Let $\{X_n\}_{n\ge 1}$ be a sequence of independent identically distributed Poisson random variables with mean $\lambda^*$. Consider a sequence of random variables $\{\hat{\lambda}_{n}\}_{n\ge 1}$ defined recursively as solutions to the following equation: $$ X_{n+1} = \hat{\lambda}_{n+1} + \left(\sum_{i=1}^n\hat{\lambda}_i\right)\ln\frac{\hat{\lambda}_{n+1}}{\hat{\lambda}_{n}},\ n=1,2,\ldots$$ with $\hat{\lambda}_{1}=X_1$. That is, given $\{\hat{\lambda}_{i}\}_{i=1}^n$ and $X_{n+1}$, we solve the above equation to find $\hat{\lambda}_{n+1}.$

Question: does $\hat{\lambda}_{n}$ converge to $\lambda^*$ in some sense (e.g., in probability)?

I am looking for a general argument if possible using tools like martingale convergence theory, stochastic approximation, etc., but the basic convergence can be seen in simulations. Here is an example for $\lambda^*=10$:

enter image description here

Moreover, if $\delta=\hat{\lambda}_{n+1}-\hat{\lambda}_{n}$ is small, then $\ln\frac{\hat{\lambda}_{n+1}}{\hat{\lambda}_{n}}\approx \frac{\delta}{\hat{\lambda}_{n}}$. If also $\{\hat{\lambda}_{i}\}_{i=1}^n$ are all "close" to a particular value, then the right-hand-side is approximately $\hat{\lambda}_{n+1} + n\delta$ and $$\hat{\lambda}_{n+1}\approx\frac{X_{n+1}+n\hat{\lambda}_{n}}{n+1},$$ which is the standard recursive formula for the sample average.


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    $\begingroup$ why would you expect it to converge to the number (constant RV) $\lambda^*$? $\endgroup$ – William May 13 '16 at 16:39
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    $\begingroup$ I have updated the question description with a rationale for convergence. $\endgroup$ – user3605620 May 13 '16 at 18:30
  • $\begingroup$ that's a really nice graphic! Sorry I don't know how to answer your question, but hopefully more people will take a look at the question now. $\endgroup$ – William May 13 '16 at 18:37

This is just a bound from below. The derivation is a bit long for a comment. Clearly, a positive solution to the equation exists (although you need to specify in more details what happens when $X_1=0$ as $\lambda_1=0$ in this case). Now, we have \begin{align*} \sum_{j=2}^nX_{j} &= \sum_{j=2}^n\lambda_j+\sum_{j=2}^n\left(\sum_{i=1}^{j-1} \lambda_i\right)\ln (\lambda_j/\lambda_{j-1})\\ &=\sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i\left(\sum_{j=i+1}^n \ln(\lambda_j/\lambda_{j-1})\right) \\ &=\sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i \ln(\lambda_n/\lambda_i) \end{align*} Now since $\lambda_i>0$ we can make use of the inequality $\ln(1+x)\le x$ to obtain \begin{align*} \sum_{j=2}^nX_{j}&\le \sum_{j=2}^n\lambda_j+\sum_{i=1}^{n-1}\lambda_i (\lambda_n/\lambda_i-1)\\ &=\lambda_n-\lambda_1+(n-1)\lambda_n \end{align*} Dividing both sides by $n$ and letting $n \to \infty$ we obtain by the Law of Large Numbers, $$ \lambda \le \liminf_{n\to\infty}\lambda_n, \quad \mbox{a.s.} $$

  • $\begingroup$ Thanks for pointing out that the sequence needs to be defined differently if initial observations are 0. A more accurate definition for that case is that $\hat{\lambda}_i=X_i$ for $i=1,\ldots,n+1$ as long as $X_i=0$ for $i=1,\ldots,n$. After that, the equation applies. I think you can also claim from you proof that $\frac1{n}\sum_{i=1}^nX_i\le\hat{\lambda}_n.$ Would it be possible to use some other bound on the $\ln$ to get an inequality in the other direction? $\endgroup$ – user3605620 Jun 14 '16 at 1:01
  • $\begingroup$ An upper bound seems to be more complicated. You can try to use the elementary bound $\ln(x)\ge 1-1/x, $ for $x>0$, but then some extra arguments are needed. $\endgroup$ – Denis Denisov Jun 14 '16 at 9:50

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