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

$X_1,...,X_n$ are independent Poisson random variables with$ X_j $having parameter$j\lambda$.What is the fisher information contained in $(X_1,...,X_n)$ about $\lambda$? BTW,What is the likelihood function in this question? $S_n$ is the log likelihood function,is the first derivative right ?i dont know if it is correctas this is the first step to get the answer. What i calculate is $\frac{\partial S_n}{\partial x}=\sum_{i=1}^{n}\frac{X_i}{\lambda i}-1$

share|cite|improve this question
up vote 3 down vote accepted

Try the following:

1) Calculate the likelihood function based on observations $x_1,\ldots,x_n$ from $X_1,\ldots,X_n$. This is just $$ L(\lambda)=L(\lambda;(x_1,\ldots,x_n))=\prod_{i=1}^n p_i(x_i), $$ where $p_i$ denotes the probability function corresponding to $X_i$. Then calculate the loglikehood function $l(\lambda)=l(\lambda;(x_1,\ldots,x_n))=\log(L(\lambda;(x_1,\ldots,x_n)))$.

2) Differentiate twice with respect to $\lambda$ and get an expression for $$ \frac{\partial^2 l(\lambda)}{\partial \lambda^2}. $$

3) Then the Fischer information is the following $$ i(\lambda)=E\left[-\frac{\partial^2 l(\lambda;(X_1,\ldots,X_n)}{\partial \lambda^2}\right]. $$

I think the correct answer must be $\frac{n(n+1)}{2}\frac{1}{\lambda}$, but please correct me if I'm wrong.

share|cite|improve this answer
What is the likelihood function?i guess i get a wrong likelihood function – Mathematics Mar 6 '12 at 10:07
It is just the product of the probability functions as we have independence. As $X_i\sim po(i\lambda)$ for $i=1,\ldots,n$ we have that $p_i(k)=\frac{(i \lambda)^k}{k!}e^{-i\lambda}$, $k\in\mathbb{N}$. So if $x_1,\ldots,x_n$ are observations from $X_1,\ldots,X_n$ the likelihood function becomes $L(\lambda)=\prod_{i=1}^n p_i(x_i)=\prod_{i=1}^n \frac{(i\lambda)^{x_i}}{x_i !}e^{-i\lambda}$. – Stefan Hansen Mar 6 '12 at 10:21
i got a question,shouldn't we differentiate w.r.t. the parameter?in this case,the parameter is $\lambda i$ – Mathematics Mar 6 '12 at 10:40
It's true that $i\lambda$ is the parameter of the Poisson distribution for $X_i$, but the only unknown parameter in your setup is $\lambda$. If you have an estimate of $\lambda$ then you automatically also have an estimate for $i\lambda$ for $i=1,\ldots,n$. So differentiation should be wrt $\lambda$. – Stefan Hansen Mar 6 '12 at 11:08
Since they're i.i.d., you could just find the Fisher information in the first one and multiply it by $n$. – Michael Hardy Mar 6 '12 at 13:47

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.