
Originally Posted by
chopet
Hi all, I need to determine the variance of the estimator for Poission mean parameter $\displaystyle \lambda$
Let $\displaystyle \theta = { 1 \over n} {\sum_{i=1}^{n} } X_i$
From our knowledge of Poission rv,
$\displaystyle E[X_i] = \lambda $
$\displaystyle E[{\sum_{i=1}^{n} }X_i] = n{\lambda} $
similarly,
$\displaystyle E[{X_i}^2] = \lambda + \lambda^2$
$\displaystyle E[ {\sum_{i=1}^{n}} {X_i}^2 ] = n{\lambda} + n{\lambda^2}$
To find the variance, we make use of the above results:
$\displaystyle V[ \theta] = {1 \over n^2} V[{\sum_{i=1}^{n}} {X_i}] = {1 \over n^2} (E[ {\sum_{i=1}^{n}} {X_i}^2] - E[{\sum_{i=1}^{n} }X_i]^2)$
$\displaystyle = { 1 \over n}{\lambda} + {1 \over n}{\lambda^2} - {\lambda^2}$
but its supposed to equal $\displaystyle {1 \over n} \lambda $ which makes it an unbiased estimator. But the above equations don't arrive at this conclusion. Where am I wrong?