large enough, since the distribution is already far away from the origin.

Admittedly I haven't investigated possible improvements from continuity correction (that using Gaussian to approximate the discrete Poisson), but I doubt there's any improvement in the asymptotic region.

In summary, as an asymptotic form, Eq.(1.b) is already as good as it gets under this framework. Depending on your tolerance on the error, maybe $\lambda = 10$ or $\lambda = 8$ can already be considered asymptotic, while there's no denying (either numerically or analytically) that both Eq.(2) and Eq.(1.b) are poor fit for $\lambda < 3$.


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You're trying to compute $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n -\lambda + x_i \log(\lambda) - \log(x_i!)$ when $x_i$ are drawn from the Poisson($\lambda$) distribution. By SLLN this is $-\lambda + \lambda \log(\lambda) - E[\log(X!)]$ where $X$ has such a distribution. The issue is that it's not obvious what $E[\log(X!)]$ is.

An approximation like yours can be obtained by estimating this by $\log(\Gamma(\lambda+1))$. This amounts to replacing $X=\lambda$ (neglecting the variation in the distribution) and interpolating for non-integer $\lambda$ with the Gamma function. For large $\lambda$ you can then apply Stirling's approximation. A three term approximation of $\log(\Gamma(\lambda+1))$ is $\lambda \log(\lambda) - \lambda + \log(\sqrt{2 \pi \lambda})$, so the first two terms cancel out with the other two terms from before, leaving $-\log(\sqrt{2 \pi \lambda})$. This has the same scaling as your version but it is offset by a constant.

A different approach is to use Stirling's approximation directly in the sum defining $E[\log(X!)]$. In fact we can give pretty nice explicit bounds:

$$E[X \log(X)]-\lambda+\frac{1}{2} E[\log(X)]+\frac{1}{2} \log(2\pi) \leq E[\log(X!)] \\ \leq E[X\log(X)]-\lambda+\frac{1}{2} E[\log(X)] + 1.$$

These two bounds differ only by a constant, namely $1-\frac{1}{2} \log(2 \pi) \approx 0.081$, but the problem of computing $E[X \log(X)]$ and $E[\log(X)]$ persists.

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  • $\begingroup$ (commenting in case the pin in my post doesn't work) I wouldn't say this approach follows the same basic idea as mine. In fact I think your approach is better (more robust, allowing further improvement), especially for the region of smaller $\lambda$. $\endgroup$ – Lee David Chung Lin Feb 3 '19 at 5:39
  • $\begingroup$ @LeeDavidChungLin In my comment on the original post, when I said "following" I meant that I could follow your reasoning, not that my own answer followed your idea. $\endgroup$ – Ian Feb 6 '19 at 18:06

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