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

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


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

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.


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

  • 1
    $\begingroup$ You are manipulating (subtracting) photon fluxes (photons/time). Should you not instead be manipulating photon counts (photons = rate x time)? $\endgroup$ Commented Mar 15, 2022 at 14:28
  • $\begingroup$ 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 $F_B$ . $\endgroup$ Commented Mar 15, 2022 at 15:13
  • $\begingroup$ 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. $\endgroup$
    – Andrea
    Commented Mar 15, 2022 at 15:51

1 Answer 1


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 $\sigma_x^2=\bar x$


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