Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

I dont know around what point to take the Taylor series please help!

share|cite|improve this question
up vote 0 down vote accepted

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 $$ 2n(\mathbb E(\eta_n)-\lambda)\to\mu,\qquad\mathrm{var}(\eta_n)\to\mu. $$

share|cite|improve this answer

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 $e^{-\lambda}$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.