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The zero-truncated poisson distribution has probability mass function:

$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{(1-e^{-\lambda})k!}$$, $k=1,2,...$

And the expectation of the truncated Poisson distribution via MLE is given as $\frac{\lambda}{(1-e^{-\lambda})}$ According to this document (pages 19-22) the Fisher Information is given by

$$I(\theta) = \frac{n}{(1-e^{-\lambda})}\left[\frac{1}{\lambda}-\frac{e^{-\lambda}}{(1-e^{-\lambda})}\right]$$

How is this derived?

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Here is my best attempt thus far (but still the wrong answer at the end):

The likelihood is given by:

$$L(\lambda) = \prod_{n=1}^n\frac{\lambda^{x_i}e^{-\lambda}}{x_i!(1-e^{-\lambda})}$$

and thus the log-likelihood is:

$$l(\lambda)=-\ln(x_i!)+\ln(\lambda)\sum_{i=1}^nx_i-n\lambda-n\ln(1-e^{-\lambda})$$

Differentiating with respect to $\lambda$ we get the first derivative:

$$l'(\lambda)=-n-n\ln(1-e^{-\lambda})+\frac{1}{\lambda}\sum_{i=1}^nx_i$$

and the second derivative with respect to $\lambda$ is:

$$l^"(\lambda)=n\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}-\frac{1}{\lambda^2}\sum_{i=1}^nx_i$$

The Fisher Information is given by

$$I(\lambda)=E[-l^"(\lambda)|\lambda]=E\left[-\left(n\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}-\frac{1}{\lambda^2}\sum_{i=1}^nx_i\right)|\lambda\right]$$

$$\Rightarrow -n\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}+\frac{1}{\lambda^2}\sum_{i=1}^nE[x_i|\lambda]$$

Which yields my incorrect Fisher Information:

$$I(\lambda)=-n\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}+\frac{n\lambda}{\lambda^2}=n\left(\frac{1}{\lambda}-\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}\right)$$

What have I done wrong with the expectation?

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In the last line of your derivation, you substituted the expectation of the Poisson distribution, rather than the expectation of the truncated Poisson distribution. Fix that, and the correct result should follow.

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  • 2
    $\begingroup$ +1 Good catch--thanks for your patience and time for looking this over so carefully. $\endgroup$ – whuber Jan 10 '13 at 14:05

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