Estimating Lambda parameter in Poisson Distribution I'm working on a problem where I am given a very small amount of information, and need to impute the median from it. Suppose I have the following
╔═══════╦═════════════════════╗
║ Value ║ P(Observed > Value) ║
╠═══════╬═════════════════════╣
║     1 ║ 0.476               ║
║     2 ║ 0.168               ║
║     3 ║ 0.069               ║
║     4 ║ 0.036               ║
╚═══════╩═════════════════════╝

I would like to find the value that corresponds to $P(\text{Observed} > \text{Value}) = 0.50$, constrained to values below $0$ having $P(\text{Observed} > \text{Value}) = 1$. The values have a Poisson distribution, and I'd like to be able to solve for $\lambda$, but not sure of the best way to estimate $\lambda$ given this data.
 A: $$P(X\le k) = \sum\limits_{i=0}^{k}\frac{\lambda^i}{i!}$$
So,
$$P(X \gt k)= 1 - P(X \le k)$$
Notice, $$P(X \gt k-1) - P(X \gt k) = \frac{i^k}{k!}e^{-\lambda}$$
Then, you get the follow estimates of $\lambda$,
\begin{align}
&\frac{\lambda^4}{4!} e^{-\lambda} = 0.033 \\
&\frac{\lambda^3}{3!} e^{-\lambda} = 0.099 \\
&\frac{\lambda^2}{2!} e^{-\lambda} = 0.308 \\
&\frac{\lambda^1}{1!} e^{-\lambda} = 0.524
\end{align}
There are several ways to estimate $\lambda$. One is by brute force, scanning through a probable range, and selecting a value that minimizes some error function. On first pass, $\lambda \approx 0.9 - 1.3$
A: In the following paper: https://link.springer.com/article/10.1007/s00184-020-00765-3
(If you don't have access, there is an arxiv version.)
the authors jitter a Poisson$(\lambda)$ random variable by a Uniform$(0,1)$ to find the asymptotics of the median. This leads to a robust estimator of $\lambda$.
This might not be so useful for small samples, but it seems highly relevant if you have a large sample.
