Q:

Maximum Likelihood Estimator for Poisson Distribution

Studying for my upcoming exams I came upon this weird MLE exercise :

Let the random variable $X$ follow the Poisson Distribution with unknown parameter $\theta >0$. In $50$ observations of $X$, you only know that $20$ of them are zero. Find the Maximum Likelihood Estimator of $\theta$ using only this fact.

How would one proceed to finding the MLE using the fact that you have $20$ zero observations ? I have never came upon such a problem of MLE.

A:

You have $X_1,\ldots,X_{50} \sim\operatorname{i.i.d. Poisson}(\lambda),$ so $\Pr(X_1=0) = e^{-\lambda}.$
The number of successes in $50$ trials with probability $e^{-\lambda}$ of success on each trial has a binomial distribution. So one has
\begin{align}
L(\lambda) & = \Pr(\text{exactly 20 such “successes'' in 50 trials)} \\[10pt]
& = \binom{50}{20} \left( e^{-\lambda} \right)^{20} \left( 1 - e^{-\lambda}\right)^{30}.
\end{align}
We seek the value of $\lambda$ that minimizes that. Letting $p= e^{-\lambda},$ we can find the value of $p$ that minimizes the expression above, by writing
\begin{align}
\log L(\lambda) & = 20 \log p + 30\log(1-p) \\[10pt]
\frac d{dp} \log L(\lambda) & = \frac{20} p - \frac{30}{1-p} = \frac{50((20/50) - p)}{p(1-p)} =\begin{cases} >0 & \text{if } 0 \le p < 20/50 \\ <0 & \text{if } 20/50 < p \le 1. \end{cases}
\end{align}

