Counting measurements in the presence of background

As an example of a different kind of systematic effect, let us think of counting experiments in the presence of background. For example we are searching for a new particle, we make some selection cuts and count $x$ events. But we also expect an average number of background events $\lambda_{B_\circ}\pm\sigma_B$, where $\sigma_B$ is the standard uncertainty of $\lambda_{B_\circ}$, not to be confused with $\sqrt{\lambda_{B_\circ}}$. What can we say about $\lambda_S$, the true value of the average number associated with the signal? First we will treat the case in which the determination of the expected number of background events is well known ( $\sigma_B/\lambda_{B_\circ}\ll 1$), and then the general case.

:

The true value of the sum of signal and background is $\lambda=\lambda_S+\lambda_{B_\circ}$. The likelihood is

$$P(x\,\vert\,\lambda) =\frac{e^{-\lambda}\lambda^x}{x!}\,.$$ (5.87)

Applying Bayes' theorem we have

$$f(\lambda_S\,\vert\,x,\lambda_{B_\circ}) = \frac{ e^{-(\lambda_{B_\circ}+\lambda_S)} \,(\lambda_{B_\circ}+\l... ...bda_S)} \,(\lambda_{B_\circ}+\lambda_S)^x \,f_\circ(\lambda_S)\,d\lambda_S }\,.$$ (5.88)

Choosing again $f_\circ(\lambda_S)$ uniform (in a reasonable interval) this gets simplified. The integral in the denominator can be calculated easily by parts and the final result is

$$f(\lambda_S\,\vert\,x,\lambda_{B_\circ}) = \frac{e^{-\lambda_S}\,(\lambda_{B_\circ}+\lambda_S)^x} {x!\,\sum_{n=0}^x\frac{\lambda_{B_\circ}^n}{n!}},$$ (5.89)

$$F(\lambda_S\,\vert\,x,\lambda_{B_\circ}) = 1 - \frac{e^{-\lambda_S}\, \sum_{n=0}^x\frac{(\lambda_{B_\circ}+\lambda_S)^n}{n!}} {\sum_{n=0}^x\frac{\lambda_{B_\circ}^n}{n!}}\,.$$ (5.90)

From ()-() it is possible to calculate in the usual way the best estimate and the credibility intervals of $\lambda_S$. Two particular cases are of interest:

• If $\lambda_{B_\circ}=0$ then formulae ()-() are recovered. In such a case one measured count is enough to claim for a signal (if somebody is willing to believe that really $\lambda_{B_\circ}=0$ without any uncertainty$\ldots$).
• If $x=0$ then

  $$f(\lambda\,\vert\,x,\lambda_{B_\circ}) = e^{-\lambda_S}\,,$$ (5.91)

  
  
  independently of $\lambda_{B_\circ}$. This result is not really obvious.

:

In the general case, the true value of the average number of background events $\lambda_B$ is unknown. We only known that it is distributed around $\lambda_{B_\circ}$ with standard deviation $\sigma_B$ and probability density function $g(\lambda_B)$, not necessarily a Gaussian. What changes with respect to the previous case is the initial distribution, now a joint function of $\lambda_S$ and of $\lambda_B$. Assuming $\lambda_B$ and $\lambda_S$ independent the prior density function is

$$f_\circ(\lambda_S,\lambda_B)=f_\circ(\lambda_S)\,g_\circ(\lambda_B)\,.$$ (5.92)

We leave $f_\circ$ in the form of a joint distribution to indicate that the result we shall get is the most general for this kind of problem. The likelihood, on the other hand, remains the same as in the previous example. The inference of $\lambda_S$ is done in the usual way, applying Bayes' theorem and marginalizing with respect to $\lambda_S$:

$$f(\lambda_S\,\vert\,x,g_\circ(\lambda_B)) = \frac{\int e^{-(\lamb... ...bda_S)^x \,f_\circ(\lambda_S,\lambda_B)\,\rm {d}\lambda_S\,\rm {d}\lambda_B}\,.$$ (5.93)

The previous case [formula ()] is recovered if the only value allowed for $\lambda_B$ is $\lambda_{B_\circ}$ and $f_\circ(\lambda_S)$ is uniform:

$$f_\circ(\lambda_S,\lambda_B) = k\,\delta(\lambda_B-\lambda_{B_\circ})\,.$$ (5.94)