Particle detecting Poisson process Problem:
We are measuring cosmic ray muons.
If we add a lead shielding over the detector, the rate decreases, $\lambda_1=0.1746s^{-1}$
We find the original detection rate is $\lambda_2=0.18 s^{-1}$ (muons per second).
We want to measure the muon rate first with shielding (for $t_1$ time) and without (for $t_2$ time).
We want to measure for as short overall time as possible ($t_1+t_2$).
We would like to have an expected significance of $5\sigma$ that the two rates differ. (So, the relative uncertainty of the difference of the two rates should be 20%.)
Question:
How long should we measure the two cases, i.e. what should $t_1$ and $t_2$ be?
Attempt of solution:
The particle rate is Poisson distributed, that means the probability of detecting $n$ particle in a $t$ time period is: $$P(n)=\frac{(\lambda t)^n}{n!}e^{-\lambda t}$$
We should wind at least one particle after $t_1$ and $t_2$:
$$P(at\,least\,one\,muon)=1-P(n=0)=1-e^{\lambda_1t_1}$$
also, we have the variance:
$$\sigma^2=\lambda t$$ so that $$\sigma^2_1-\sigma_2^2=\lambda_1 t_1-\lambda_2 t_2=0.2$$
I still need help to determine $t_1$ and $t_2$.
edit: typos
 A: If $t_1 = t_2 = t$
Below you see an example with the distribution of
$$x \sim Poisson(17460) \qquad y \sim Poisson(18000)$$

You can make several approximations

*

*The standard error of $y-x$ will be $\sigma_{y-x} = \sqrt{x+y}$ and you can see that the $\pm 5 \sigma$ boundary is roughly a straight line.

*You could approximate the $y-x$ as normal distributed.

The distribution of $y - x$ is approximately:
$$x-y \sim N(\mu,\sigma^2) \\\text{with}\\
\begin{array}{}\mu &=& (0.18-0.1746)\, t \\
 \sigma& =& (0.18+0.1746) \sqrt{t}\end{array}$$
Then you can compute the power $P(y-x > 5 \sigma)$ based on $$\mu/\sigma = \frac{0.18-0.1746}{\sqrt{0.18+0.1746}} \sqrt{t}$$

If $t_1 \neq t_2$
This time you can not look so easily $x - y$ but you can look at the estimates $\hat{\lambda}_i = x_i/t_i$ and their difference, which will be again approximately normal distributed.
The mean will be
$$\mu_{\hat{\lambda}_2 - \hat{\lambda}_1} = 0.18-0.17460$$
The standard deviation
$$\sigma_{\hat{\lambda}_2 - \hat{\lambda}_1} = \sqrt{0.18/t_2+0.17460/t_1}$$
and the ratio of $\mu/\sigma$ is
$$\frac{\mu}{\sigma} = \frac{0.18-0.17460}{\sqrt{0.18/t_2+0.17460/t_1}}$$
You can decide on a certain optimum/desired ${\mu}/{\sigma}$ and then compute the $t = t_1+t_2$ and optimal distribution of $t_1$ and $t_2$ by writing
$$t  = t_1 + 0.18 \left(\left(\frac{0.18-0.17460}{{\mu}/{\sigma}}\right)^2 -\frac{0.17460}{t_1}\right)^{-1}$$
and minimize the right hand side by finding the right $t_1$.
