[Math] Sampling Distribution of sample mean for Poisson Distribution

Question

I am particularly struggling with part b, I don't know where to begin. For part a, I think the answer is that the sampling distribution is a Poisson(n$\lambda$).

Answer 1

The sampling distribution of a Poisson(λ) distributed random variable is given by:

$$ P(X_i = x_i) = f(x_i) = \frac{e^{-\lambda}\lambda^{x_i}} {x_i!} $$ where $x_i \in \{0, 1, 2 ...\}$

As you have correctly suggested the sum is Poisson(nλ) and therefore, substituting nλ for λ and $\Sigma x_i$ for $x_i$ the sampling distribution for $\Sigma X_i$ is given by:

$$ P(\Sigma X_i = \Sigma x_i) = g(\Sigma x_i) = \frac{e^{-n\lambda}(n\lambda)^{\Sigma x_i}} {(\Sigma x_i)!} $$ where $\Sigma x_i \in \{0, 1, 2 ...\}$

the sampling distribution for the sample mean, $\bar{X}$, is derived using the relation $\Sigma X_i =n\bar{X} $ as follows:

$$ h(\bar{x}) = P(\bar{X} = \bar{x}) = P(\Sigma X_i = n\bar{x}) = g(n\bar{x}) $$

$$ = \frac{e^{-n\lambda}(n\lambda)^{n\bar{x}}} {(n\bar{x})!} $$ where $\bar{x} \in \{0, \frac{1}{n}, \frac{2}{n} ...\}$

For $ n \ne 1$ This is not a poisson distribution as the pdf is not of the form $ \frac{e^{-\lambda}\lambda^{\bar{x}}} {\bar{x}!}$.

The sum distribution is stretched by factor 1/n as can be seen in these plots

(Note that the probabilities are only defined for the plotted circles, trendlines are just for visualising).

N = 2, λ = 3

N = 10, λ = 2