# Show $\mu$ is unbiased and find $\mathsf{Var}(\mu)$

We have $n$ independent observations, $\{Y_1,...,Y_n\}\overset{iid}\sim\mathcal{Poissson}(\mu)$. Consider the estimator $\tilde\mu = \bar Y$. I have to show that $\tilde\mu$ is unbiased and find the variance. Did I do this correctly?

\begin{align}\mathbb{Bias}(\tilde\mu) &= \mathbb E\tilde\mu-\mu\\ &= \mathbb E(\tfrac 1n\sum_{i=1}^n\mathbb E(Y_i))-\mu\\ &= \tfrac 1n\sum_{i=1}^n\mathbb E(Y_i)-\mu\\&=\mu-\mu\\&=0 \\ \\\therefore\quad \mathbb{Var}(\mu) &=\mu&&\text{Since }\{Y_1,...,Y_n\}\overset{iid}\sim\mathcal{Poissson}(\mu)\end{align}

The variance is incorrect. Note that $$\text{Var}(\bar{Y})=\frac{1}{n^2}\sum \text{Var}(Y_i)=\frac{1}{n^2}\times n\mu=\frac{\mu}{n}$$ since the $Y_i$ are independent.
• @ChanceGordon You've shown that $E\bar{Y}=\mu$ when you showed that the bias is zero. Hence $\bar{Y}$ is a consistent estimator of $\mu$. – Foobaz John Feb 3 '18 at 1:35
• Make sure you can distinguish the terms consistent estimator and unbiased estimator. For unbiasedness, John has already argued in the above comment. For consistency, you need to show that $\bar{Y} \to \mu$ in probability, which is easy after you calculating the variance as you can now use the Chebyshev's inequality and the variance term will squeeze the probability to $0$. – BGM Feb 3 '18 at 2:46