# Error propagation in measurement of photon rate

A detector is measuring photons coming from a known source $$D$$ and a background $$B$$ which produce photons respectively with rates $$F_D$$ and $$F_B$$.

Suppose we want to measure $$F_D$$ by performing two measurements: (1) the number of photons from both sources for a time $$t_1$$ and (2) the number of photons from $$F_B$$ only (by covering $$D$$) for a time $$t_2$$.

How much should the total time $$T=t_1+t_2$$ be so that the error on $$F_D$$ is less than $$1\%$$? Assume $$t_1$$ and $$t_2$$ are known exactly and Poisson statistics.

What I have tried so far

If we measure $$x$$ photons in $$t_1$$ (from both sources) and $$y$$ photons in $$t_2$$ from $$B$$ only, then since $$x=(F_D+F_B)t_1$$ and $$y=F_Bt_2$$

$$F_D = \frac{x}{t_1}-\frac{y}{t_2}$$

now what I want is that $$\frac{\delta F_D}{F_D}\leq 0.01$$.

$$\delta F_D$$ can be found from the standard error propagation formula, namely

$$\delta F_D = \sqrt{\Big(\frac{\partial F_D}{\partial x}\Big)^2 \delta x^2+\Big(\frac{\partial F_D}{\partial y}\Big)^2\delta y^2} = \sqrt{\frac{\delta x^2}{t_1^2}+\frac{\delta y^2}{t_2^2}}$$

then

$$\frac{\sqrt{\frac{\delta x^2}{t_1^2}+\frac{\delta y^2}{t_2^2}}}{\frac{x}{t_1}-\frac{y}{t_2}} \leq 0.01 \Longrightarrow \sqrt{\frac{\delta x^2}{t_1^2}+\frac{\delta y^2}{t_2^2}} \leq 0.01 \Bigg(\frac{x}{t_1}-\frac{y}{t_2}\Bigg)$$

Assuming a Poisson statistics I also know that the sigma is proportional to the average value, therefore $$\delta x = x$$ and $$\delta y = y$$, therefore

$$\sqrt{\frac{x^2}{t_1^2}+\frac{y^2}{t_2^2}} \leq 0.01 \Bigg(\frac{x}{t_1}-\frac{y}{t_2}\Bigg)$$

by taking the square of both sides, this equation leads to an impossible result.

Question

What did I do wrong? Is there another way? Thanks

• You are manipulating (subtracting) photon fluxes (photons/time). Should you not instead be manipulating photon counts (photons = rate x time)? Commented Mar 15, 2022 at 14:28
• At first glance, I find it a bit odd that the problem asks only for $T$ . What if $t_2 = 0$ or $t_1 = 0$ ? Then you'll never be able to separate $F_D$ from $F_B$ . Commented Mar 15, 2022 at 15:13
• I am subtracting rates just to get back $F_D$ basically. Regarding the time, I also find it strange but that's a problem I just copied from an old exam. Commented Mar 15, 2022 at 15:51

## 1 Answer

So you have two Poisson-distributed random variables, $$x$$ and $$y$$, such that $$x \sim \text{Po}(F_D t_1+ F_B t_1)$$ and $$y\sim\text{Po}(F_B t_2)$$ where $$F_B$$ and $$F_D$$ are to be estimated from the values of $$x$$ and $$y$$ and $$t_1$$ and $$t_2$$ are known variables.

So, by the properties of the Poisson distribution the expected value of $$x$$ is $$\bar x= F_D t_1+F_B t_1$$ and similarly $$\bar y = F_B t_2$$ then we can solve those two equations in two unknowns and get $$F_B = \frac{y}{t_2}$$ $$F_D=\frac{t_2 x-t_1 y}{t_1 t_2}$$ So then by the propagation of errors formula $$\sigma_{F_B}^2 = \frac{\sigma_y^2}{t_2^2}$$ $$\sigma_{F_D}^2=\frac{\sigma_x^2}{t_1^2} + \frac{\sigma_y^2}{t_2^2}$$ From there we simply set $$\sigma_{F_D}/F_D=0.01$$ and solve, recalling that for the Poisson distribution $$\sigma_x^2=\bar x$$