Maximum Likelihood Problem of a variation of a Poisson random variable

Question

I currently have to solve exercise 3.2.

Can someone give me a hint on how to come up with a solution for 3.2? It should be similar to 3.1, which I have solved, not?

3.1. Given a realization (observation) $x_1,\dots,x_n$ of these (Poisson $(\lambda)$ i.i.d. random variables $X_1,\dots,X_n$) find the maximum likelihood estimate of $\lambda.$

$$\begin{align} L(\lambda) &= P(x_1,\dots,x_n)\\ &=\prod_{i=1}^n P(x_i)\\ &=\prod_{i=1}^n \frac{e^{-\lambda}\lambda^{x_i}}{x_i!}\\ &=e^{-n\lambda} \lambda^{\sum x_i}\frac 1 {\prod x_i!} \end{align}$$

Hence,

$$\begin{align} \log(L(\lambda))&=\log\left(e^{-n\lambda}\right)\log\left( \lambda^{\sum x_i}\right)\log\left(\frac 1 {\prod x_i!}\right)\\ &=-n\lambda +\sum x_i \log \lambda + \log \frac{1}{\prod x_i!} \end{align}$$

Getting the maximum,

$$\begin{align} \frac{d\log(L(\lambda))}{d\lambda}&=-n+\frac{\sum x_i} \lambda + 0\\ &\implies \lambda= \frac{\sum x_i} n \end{align}$$

3.2. Assume that we only observe events $X_i=0$ or $X_i>0$ for each random variable. Given a realization $y_i,\dots, y_n$ of these events ($y_i=0$ if $x_i=0$ and $y_i=1$ if $x_i>0$), find the MLE of $\lambda.$

Answer 1

So the probability of getting $y_i=1$ can be considered a success of a Bernoulli experiment. And this $\Pr(Y_i=1)=1- \Pr(X_i=0) = 1-\exp(-\lambda).$

The MLE of a Bernoulli rv is the same as for a Poisson - the mean of the sample, which corresponds to the estimated probability of success.

Therefore,

$$\begin{align} \bar y&=1-\exp(-\lambda)\\[2ex] \lambda&=-\log \left( 1 -{\bar y}\right). \end{align}$$

This seems to hold on a quick simulation:

set.seed(561)

m = matrix(0,12,2)

for(i in 1:12){
l = i
n = 1e6
sam = rpois(n,l)
ybar = mean(sam!=0)
m[i,1] <- i
m[i,2] <- round(-log(1-ybar), 3)
}
plot(1:12, m[,2], xlab="λ", ylab= "Estimated λ", main="MLE", pch=19, col='tan4')