How to find $V(\bar y)$ in a Poisson distribution?

Question

Consider two estimators for the parameter $\lambda$ of a Poisson distribution, based on $n$ independent observations $y_1,\dots,y_n$

$\hat \lambda_1 = \bar y$

$\hat \lambda_2 = s^2 = \sum_{i=1}^{n}{(y_i-\bar y)^2/(n-1)}$

In a Poisson distribution both mean and variance are equal to $\lambda$, so both estimators are unbiased.

How do I find $V(\hat \lambda_1)$?

Answer 1

$Extended\; comment.$

Your question is the first part of an interesting story. Both estimators are unbiased, but the first is better than the second, because the first has a smaller variance. That is $V(\hat \lambda_1) < V(\hat \lambda_1).$ In your question, you will have found the first, but not the second.

Below are results of a simulation of 100,000 repetitions of an experiment in which we have five independent observations from $Pois(\lambda = 4).$ For this example, you will have found that $SD(\hat \lambda_1) = \sqrt{4/5} = 0.8944.$ (For a better comparison in the graph, about 20 very large simulated values out of 100,000 for $V(\hat \lambda_2)$ have been ignored; largest about 40.)

 m = 10^5;  n = 5;  lam = 4
 DTA = matrix(rpois(m*n, lam), nrow=m)
 est.1 = rowMeans(DTA)  # each row of DTA matrix has 5 obs.
 est.2 = apply(DTA, 1, var)
 mean(est.1);  mean(est.2)
 ## 3.996712  # both estimates unbiased
 ## 4.004557
 sd(est.1); sd(est.2)
 ## 0.8926094 # but first has the smaller SD
 ## 2.975659

Answer 2

For independent, identically distributed $X_i,$ we have $$V(\bar X) = V\left(\frac{1}{n} \sum_{i=1}^n X_i\right) = \frac{1}{n^2}V\left(\sum_{i=1}^n X_i\right) = \frac{1}{n}V(X_i).$$

Now you should be able to finish it for $X_i$ iid $Pois(\lambda).$