Poisson background on the observed number of `trials' and of `successes'

Let us know move to problem b) of the introduction. Again, we consider only the background parameters are well known, and refer to the previous subsection for treating their uncertainty. To summarize, that is what we assume to know with certainty:

$n$

: the total observed numbers of `objects', $n_s$ of which are due to signal and $n_b$ to background; but these two numbers are not directly observable and can only be inferred;

$x$

: the total observed numbers of the `objects' of the subclass of interest, sum of the unobservable $x_s$ and $x_b$;

$\lambda_b$

: the expected number of background objects;

$p_b$

: the expected proportion of successes due to the background events.

As we discussed in the introduction, we are interested in inferring the number of signal objects $n_s$, as well as the parameter $p_s$ of the `signal'. We need then to build a likelihood that connects the observed numbers to all quantities we want to infer. Therefore we need to calculate the probability function $f(x\,\vert\,n,\ n_s,\, p_s,\, \lambda_b,\, p_b)$.

Let us first calculate the probability function $f(x\,\vert\,n_s,\,p_s\,n_b,\,p_b)$ that depends on the unobservable $n_s$ and $n_b$. This is the probability function of the sum of two binomial variables:

$$\hspace{-5mm}f_{2{\cal B}}(x\,\vert\,n_s,\,p_s\,n_b,\,p_b) \! = \sum_{x_s,\,x_b} \delta_{x,\,x_s+x_b}\, f(x_s\,\vert\,{\cal B}_{n_s,\,p_s}) \cdot f(x_b\,\vert\,{\cal B}_{n_b,\,p_b})\,,$$ (35)

where $x_s$ ranges between $0$ and $n_s$, and $x_b$ ranges between $0$ and $n_b$. $x$ can vary between 0 and $n_s+n_b$, has expected value $\mbox{E}(x)=n_s\,p_s+n_b\,p_b$ and variance $\mbox{Var}(x) = n_s\,p_s\,(1-p_s)+n_b\,p_b\,(1-p_b)$. As for Eq. (32), we need to evaluate Eq. (35) only for the observed number of successes. Contrary to the implicit convention within this paper to use the same symbol $f(\cdot)$ meaning different probability functions and pdf's, we name Eq. (35) $f_{2{\cal B}}$ for later convenience.

In order to obtain the general likelihood we need, two observations are in order:

• Since $x$ depends from $\lambda$ only via $n_b$, then $f(x\,\vert\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b)$ is equal to $f_{2{\cal B}}(x\,\vert\,n_s,\,p_s\,n_b,\,p_b)$.
• The likelihood that depends also on $n$ can obtained from
  $f(x\,\vert\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b)$ by the following reasoning:
  • if $n=n_s+n_b$, then
    
    $$f(x\,\vert\,n,\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b) = f(x\,\vert\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b)\,;$$
    
    

  • else
    
    $$f(x\,\vert\,n,\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b) = 0\,.$$
    
    

It follows that

$$f(x\,\vert\,n,\,n_s,\,p_s,\,n_b,\,p_b,\,\lambda_b) = f(x\,\vert\,n_s,\,p_s\,n_b,\,p_b,\,\lambda_b) \,\delta_{n,\,n_s+n_b}$$ (36)

$$= f_{2{\cal B}}(x\,\vert\,n_s,\,p_s\,n_b,\,p_b) \,\delta_{n,\,n_s+n_b}\,.$$ (37)

At this point we get rid of $n_b$ in the conditions, taking account its possible values and their probabilities, given $\lambda_b$:

$$\hspace{-13mm}f(x\,\vert\,n,\,n_s,\,p_s,\,p_b,\,\lambda_b) = \sum_{n_b} f(x\,\vert\,n,\,n_s,\,p_s,\,n_b,\,p_b,\,\lambda_b) \, f(n_b\,\vert\,{\cal P}_{\lambda_b}) \,,$$ (38)

i.e.

$$\hspace{-7mm}f(x\,\vert\,n,\,n_s,\,p_s,\,p_b,\,\lambda_b) = \sum_{n_b} f_{2{\cal B}}(x\,\vert\,n_s,\,p_s\,n_b,\,p_b) \, f(n_b\,\vert\,{\cal P}_{\lambda_b})\,\delta_{n,\,n_s+n_b}\,,$$ (39)

where $n_b$ ranges between 0 and $x$, due to the $\delta_{n,\,n_s+n_b}$ condition. Finally, we can use Eq. (39) in Bayes theorem to infer $n_s$ and $p_s$:

$$f(n_s,\,p_s\,\vert\,x,\,n,\,\lambda_b,\,p_b) \propto f(x\,\vert\,n,\,n_s,\,p_s,\,p_b,\,\lambda_b) \, f_0(n_s,\,p_s)$$ (40)

$$f(p_s\,\vert\,x,\,n,\,\lambda_b,\,p_b) = \sum_{n_s}f(n_s,\,p_s\,\vert\,x,\,n,\,\lambda_b,\,p_b)$$ (41)

$$f(n_s\,\vert\,x,\,n,\,\lambda_b,\,p_b) = \int f(n_s,\,p_s\,\vert\,x,\,n,\,\lambda_b,\,p_b) \, \mbox{d}p_s$$ (42)

$$f(p_s\,\vert\,x,\,n,\,n_s,\,\lambda_b,\,p_b) = \frac{f(n_s,\,p_s\,\vert\,x,\,n,\,\lambda_b,\,p_b)} {f(n_s\,\vert\,x,\,n,\,\lambda_b,\,p_b)}\,.$$ (43)

We give now some numerical examples. For simplicity (and because we are not thinking to a specific physical case) we take uniform priors, i.e. $f_0(n_s,\,p_s) = \mbox{\it const}$. We refer to section 3.1 for an extensive discussion on prior and on critical `frontier' cases.