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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!

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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 $N$".

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