# Taylor series approximation statistics

how can I show the following:

Let $X_1, X_2,\ldots, X_n$ be i.i.d Poisson with mean $\lambda$. Let $Y = |\{i: X_i =0\}|$. Then $\lambda$ is estimated by

$$\eta = - \log(Y/n)$$

Use Taylor series to find approximation for E($\eta$) and Var($\eta$)

Thank you!

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Let $Y_i=\mathrm e^\lambda\mathbf 1_{X_i=0}-1$ and $Z_n=\frac1n\sum\limits_{i=1}^nY_i$ then $\mathbb E(Y_i)=0$, $\mathrm{var}(Y_i)=\mu$ with $\mu=\mathrm e^\lambda-1$ and, by the central limit theorem, $Z_n\to0$ roughly like $\frac1{\sqrt{n}}$.
When $z\to0$, $\log(1+z)=z-\frac12z^2+o(z^2)$ hence $$\eta_n=\lambda-\log(1+Z_n)=\lambda-Z_n+\frac12Z_n^2+\varepsilon_n,$$ for some error term $\varepsilon_n$ roughly of order $\frac1{n\sqrt{n}}$. In particular, $$\mathbb E(\eta_n)=\lambda+\frac{\mu}{2n}+o\left(\frac1n\right).$$ Likewise, $$\eta_n^2=\lambda^2-2\lambda Z_n+(\lambda+1)Z_n^2+\varepsilon'_n,$$ for some error term $\varepsilon'_n$ roughly of order $\frac1{n\sqrt{n}}$. In particular, $$\mathbb E(\eta_n^2)=\lambda^2+(\lambda+1)\frac{\mu}n+o\left(\frac1n\right),$$ and $$\mathrm{var}(\eta_n)=\frac{\mu}n+o\left(\frac1n\right).$$ The two results above can be summarized as $$2n(\mathbb E(\eta_n)-\lambda)\to\mu,\qquad\mathrm{var}(\eta_n)\to\mu.$$
We have that $P(X_i = 0) = e^{-\lambda}$. So when $n$ is large we have that $Y/n \approx e^{-\lambda}$, you can see this from the ergodic theorem or equivalently applying the law of large numbers to the Bernoulli variables $1_{\{X_i=0\}}$. Hence, you should do the expansion around $e^{-\lambda}$
I guess it should be around $\lambda$ then? What is the corresponding answer in that case? Thank you?? – Salih Ucan Mar 4 '13 at 22:32
i guess it is the taylor expansion around lambda of the function $-\log(z)$ – Salih Ucan Mar 4 '13 at 22:36
$Y/n \approx e^{-\lambda}$ so you can say that $Y/n = e^{-\lambda} + \text{error}$. So you want to write the expansion $-\log(Y/n) = -\log(e^{-\lambda} + \text{error}) = \ldots$ . Alternatively you can center the variables and do an expansion around zero like @Did did. – Bunder Mar 5 '13 at 13:49