Estimating Lambda parameter in Poisson Distribution

Question

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.

Answer 1

Given the data, we have $$\Pr[X \le 1] = 1 - \Pr[X > 1] = 1 - 0.476 = 0.524, \\ \Pr[X = 2] = \Pr[X > 1] - \Pr[X > 2] = 0.476 - 0.168 = 0.308, \\ \Pr[X = 3] = \Pr[X > 2] - \Pr[X > 3] = 0.168 - 0.069 = 0.099, \\ \Pr[X = 4] = \Pr[X > 3] - \Pr[X > 4] = 0.069 - 0.036 = 0.033, \\ \Pr[X > 4] = 0.036.$$ We can multiply these probabilities by $1000$ to obtain a hypothetical sample $\boldsymbol x$ of size $n = 1000$, which has likelihood function $$\mathcal L(\lambda \mid \boldsymbol x) = \left(e^{-\lambda} (1+\lambda)\right)^{524} \left(e^{-\lambda} \frac{\lambda^2}{2!}\right)^{308} \left(e^{-\lambda} \frac{\lambda^3}{3!}\right)^{99} \left(e^{-\lambda} \frac{\lambda^4}{4!}\right)^{33} \left(1 - \sum_{x=0}^4 e^{-\lambda} \frac{\lambda^k}{k!}\right)^{36}.$$ Then after dropping out the constant multiplicative factors, the log-likelihood simplifies to $$\ell(\lambda \mid \boldsymbol x) = -964\lambda + 524 \log (1 + \lambda) + 1045 \log \lambda + 36 \log \left(1 - \sum_{x=0}^4 e^{-\lambda} \frac{\lambda^k}{k!}\right).$$ We will need to use a computer to numerically search for the global extrema; we get $$\hat \lambda = 1.555229077495491498 \ldots.$$

That said, there are two issues. First, it is not clear what you are trying to do with the parameter estimate; if you only wanted to compute the median, there are easier ways; one simply notes that $\Pr[X \le 1] > 0.5$ and $\Pr[X \ge 1] > 0.5$, so the median is $1$. Second, the numeric computation is not trivial. The most sound approach is to establish a Newton's method recursion relation on the derivative of the log-likelihood to find the critical point: $$\lambda_{m+1} = \lambda_m - \frac{\ell'(\lambda_m \mid \boldsymbol x)}{\ell''(\lambda_m \mid \boldsymbol x)}, \quad \lambda_0 = 1.$$ Here I have used a suitable initial guess $\lambda_0 = 1$ informed by the fact that the sample mean is bounded below by $\bar X \ge 1.045$.

If this method is not appealing, we can use a recursive method of moments calculation. Suppose $\tilde \lambda_{m+1} = \bar X_m$ is our $(m+1)^{\rm th}$ guess for $\lambda$, which is the sample mean based on the previous estimate $\lambda = \tilde \lambda_m$. Then we would expect that $$524 \cdot \frac{\Pr[X = 1]}{\Pr[X \le 1]} = 524 \frac{\lambda}{1+\lambda}$$ of the observations are equal to $1$, and the others are zero; similarly, for each $x \ge 5$, $$36 \cdot \frac{\Pr[X = x]}{\Pr[X > 4]}$$ of the observations are equal to $x$, approximately. So for an initial guess $\tilde \lambda_0 = 1$, we have $$\tilde \lambda_1 = \bar X_0 = \frac{262(1) + 308(2) + 99(3) + 33(4) + 30(5) + 5(6) + 1(7)}{1000} = 1.494, \\ \tilde \lambda_2 = \bar X_1 = \frac{314(1) + 308(2) + 99(3) + 33(4) + 27(5) + 7(6) + 1(7) + 1(8)}{1000} = 1.551, \\ \tilde \lambda_3 = \bar X_2 = \frac{319(1) + 308(2) + 99(3) + 33(4) + 27(5) + 7(6) + 2(7)}{1000} = 1.555, \\ $$ and subsequent iterations are stable.

Answer 2

$$P(X\le k) = \sum\limits_{i=0}^{k}\frac{\lambda^i}{i!}$$ 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, $\lambda \approx 0.9 - 1.3$

Answer 3

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.