A day without calls

Idea. Estimate $\lambda$ by maximum likelihood, then plug it into $e^{-\lambda }$. The invariance property of the MLE says this is the MLE of $e^{-\lambda }$ itself, which is not the same as the fraction of days with no calls.

Step 1: likelihood. For counts $x_1,\dots ,x_n$, $$\begin{array}{rl}\log L(\lambda )& =\sum _i\left(x_i\log \lambda -\lambda -\log x_i!\right)\\ & =\left(\sum _i{x}_i\right)\log \lambda -n\lambda +\text{const}.\end{array}$$

Step 2: maximise. Setting the derivative to zero gives $\frac{1}{\lambda }{\sum }_i{x}_i=n$, so $\hat{\lambda }=\bar{x}=\frac{2+0+3+1+4}{5}=2$.

Step 3: invariance. $\mathbb{P}(X=0)=e^{-\lambda }$ is a one-to-one function of $\lambda$, so its MLE is $$e^{-\hat{\lambda }}=e^{-2}\approx \mathrm{0.135.}$$

Comparison. The naive estimate, the fraction of observed days with no calls, is $\frac{1}{5}=0.2$. It ignores the Poisson model. The MLE uses all five counts and is far less noisy, though it is slightly biased (by Jensen's inequality, $\mathbb{E}[e^{-\bar{X}}]>e^{-\lambda }$).