# Thread: Deriving method of moments

1. ## Deriving method of moments

Let's hypothesize that I have n iid observations from a Poisson distribution with parameter lambda, how do I derive the method of moments estimator of lambda?

$\displaystyle X_1,...,X_n \sim Poisson(\lambda)$

$\displaystyle \mbox{Derive the method of moments estimator of } \lambda$

Here's what I have.

$\displaystyle Pr\{X_i = k\} = \frac{\lambda^k}{k!}e^{-\lambda}$

$\displaystyle \mu_1 = E[X] = \sum_{i=1}^{k} k \frac{ \lambda^k}{k!}e^{- \lambda} = e^{- \lambda} \sum_{i=1}^{k}k \frac{ \lambda^k}{k!}$

How do I simply this more?

2. For a Poisson Distribution:

$\displaystyle \displaystyle{E[X] = \sum_{x=0}^{\infty} x \frac{e^{-\lambda}\lambda^{x}}{x!}}$

$\displaystyle \displaystyle{= \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{x}}{(x-1)!}}$

$\displaystyle \displaystyle{=\lambda \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{(x-1)}}{(x-1)!}}$

substitute y = x-1

$\displaystyle \displaystyle{=\lambda \sum_{y=0}^{\infty} \frac{e^{-\lambda}\lambda^{y}}{y!}}$

$\displaystyle \displaystyle{= \lambda}$

Note that $\displaystyle \displaystyle{\frac{e^{-\lambda}\lambda^{y}}{y!}}}$ is the pmf of a Poisson distribution, and the pmfs of all discrete distributions sum up to 1

3. Originally Posted by financial
Let's hypothesize that I have n iid observations from a Poisson distribution with parameter lambda, how do I derive the method of moments estimator of lambda?

$\displaystyle X_1,...,X_n \sim Poisson(\lambda)$

$\displaystyle \mbox{Derive the method of moments estimator of } \lambda$

Here's what I have.

$\displaystyle Pr\{X_i = k\} = \frac{\lambda^k}{k!}e^{-\lambda}$

$\displaystyle \mu_1 = E[X] = \sum_{i=1}^{k} k \frac{ \lambda^k}{k!}e^{- \lambda} = e^{- \lambda} \sum_{i=1}^{k}k \frac{ \lambda^k}{k!}$

How do I simply this more?
First show that $\displaystyle $$\lambda is the mean of a Poisson RV with parameter \displaystyle$$ \lambda$

Since the $\displaystyle $$Xs are iid each with mean \displaystyle$$ \lambda$ the sample mean of the $\displaystyle $$Xs has expectation equal to \displaystyle$$ \lambda$

CB