$\newcommand{\eff}{\operatorname{eff}}$
I am asked to derive the efficiency of the estimator $\hat{\lambda}_1 = \frac{1}{2}(Y_1+Y_2)$ relative to $\hat{\lambda}_2=\bar{Y}$, where $Y_1,Y_2,\ldots,Y_n$ denotes a random sample of size n from a Poisson distribution with mean $\lambda$.
I know that $\eff(\hat{\theta}_1,\hat{\theta}_2)=\frac{V(\hat{\theta}_2)}{V(\hat{\theta}_1)}$ for two unbiased estimators $\hat{\theta_1}$ and $\hat{\theta_2}$.
So far I have:
$$ \begin{align} \eff(\hat{\lambda}_1,\hat{\lambda}_2) & =\frac{V(\bar{Y})}{V(\frac{1}{2}(Y_1+Y_2))} \\[8pt] & =\frac{\lambda/n}{\frac{1}{4}V(Y_1)+\frac{1}{4}V(Y_2)} \\[8pt] & =\frac{\lambda/n}{\lambda/2} \\[8pt] & =n/2 \end{align} $$
Is this correct, or am I missing something? I have to do a short presentation of this problem in class and I'd like to make sure I didn't screw up somewhere.
You should have $$\frac{\lambda/n}{\lambda/2} = \frac{2}{n},$$ not $\frac{n}{2}$. Thus the estimator $\hat \lambda_2 = \bar Y$ is a more efficient estimator (and dominates $\hat \lambda_1$) for all $n > 2$, which makes sense, because $\hat \lambda_1$ discards information about the sample if there are more than two observations.