# standard error & standard deviation (Poisson counts)

I observe 25 events over a period of 50 seconds. I assume that the events are generated by an underlying Poisson process.

1. What is the standard deviation of the count?

2. What is the standard error of the count?

Please note that I understand the concept of a standard error for a sample mean, but am having difficulty understanding the concept of a standard error for an observed count (i.e., the observation of 25 counts is just one observation).

My understanding of standard error is thus: given an infinite population with a known mean ($\mu$) and known standard deviation ($\sigma$), then $\sim 68$% of random samples will have $\bar{x}$ within $\pm 1 \times$ standard error of the true mean $\mu$. The standard deviation of the sample is the best estimate of $\sigma$. So standard error on the sample mean is given by $SE_\bar{x}=s/\sqrt{n}$ - intuitively, I can see why it's proportional to $\sigma$ and inversely proportional to $n$, but my knowledge goes no further (e.g., why $\sqrt{n}$, why not proportional to $\sigma^2$, etc).

• What, then, is your understanding of a standard error of the mean? If it's related to the concept of variation under repeated sampling, then contemplate what the outcome of your count would be under repeated sampling. If variation under repeated sampling is not part of your understanding, then it would be best for you to edit this post to include a statement of what your understanding is. – whuber Feb 16 '18 at 16:44
• I think a good way to approach this question (assuming independence) is that the seconds are an exposure. This allows you to estimate the mean and variance of the Poisson process. The sum of 50 Poissons is also a Poisson, which allows you to get the mean and standard deviation of the count, which will pin down the standard error. – Dimitriy V. Masterov Feb 16 '18 at 19:34
• @whuber - have edited as per suggestion. Given my single observation, my best estimate is that further 50-second observations will yield 25 events (on average). And given that the events are generated by a Poisson process, the best estimate for the standard deviation for each 50-second interval will be $s=\sqrt{25}$. Given that I have taken a single observation ($n=1$), I think the standard error should be $SE = s / \sqrt{n} = \sqrt{25} / \sqrt{1}$. So my conclusion is that standard error and standard deviation are the same. But I feel like I've totally misunderstood somewhere along the lines. – Ben Feb 16 '18 at 21:39
• @Dimitriy - not sure I follow - are you saying I can treat this as 50 samples of a Poisson process parameterised by $\mathsf{Pois}(\lambda=0.5)$? – Ben Feb 16 '18 at 21:42
• @Ben Yes, that is what I had in mind. – Dimitriy V. Masterov Feb 16 '18 at 22:06