Compute maximum likelihood estimator and confidence interval

Let $X$ have a Poisson distribution with parameter $\lambda$. Find the maximum likelihood estimator of $\alpha=\mathbb P(X=0)$. In a sample of size 100 from a Poisson distribution, it is found that the mean is 0.75. Calculate an approximate 95% confidence interval for ${\alpha}$.

For the MLE, we want to maximise $e^{-\lambda}$, but isn't that a constant?

For the confidence interval, I think it is an interval of radius $\frac{0.075}{0.075/100}$, but I can't quite remember. A few years since A levels...

• For the MLE, certainly $\lambda$ is a constant: but what value does it have? MLE tells you to select the value that maximizes the likelihood. – whuber Apr 30 '14 at 18:07
• @whuber we would like ${\lambda}$ as small as possible. So what is that? – user134724 Apr 30 '14 at 19:10
• You are not trying to maximize the function $e^{-\lambda}$, you are trying to maximize the likelihood with respect to the parameter $e^{-\lambda}$ – jsk Apr 30 '14 at 19:15
• Do you know how you would find the MLE for $\lambda$? What function do you maximize to find the MLE for $\lambda$? – jsk Apr 30 '14 at 19:20
• – jsk Apr 30 '14 at 19:43

There are several things you need to recall in order to do this problem.

1) Invariance of the MLE to transformations.

2) Approximate normality of the MLE.

3) Delta method to create interval for $g(\lambda ) = e^{-\lambda}$.

4) Instead of using 3, you might want to recall that for monotonic functions $g(\lambda)$, you can start with a CI for $\lambda$, call it $(L,U)$. Then a confidence interval for $g(\lambda)$ would be $( g(L), g(U) )$ if g is monotonically increasing, and $( g(U), g(L) )$ if monotonically decreasing.

As a sidenote, the method listed in 4 may be preferred to the method in 3 since it would be guaranteed to be a range respecting CI for $e^{-\lambda}$.

• I'm confused about the first one. I don't know what I'm meant to be maximising – user134724 Apr 30 '14 at 19:22
• It's a useful property to recall that if you have the MLE for a parameter $\lambda$, call it $\hat{\lambda}$, then if you want the MLE for any function $g(\lambda)$ is $g(\hat{\lambda})$ – jsk Apr 30 '14 at 19:26
• Alternatively, if you're not familiar with that property of MLEs, you should define $\theta=e^{-\lambda}$, then rewrite the likelihood in terms of $\theta$ instead of $\lambda$, then maximize the likelihood with respect to $\theta$. – jsk Apr 30 '14 at 19:29

As this question is tagged as self-study, I will only sketch an answer.

First, you need to fix some notations. Denote $x_1, \dots, x_n$ the observed values of $X_1, \dots, X_n$, iid random variables of law $\mathcal P(\lambda)$.

You have $P( X_i = x_i ) = { \lambda^{x_i} \over x_i! } e^{-\lambda}$, hence a log-likelihood for the $i^{th}$ observation $\ell_i(\lambda ; x_i) = x_i \log \lambda - \lambda + \text{cste}$, and the log-likelihood for the whole sample is $$\ell(\lambda; x_1, \dots, x_n) = \left(\sum_i x_i \right) \log \lambda - n \lambda,$$ from which you could derive the MLE for $\lambda$, $\hat\lambda = \overline x$.

Yes, this is fine, but you are interested in $\alpha = e^{-\lambda}$, which gives an alternate parametrization of the Poisson distribution. The log-likelihood of $\alpha$ for the whole sample is $$\ell(\alpha ; x_1, \dots, x_n) = \left(\sum_i x_i \right) \log( -\log \alpha ) + n \log \alpha.$$

You can now find $\hat \alpha$ which is of course $e^{-\hat\lambda}$, but more importantly, you can compute the Fisher information $- {\partial ^2 \over \partial \alpha^2}\ell(\alpha ; X_1, \dots, X_n)$, and its expected value, and the variance of the (asymptotically normal) MLE of $\alpha$...

Hope this helps.

PS. Suggestion 4 from jsk’s answer is surely better (easier to handle): compute a CI for $\hat\lambda$, deduce a CI for $\hat\alpha = e^{-\hat\lambda}$...