Poisson Distribution

• Feb 7th 2009, 04:22 PM
eigenvector11
Poisson Distribution
Given a random sample of size n from a Poisson distribution, lambda hat_1 = X_1 and lambda hat_2 = Xbar are two unbiased estimators for lambda. Calculate the relative efficiency of lambda hat_1 to lambda hat_2.

I know this means to campare the two variances. In other words,
Var(lambda hat_2)/Var(lambda hat_1). I'm not sure how to find the variances though. According to my formula sheet, the variance of a Poisson distribution is just lambda, and the mean of a Poisson distribution is also lambda. I guess I'm not sure how to put it all together. Any ideas?
• Feb 7th 2009, 09:01 PM
mr fantastic
Quote:

Originally Posted by eigenvector11
Given a random sample of size n from a Poisson distribution, lambda hat_1 = X_1 and lambda hat_2 = Xbar are two unbiased estimators for lambda. Calculate the relative efficiency of lambda hat_1 to lambda hat_2.

I know this means to campare the two variances. In other words,
Var(lambda hat_2)/Var(lambda hat_1). I'm not sure how to find the variances though. According to my formula sheet, the variance of a Poisson distribution is just lambda, and the mean of a Poisson distribution is also lambda. I guess I'm not sure how to put it all together. Any ideas?

$Var(\hat{\lambda_1}) = \lambda$.

$Var(\hat{\lambda_2}) = Var\left( \frac{X_1}{n}\right) + Var\left( \frac{X_2}{n}\right) + \, .... + Var\left( \frac{X_n}{n}\right)$ $= \frac{1}{n^2} Var(X_1) + \frac{1}{n^2} Var(X_2) + \, .... \, + \frac{1}{n^2} Var(X_n) = \frac{\lambda}{n}$.