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In Example 4.4.2 on page 163 of Statistical Inference, the following is implied:

$\sum_{t=0}^{\infty}\frac{((1 - p)\lambda)^t}{t!} = e^{(1-p)\lambda}$

with a note that says "[the] sum is a kernel for a Poisson distribution". I understand the Poisson distribution, but I don't fully grasp the kernel concept yet. I tried searching for this identity, but couldn't find it. Can someone help me understand this, or provide references that I can use for background information?

Update: I initially had the incorrect equation above. I corrected it after Anand's comments.

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I checked the book and there is a typo in your question. It should be

\begin{align*} \sum_{t=0}^{\infty} \frac{((1-p) \lambda)^t}{t!} & = e^{(1-p) \lambda} \end{align*}

The above formula basically follows from the definition of an exponential function \begin{align*} \sum_{t=0}^{\infty} \frac{x^t}{t!} & = e^{x} \end{align*}

and once you substitute $x = (1-p) \lambda$ you get your desired result.

The above result can be derived using the Poisson probability mass function (pmf) which the authors refer to "[the] sum is a kernel for a Poisson distribution". This follows from the fact the pmf should sum to $1$.

\begin{align*} \sum_{x} \Pr(X = x) &= 1 \end{align*}

The pmf of Poisson is given by \begin{align*} \Pr(X = x) &= \frac{e^{-\lambda} \lambda^{x}}{x!} \end{align*}

Using the above two equations, we get \begin{align*} \sum_{x} \Pr(X = x) &= \sum_{x} \frac{e^{-\lambda} \lambda^{x}}{x!} = 1 \end{align*}

Multiply both sides by $e^{\lambda}$ we get \begin{align*} \sum_{x} \frac{\lambda^{x}}{x!} = e^{\lambda} \end{align*}

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  • $\begingroup$ Thanks for the answer and the corrections! I updated my question with the proper equation. $\endgroup$ – Alec Jan 16 '12 at 19:22

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