TITLE: Convergence of an implicitly defined sequence of random variables
QUESTION [12 upvotes]: 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 $\lambda^*=10$:
 
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.

REPLY [2 votes]: 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, 
$$
\lambda \le \liminf_{n\to\infty}\lambda_n, \quad \mbox{a.s.}
$$