How Does the Rate of False Signals Change With Three Sensors?

[McCoy13]

Homework Statement

Suppose you have two sensors with rates of noise $R_{A}^{noise}$ and $R_{B}^{noise}$. Suppose an erroneous signal occurs if the two sensors have a noise trigger within time $\tau$ of each other. Show that the rate of false signals is $R_{AB}^{noise}=2R_{A}^{noise}R_{B}^{noise}\tau$. How does this change if a third sensor is introduced?

Homework Equations

N/A

The Attempt at a Solution

I will now drop the noise superscripts for convenience. Also note that the rates above are constants, and do not vary in time.

The probability that each sensor has a false signal in a time dt is obviously R*dt. So the probability that they both fire ought to be $R_{A}R_{B}dtds$. We are interested in number of counts per time, so we will need to take an integral to bring probability to counts and a time derivative to get a rate.

$$R_{AB}=\frac{d}{dt}\int^{\tau}_{-\tau}R_{A}R_{B}dsdt = R_{A}R_{B}\int^{\tau}_{-\tau}ds = R_{A}R_{B}2\tau$$

So for three sensors I do the analogous thing, presuming I got the above right. Here's the catch - I'm not sure what do with the integral limits.

$$R_{ABC}=\frac{d}{dt}\int \int R_{A}R_{B}R_{C}dudsdt$$

My only guess is that the first integral is integrated to the next integral's variable.

$$R_{ABC}=\frac{d}{dt}\int^{\tau}_{-\tau} \int^{s}_{-s} R_{A}R_{B}R_{C}dudsdt = R_{A}R_{B}R_{C}\frac{d}{dt}\int^{\tau}_{-\tau}2sdsdt = R_{A}R_{B}R_{C}[s^{2}]^{\tau}_{-\tau} = 0$$

This is obviously not the correct answer. Any help would be appreciated.

 

[McCoy13]

Maybe I didn't really do the analogous thing. I still have 3 pure numbers with $R_{A}dtR_{B}dsR_{C}du$, so by dimensional analysis I still only need one integral and one derivative, but what happens to the remaining differential time?