The level of radiation in the control room of a nuclear reactor is to be automatically monitored by a Geiger counter. The monitoring device works as follows:

Every tenth minute the number (frequency) of "clicks" occurring in t seconds is counted automatically. A scheme is required such that if the frequency exceeds a number c an alarm will sound. The scheme should have the following properties:

  1. If the number of clicks per second is less than 4, there should be only l chance in 500 that the alarm will sound.
  2. If the number reaches 16, there should be only about 1 chance in 500 that the alarm will not sound.

What values should be chosen for t and c? Hint: Utilize the Poisson distribution and normal approximation.

My thoughts so far:

The number of clicks we measure, ${N}$, in ${t}$ seconds can be assumed to be approximately Poisson with true rate ${\eta t}$. With time ${t}$ being a constant, the scheme relates to ${P(\frac{N}{t}>c)=P(N >ct)}$. Hence we need to choose constants ${c}$ and ${t}$ such that the alarm sounds with probability ${\frac{1}{500}}$ when the true rate per second ${\eta}$ is less than 4, and the alarm should sound with probability ${\frac{499}{500}}$ when the true rate per second is greater than 16. Thus we need ${P(N > ct|\eta<4)=1/500}$ and ${P(N > ct |\eta \ge 16)=499/500}$. We can approximate ${N}$ as ${Normal(\eta t, \sqrt{\eta t})}$. In the first case we have ${P(Z > \frac{ct-\eta t}{\sqrt{\eta t}})=0.002}$ and ${\eta < 4}$. Because we're only given a bound for ${\eta}$ rather than the true value for ${\eta}$, I'm at a loss of what to do next. Am I going about this the right way?

  • $\begingroup$ Incidentally, the way the numbers are given, and the nice way they work out, make me think that condition $1$ should really read "if the number of clicks per second equals $4$..." (to match the wording of "reaches" in condition $2$). $\endgroup$ – Brian Tung Mar 12 at 17:10

Basic approach. The symmetry between the probabilities of triggering and not triggering the alarm suggest a ratio between $c$ and $t$ of $m$ to $1$ (it is some kind of mean of $4$ and $16$). That is to say, we set $c = mt$.

How does this work? Suppose the actual rate is $4$ clicks per second. Then over a period of $t$ seconds, the distribution of the number of clicks is Poisson with mean $4t$. For large-ish means $\lambda$, the Poisson distribution is roughly normal with variance $\lambda$ (and therefore standard deviation $\sqrt\lambda$); in this case, that means a mean of $4t$ and a standard deviation of $2\sqrt{t}$. The threshold of $c = mt$ would then be $\frac{(m-4)t}{2\sqrt{t}} = \frac{m-4}{2}\sqrt{t}$ standard deviations above the mean.

Similarly, for an actual rate of $16$ clicks per second, a threshold of $c = mt$ would be $\frac{(16-m)t}{4\sqrt{t}} = \frac{16-m}{4}\sqrt{t}$ standard deviations below the mean. By symmetry, those two $z$-scores should be of equal magnitude. Find the value of $m$ that satisfies this.

Lastly, find the value of $t$ for which the common $z$-score gives the appropriate probability $1/500$ for the mass in the tail.

  • 1
    $\begingroup$ Incidentally, I am not sure that for the numbers you actually get, the normal approximation is valid enough to support a tail mass as small as $1/500$. (I don't know it isn't; I'm just not sure.) But they told you to use the normal approximation, so... $\endgroup$ – Brian Tung Mar 12 at 17:43
  • $\begingroup$ The book says the true values (without normal approximation) are c = 20 and t=2.25. Using the normal approximation here, we get c=16.6 and t=2.07. Based on how close these are to the true values I suspect that you are correct that the assumption should be the true mean is equal to 4 or equal to 16 in each of the two cases and that your apx method is spot on. Thanks for the help! $\endgroup$ – James Bender Mar 12 at 18:34
  • $\begingroup$ @JamesBender: Yeah, it occurs to me that you probably need to introduce some kind of continuity correction. I didn't address that, but it's probably not too difficult. $\endgroup$ – Brian Tung Mar 12 at 19:40

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