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.

  • 1
    $\begingroup$ why would you expect it to converge to the number (constant RV) $\lambda^*$? $\endgroup$ – William May 13 '16 at 16:39
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.