$X_1,...,X_n$ 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 $\frac{\partial S_n}{\partial x}=\sum_{i=1}^{n}\frac{X_i}{\lambda i}-1$


1 Answer 1


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.

  • $\begingroup$ What is the likelihood function?i guess i get a wrong likelihood function $\endgroup$ Mar 6, 2012 at 10:07
  • $\begingroup$ 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 $L(\lambda)=\prod_{i=1}^n p_i(x_i)=\prod_{i=1}^n \frac{(i\lambda)^{x_i}}{x_i !}e^{-i\lambda}$. $\endgroup$ Mar 6, 2012 at 10:21
  • $\begingroup$ i got a question,shouldn't we differentiate w.r.t. the parameter?in this case,the parameter is $\lambda i$ $\endgroup$ Mar 6, 2012 at 10:40
  • $\begingroup$ 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 $\lambda$. $\endgroup$ Mar 6, 2012 at 11:08
  • $\begingroup$ Since they're i.i.d., you could just find the Fisher information in the first one and multiply it by $n$. $\endgroup$ Mar 6, 2012 at 13:47

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