4
$\begingroup$

So I have a question in the exam of Statistics that I dont know how to do it:

There is $X\sim \text{Poisson}(\theta)$.

I want to estimate $\exp[-\theta] = P(X=0)$.

I know by the maximum likelihood method that a consistent estimator is $\exp[-\bar{X}]$.

Now is asked me to use R commands to calculate an approximation of $\text{Var}(\exp[-\bar{X}])$, with $n=20$.

I am really not getting how I should do this if I don't have any knowledge about the parameter $\theta$. How I should simulate the data?

$\endgroup$
  • $\begingroup$ Add the self study tag since this is calss work. $\endgroup$ – Michael R. Chernick Jan 26 '17 at 23:28
  • 1
    $\begingroup$ You are allowed to choose a value of $\theta$ to work with. Otherwise it's impossible to simulate anything. $\endgroup$ – whuber Jan 26 '17 at 23:38
  • $\begingroup$ Okay. I will assume that. I was a little bit confuse once in other exams with similar problems the value of θ is given. Thanks @whuber $\endgroup$ – Fernando Jan 27 '17 at 0:00
  • 1
    $\begingroup$ do you want to simulate, or do you want to use something like the delta method? $\endgroup$ – Taylor Jan 27 '17 at 0:21
  • 1
    $\begingroup$ The question is: "Suposing n=20, write the necessary commands in R to obtain an aproximate estimative of the variance of the sampling distribution of $\exp[-\bar{X}]$.” Nothing more is given in addition of what I already mention. So I assume that it is not by the delta method once I don't have any value. What would you do @Taylor? $\endgroup$ – Fernando Jan 27 '17 at 0:58
4
$\begingroup$

Here is a short R code comparing the estimators $$\frac{1}{n}\sum_{t=1}^n \mathbb{I}_0(X_t)$$ (in yellow) and $$\exp\left\{-\sum_{t=1}^nX_t\big/n\right\}$$ (in brown) for $n=20$ for a range of values of $\theta$

compest<-function(T=1e3,tmin=.1,tmax=10){

 varz=matrix(0,50,2)
 theta=seq(tmin,tmax,le=50)
 for (t in 1:50){
  sampl=matrix(rpois(n=20*T,lambda=theta[t]),ncol=20)
  varz[t,]=c(var(apply(sampl==0,1,mean)),
             var(exp(-apply(sampl,1,mean))))}
 return(varz)}

This shows that the estimator based on $\bar{X}_n$ is leading to a smaller variance than the one based on the frequency of zero draws.

The continuous curves are the theoretical values of the variances, namely $e^{-\theta}(1-e^{-\theta})/n$ for the Binomial proportion of zero draws and $e^{-2\theta}\theta/n$ for the exponential of the average. (This variance is a delta-method approximation of the exact variance, but the fit is very good!)

comparison of two estimators of exp(-theta) in terms of variance

Actually, as pointed out by George Henry on my blog, the derivation of the mean and variance of $$\exp\left\{-\sum_{t=1}^nX_t\big/n\right\}$$ is quite manageable: since $n\bar{X}_n$ is a Poisson $\mathscr{P}(n\theta)$ variable \begin{align*} \mathbb{E}[\exp\{-\bar{X}_n\}]&=\sum_{i=0}^\infty \exp\{-i/n\}\frac{(n\theta)^i}{i!}\exp\{-n\theta\}\\ &=\sum_{i=0}^\infty \left(\exp\{-1/n\} n\theta \right)^i \frac{\exp\{-\theta\}}{i!}\\ &=\exp\left\{-n\theta+n\theta\exp\{-1/n\} \right\}\\ &=\exp\left\{-n\theta[1-\exp\{-1/n\}]\right\} \end{align*} and \begin{align*} \mathbb{E}[\exp\{-\bar{X}_n\}^2]&=\sum_{i=0}^\infty \exp\{-2i/n\}\frac{(n\theta)^i}{i!}\exp\{-n\theta\}\\ &=\sum_{i=0}^\infty \left(\exp\{-2/n\} n\theta \right)^i \frac{\exp\{-\theta\}}{i!}\\ &=\exp\left\{-n\theta+n\theta\exp\{-2/n\} \right\}\\ &=\exp\left\{-n\theta[1-\exp\{-2/n\}]\right\} \end{align*} Hence $$\text{var}(\exp\{-\bar{X}_n\})=\exp\left\{-n\theta[1-\exp\{-2/n\}]\right\}-\exp\left\{-2n\theta[1-\exp\{-1/n\}]\right\}$$ As shown on the plot below, the difference with the approximation is hard to spot! enter image description here

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.