Sampling Distribution of sample mean for Poisson Distribution
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$\lambda$).
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$\begingroup$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).
$\endgroup$ $\begingroup$Your answer is related to this subject
Since each variable in the sample is Poisson($\lambda$) distributed(and they are essentially independent), the sum would be, as you said, Poisson($n \lambda$).
The proof is simply deduced from the fact that the characteristic function of sum of independent random variables is the multiplication of their characteristic function. And using the inverse formula gives its density function(in case of discrete random variables their mass probability function).
Well the section(b) is not much more different.
$\endgroup$ $\begingroup$I think the standard way of solving this is using induction. I.e. computing each probability and inducing the pattern. e.g.
$P(\sum{x_i = 0}) = \binom{n}{0}(e^{-\lambda})^n = e^{-n\lambda} ; \;$ every x has to be 0.
$P(\sum{x_i = 1}) = \binom{n}{1}(e^{-\lambda})^{n-1}(\frac{\lambda}{1!}e^{-\lambda}) = n\lambda e^{-n\lambda} ; \;$ exactly one x has to be 1.
$P(\sum{x_i = 2}) = \binom{n}{2}(e^{-\lambda})^{n-2}(\frac{\lambda}{1!}e^{-\lambda})^2 + \binom{n}{1}(e^{-\lambda})^{n-1}(\frac{\lambda^2}{2!}e^{-\lambda}) = \frac{(n\lambda)^2}{2!} e^{-n\lambda} ; \;$ either 2 x's are 1, or 1 x is 2.
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And so by induction $\sum{x_i} \sim Pois(n\lambda)$
An "easier" way is using the moment generating function (MGF) (or alternatively the characteristic function). For this you need a small conjecture:
$M_{\sum{x_i}}(t) = \mathbf{E}(e^{\sum{x_i} t}) = \mathbf{E}(\prod e^{x_i t}) = \prod\mathbf{E}(e^{x_i t}) = \prod\mathbf{E}(e^{x t}) = \prod M_{x}(t) = M_{x}(t)^n$
The MGF of Poisson is $M_x(t) = e^{\lambda(e^t-1)}$, so the MGF of the sum is $M_{\sum x_i}(t) = e^{n\lambda(e^t-1)}$, so by this we can say that $\sum{x_i} \sim Pois(n\lambda)$.
Since $\sum{x_i} = n \bar{x}$, then $\bar x \sim \frac{1}{n}Pois(n\lambda)$
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