Maximum likelihood estimator of $\operatorname{Poisson}(\lambda)$ with restricted $\lambda$

Question

Consider $X_1, X_2, \ldots,X_n$ iid $\operatorname{Poisson}(\lambda)$ random variables, where $\lambda \in [a,b]$, $0<a<b$.

How do you find the maximum likelihood estimator of the restricted $\lambda$?

I know that the maximum likelihood estimator of the unrestricted $\lambda$, $(\lambda \in [0, \infty])$, is

$$\widehat{\lambda}_\text{MLE}=\bar{X}_n = \dfrac{X_1+X_2+\cdots+X_n}{n}.$$

I was thinking of finding the behavior as the maximum likelihood estimator at as it moves between $0$ and $\infty$ and substituting $a$ and $b$ respectively, but I am unsure if that is correct.

Thank you in advanced.

Answer 1

\begin{align} L(\lambda) & = \prod_{i=1}^n \frac{\lambda^{x_i} e^{-\lambda}}{x_i!} \\[10pt] \ell(\lambda) = \log L(\lambda) & = (\log\lambda)\left(\sum_{i=1}^n x_i\right) - n\lambda + \text{constant} \\ & = n\bar x\log\lambda - n\lambda + \text{constant} \\[10pt] \ell\,'(\lambda) & = n\left( \frac{\bar x - \lambda}\lambda\right) \quad \begin{cases} >0 & \text{if } 0\le\lambda < \bar x, \\[6pt] <0 & \text{if } \lambda > \bar x. \end{cases} \end{align}

So $L$ increases on the interval $[0,\bar x]$ and decreases on the interval $[\bar x,\infty)$ (and you'll notice I used a square bracket at the endpoint $\bar x$ in both intervals.

If $0\le a < b \le \bar x$ then $L$ increases on the interval $[a,b]$ and the maximum occurs at the right endpoint $b.$

If $0\le \bar x \le a < b$ then $L$ decreases on the interval $[a,b]$ and the maximum is at the left endpoint $a.$

If $0\le a<\bar x<b$ then $L$ increases on $[a,\bar x]$ and decreases on $[\bar x,b].$

Answer 2

One gets to the usual formula for $\hat\lambda$ by finding where the likelihood function $L(\lambda)=\exp(-n\lambda)\lambda^{n\overline X_n}$ is maximized over $\mathbb R^+$. In your case: find out where $L(\lambda)$ is biggest on $[a,b]$.