Radio-active decay: probability of N particles after K halflives?

[kells]

Hi guys,

I'm working on a problem involving electrical signals which have components which appear to be decaying exponentially.

I think my problem is analogous to radio-active decay. I would like to find the probability that there are N particles remaining after K half-lives. To clarify, for example if a population started out with 100 particles and a half-life of 1 hour, what is the probability that after 2 hours there were 60 particles remaining?

I started out looking at the exponential and binomial distributions but I don't think they're what I'm looking for. Any suggestions would be gratefully received.

Thanks,
James

 

[SW VandeCarr]

Hi guys,

I'm working on a problem involving electrical signals which have components which appear to be decaying exponentially.

I think my problem is analogous to radio-active decay. I would like to find the probability that there are N particles remaining after K half-lives. To clarify, for example if a population started out with 100 particles and a half-life of 1 hour, what is the probability that after 2 hours there were 60 particles remaining?

I started out looking at the exponential and binomial distributions but I don't think they're what I'm looking for. Any suggestions would be gratefully received.

Thanks,
James

Since the half-life is one hour and your time period is two hours, you would expect the amount remaining in the original state, out of 100 particles, to be 25 particles. If you use the normal distribution to calculate the probability of observing 60 given the expectation of 25, you would need to know the variance or more typically the standard deviation. However, it seems with such a large difference, the probability is likely close to zero. It seems clear that the example or the analogy is incorrect.

If you don't know the standard deviation, you can use Chebyshev's Theorem to estimate the probability.

http://www.statisticshowto.com/articles/category/normal-distribution/

 

[kells]

The example I give is an extreme case, I'm just trying to quantify the probability that the signal intensity at a given time is due to the exponential decay or an underlying signal modulated on top of it.
I'm not sure that I can assume the probability of a number of particles at a given time is normally distributed so I didn't want to make the assumption that it was, can anyone confirm that this is the case?

 

[SW VandeCarr]

The example I give is an extreme case, I'm just trying to quantify the probability that the signal intensity at a given time is due to the exponential decay or an underlying signal modulated on top of it.
I'm not sure that I can assume the probability of a number of particles at a given time is normally distributed so I didn't want to make the assumption that it was, can anyone confirm that this is the case?

In physics radioactive decay is a Poisson process, but obviously involves very large numbers of particles, with test samples in the order of $10^{22}$ to $10^{23}$. So for practical purposes, it's a deterministic calculation subject to the uncertainty in the precision of the half-life, especially for more stable isotopes: $N_t = N_0 e^{-0.693 t/t_{1/2}}$.

For a small number of particles such as you described, there would be random variation around the expected mean. Because it is a Poisson process you should be able to use the one parameter Poisson distribution $\lambda$ where $\sqrt {\lambda}$ would be a reasonable estimate of the standard deviation (sd).

For an expectation of 25 the sd would be 5. Three sd from the mean would contain approximately 99% of the variation. You can see that 60 is seven sd from the mean, so the probability is effectively zero.

Note for a small real world sample there might be $10^{22}$ particles, so the sd would be $10^{11}$. This seems like a large number until you compare it to the original quantity.

I'm not a physicist, but nuclear decay is a random Poisson process (ie constant rate of decay), so this should be a reasonable approximation for any such process. However, I can't speak to your application.