An archaeologist has two old pieces of wood and shall decide which piece of wood is the oldest. Radioactivity from the pieces of wood are recorded by a counter. Number of registrations per unit time is Poisson distributed with rates 120 per hour for the oldest piece of wood and 150 per hour for the youngest piece of wood. The wood piece that provides one more count on the counter compared to the other, will be categorized as the youngest.

What is the probability that the archaeologist draws the wrong conclusion about which piece of wood is the oldest, when measured for 10 minutes?

  • A = 20 rate per 10 minute
  • B = 25 rate per 10 minute

The conclusion can be drawn wrong in two separate ways. This is a normal distribution. Either B can only let out 19 or less radioactive signals in the 10 minute period, or A can let out 26 or more radio active signals. Adding these two probabilities together I get 0.1151+0.0901 = 0.2052 or 20.52%. I am not sure if I have done this correctly, and confirmation is appreciated. I have used a different procedure than what is written in the answers, but the answer is roughly the same (off by a few decimals).

How long must the registration go on for the archaeologist to be 99% sure to draw right conclusion about which piece of wood is the oldest?

Even though I have the answer (and procedure) I do not understand this at all, and would very much like an explanation.

The answer is 1.59 hours.

  • $\begingroup$ I don't understand your derivation in the first part. It seems to me that you have one Poisson random variable $X$ with mean $20$ and another Poisson random variable $Y$ with mean $25$, and you're asking for $P(X>Y)$ (the probability that the one that "should" be smaller is in fact larger). This is just a sum over the joint pmf: $P(X>Y)=\sum_{n=0}^\infty \sum_{m=n+1}^\infty P(Y=n) P(X=m)$ (since they are independent). $\endgroup$ – Ian Jul 27 '16 at 0:00
  • $\begingroup$ Now you can replace $25$ and $20$ by $25t$ and $20t$, do the calculation again, set the result equal to $0.01$ and solve for $t$. The result will be the desired time for the second part, in units of 10 minutes. $\endgroup$ – Ian Jul 27 '16 at 0:00
  • $\begingroup$ Am I misunderstanding the model you are using? $\endgroup$ – Ian Jul 27 '16 at 0:00
  • $\begingroup$ @Ian am not sure. Because I am rather confused myself, and I am not sure if I have done the first part correctly. I will add a picture of my calculation tomorrow. I appreciate you trying to help :) $\endgroup$ – David Lund Jul 27 '16 at 0:06

If $X$ and $Y$ are independent Poisson with parameters $\lambda, \mu$ respectively, and $z \ge 0$, $$P(X - Y = z) = e^{-\lambda-\mu} (\lambda/\mu)^{z/2} I_{|z|}(2\sqrt{\lambda \mu})$$

where $I_z$ is the modified Bessel function of the first kind of order $z$.

For $z\ge 0$ this can be obtained from the series $$ e^{-\lambda-\mu} \sum_{y=0}^\infty \dfrac{\lambda^{y+z} \mu^y}{(y+z)!y!}$$ and similarly for $z\le 0$.

I don't know if there is a closed form for $P(X-Y>0)$.

In the problem at hand, I suspect it may be intended that you approximate the distribution of $X-Y$ by a normal distribution with the same mean and variance.

  • $\begingroup$ In case this could shed some light on the existence of a closed form: this is a Skellam distribution, and while there are bounds, no closed-form expression can be found on Wikipedia -- which is strong hint there is none. $\endgroup$ – Clement C. Jul 27 '16 at 0:52
  • $\begingroup$ You are right, I am supposed to do a normal distribution but I do not understand how I can do this. It doesn't make sense to me that when I have two independent events I can put them together in one formula somehow and then get the result. At the same time, I would have wanted to treat them differently, and then find the result, because that makes more sense to me but maybe that is not possible. I am going to google "sum over joint normal distribution" to see if i can get some clearance. $\endgroup$ – David Lund Jul 27 '16 at 10:34
  • $\begingroup$ @DavidLund Note that a Poisson r.v. $X$ with parameter $\lambda$ has mean $\lambda$ and variance $\lambda$. For large $\lambda$ this suggests that $\frac{X-\lambda}{\sqrt{\lambda}}$ is approximately $N(0,1)$ (and in fact this is the case). As it happens, $20$ is not all that large; the maximum error in this approximation in that case exceeds 0.05 (without any continuity correction; it appears to be below 0.02 if you use a continuity correction). But it improves as you increase the mean. $\endgroup$ – Ian Jul 27 '16 at 10:47
  • $\begingroup$ @DavidLund So in your case you have a bivariate normal distribution with mean $(20,25)$ and covariance $\begin{bmatrix} 20 & 0 \\ 0 & 25 \end{bmatrix}$. This has a known pdf; integrate that pdf over an appropriate region. $\endgroup$ – Ian Jul 27 '16 at 10:47
  • 1
    $\begingroup$ No need to bring in bivariate distributions: the sum of independent normal random variables is normal. All you need is $\mathbb E[X-Y] = \mathbb E[X] - \mathbb E[Y]$, and $\text{Var}(X-Y) = \text{Var}(X)+\text{Var}(Y)$ when $X$ and $Y$ are independent. $\endgroup$ – Robert Israel Jul 27 '16 at 15:33

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