# Error analysis for counting experiments

i am trying to refresh my knowledge in error analysis and stumbled over an interesting question.

Suppose i have a radioactive compound and i want to measure the standard deviation of the decay count. To do that, i have two possible strategies

Strategy 1: I measure the decay count $$\nu$$ for a fixed amount of time $$N$$ times. I assume the timing to be exact. I then calculate the sample mean $$\bar{\nu}$$ and the SDOM using $$\bar{\nu} = \frac{\sum_{i=1}^{N}{\nu_i}}{N} \qquad \text{and} \qquad \sigma_{\bar{\nu}} = \sqrt{\frac{1}{N(N-1)} \sum_{i=1}^{N}{(\nu_{i}-\bar{\nu})^{2}}}$$

Strategy 2: I treat the $$N$$ measurements as one big measurement and derive the error by taking the square root of the total count and deviding by $$N$$. See below $$\sigma = \frac{\sqrt{\sum_{i=1}^{N}{\nu_i}}}{N}$$

My question now is: Shouldn't $$\sigma_{\bar{\nu}}$$ in strategy 1 and $$\sigma$$ in strategy 2 ideally lead to one and the same solution? And shouldn't the two expressions for the errors then be equal algebraically? I tried deriving one expression using the other but did not get anywhere really. Maybe it only holds true for $$N \rightarrow \infty$$? Thank you for your help in advance!

The data follows a Poisson distribution. This is important, because the Poisson distribution is characterised by a single parameter, the so called rate $$\lambda$$. The rate is equal to the mean value $$\lambda = \mu$$, as well as to the variance $$\lambda = \sigma^2$$.
The maximum likelihood, method of moments, and minimum variance unbiased estimator of $$\lambda$$ is given by the total count divided by the total time interval, $$\hat \lambda = \frac{\textrm{total count}}{T_N} = \frac{\textrm{total count}}{N \cdot T_1}$$. By rearranging the terms we see that it is equal to the average count divided by a single time interval, $$\hat \lambda = \frac{\sum_{i=1}^N x_i}{N\cdot T_1} = \frac{\frac{1}{N}\sum_{i=1}^N x_i}{T_1} = \frac{\bar x}{T_1}$$.
In your question you discuss methods for estimating the standard deviation of $$\hat \lambda$$. I feel it makes more sense to calculate its confidence interval. Two commonly used approximations for the confidence interval of $$\hat \lambda$$ are (1) the Pearson-Hartley approximation, which uses the $$\chi^2$$ distribution, and (2) the normal distribution, which uses the central limit result $$\hat \sigma_{\hat \lambda}=\sqrt{\hat \lambda/N}$$, which holds for "large $$N$$".