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Suppose I take $n$ measurements of a discrete random variable in an experiment, and from that get a mean:
$ \mu = \frac{1}{n}\sum{x_i}$
I take this to be the mean in a Poisson distribution. Is the uncertainty in that mean the standard error:
$s_n = (\frac{\mu}{n})^\frac{1}{2}$ ?
Yes if the sample is independent and identically distributed, this is an appropriate expression for the standard error of the mean. It is a direct result of both 1) the law of total variance which states that $\mbox{var}(X+Y) = \mbox{var}(X) + \mbox{var}(Y) + 2 \mbox{cov} (X, Y)$ (where the last term is 0 if $X,Y$ are independent) and 2) appealing to the closure of Poisson distributions under a sum: so that $\sum_{i=1}^n X_i$ has a Poisson($n\mu$) distribution with variance $n\mu$ and the sample mean $\bar{X}$ is $$\mbox{var}(\frac{1}{n} \sum X) = \frac{1}{n^2}n\mu=\frac{\mu}{n}$$
And the variance of the sum is transformed to a standard deviation through the square root.
