# Radioactive decay, relation between binomial to expon. dist

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1. Jul 9, 2016

### laserbuddha

You can model the probability for radioactive decay as a Poisson distribution. This is the probability for radioactive decay within a specific time interval. (I probably got some of it wrong here).

P(k,μ)=λ^k⋅exp(-μ)/k!

Is there a way to use this formula to derive the other formula for radioactive decay?

N(t)=N0⋅exp(-λt)

The only thing I've found, is all connected to differential equation, but nothing about how it is connected to Poisson distribution.

2. Jul 9, 2016

Staff Emeritus
3. Jul 9, 2016

### laserbuddha

Yeah, I've read it. Don't know if you get my question.

What I'm wondering is, what is the reasoning (physics) or mathematical connection between these two formulas. I get the basic premise and reasoning for both formulas, but I don't get the relationship between the two.

P(k,μ)=λ^k⋅exp(-μ)/k!

What is the reasoning for endup with

N(t)=N0⋅exp(-λt)

4. Jul 9, 2016

### Ray Vickson

The answer to your question AS WRITTEN is no, you cannot use one formula to derive the other. The reason is that they are describing different aspects of the situation.

The second one has no randomness in it; it gives the average number of particles remaining in a sample (where "average" = average in the statistical sense, which is the mean of a probability distribution).

What do you think the first formula represents? In other words, the quantity $\lambda^k e^{-\mu}/k!$ represents the probability of some event; what IS that event? What is the relation between $\lambda$ and $\mu$, if any?

5. Jul 9, 2016

### laserbuddha

First of all, don't know if this is the right forum (someone moved this from the physics forum). So perhaps I need to give some background. Both functions are related to radioactive decay.

I've also written the first expression wrong, should be:
P(k,μ)=μ^k⋅exp(-μ)/k!

This is the probability for k atoms to decay, μ=np, n total number of atoms p probability of decay.

The second expressions is how many atoms are left, with regard to time. λ decay rate and N0 the number of atoms from start.

My speculation is that the two functions could be related somehow, e.g. something similar to how poisson distribution and exponential distribution are related.
From the first expression you get the number of atoms left, by e.g. calculating the number of decays for the mean. And this is for a specific time interval.
While the second function is related to the time between.

Don't know if any of this makes sense. I'm just interested to see if someone has seen something similar about this.

6. Jul 9, 2016

### Ray Vickson

Basically, the formula $P(k) = (\mu t)^k e^{-\mu t}/k!, \: k = 0, 1, 2, \ldots$ is supposed to be the probability that $k$ decays occur in a time interval of length $t$; here $\mu = n \lambda$ is the decay rate parameter when $n$ radioactive (un-decayed) particles are present and $\lambda$ is decay rate parameter for a single particle. That means that for a small time interval of length $\Delta t > 0$ a single un-decayed particle has decay probability of $\lambda \Delta t + O((\Delta t)^2) \approx \lambda \Delta t$ to first order in small $\Delta t$. The probability that one of the $n$ particles decays during $\Delta t$ is $n \lambda \Delta t$ to first order in $\Delta t$. By making $\Delta t$ small enough, the probability of more than one decay during $\Delta t$ is negligible, it being of higher than first order in $\Delta t$.

However, all that is merely an approximation; it may be a good one in many cases, but in principle is not exact. Why? Well, after the first decay, you will be left with $n-1$ particles, so the relevant decay rate for the next decay has a slightly smaller rate parameter $(n-1) \lambda$. Then, after the second decay the new rate is $(n-2) \lambda$, etc.

Working out the exact expression for the probability of $N-k$ decays in an interval of length $t$ has been done be several authors; an example is in
wwwhome.math.utwente.nl/~boucherierj/onderwijs/158052/158052sheetshc2.ppt
(which, unfortunately, a powerpoint slide show). The formula for the probability there are $j$ particles left at time $t$, given $N$ particles at time 0, is given on slide 8 (which uses the symbol $\mu$ instead of what I called $\lambda$). You will see that the distribution is NOT Poisson: it is binomial with parameters $N$ and $p = e^{-\lambda t}$ (I use $\lambda$ instead of the author's $\mu$). So, in the large-$N$ small $\lambda t$ limit, with $\mu = \lambda t N$ held constant, we obtain essentially the Poisson distribution for $k=N-j$ itself; that is, the probabilities for the number of decays follow pretty closely to a Poisson distribution! However, that is NOT EXACT, just very often a really good approximation (but sometime not so good at all).

What about the other issue: that of times? In a pure Poisson process with constant rate $\mu$ the times between successive events (individual decays) are independent and identically-distributed random variables with exponential distribution of rate $\mu$. That means that If $T_i$ is the time between the $i$th and $(i+1)$th decay, then
$$P\{ t < T_i < t + \Delta t \} = \mu e^{- \mu t} \: \Delta t$$
to first order in small $\Delta t$. In other words, the probability density function of $T_i$ is $f(t) = \mu e^{-\mu t}$ for $t > 0$ and is $0$ for $t < 0$. The expected time to the next event is $E(T_i) = 1/\mu$.

Why does that work? Well, look at the Poisson probability of no decay in time interval $0 \longrightarrow t$; that is $P_0(t) = e^{=\mu t}$ (obtained by putting $k = 0$ in the formula $(\mu t)^k e^{-\mu t}/k!$). However, 0 arrivals occur between times 0 and $t$ if and only if the first arrival occurs after $t$, so
$$P_0(t) = P\{ T_i > t \} = e^{-\mu t}.$$
That implies that the probability density function of $T_i$ is $f(t) = -\frac{d}{dt} P \{ T_i > t \} = \mu e^{-\mu t}$. Voila! The exponential distribution!

To prove that the successive inter-event times are independent takes a bit more work and must look at some additional properties of Poisson distribution, but it all works out eventually.

However: none of that applies exactly to radioactive decay! Again, the reason is the variable rates involved. The time to the first decay is, indeed, exponentially distributed with rate $\lambda n$ and expected value $1/ \lambda n$. However, after the first decay the time to the next decay is exponentially distributed with rate $\lambda (n-1)$, etc. The two times are, indeed, independent, but are no longer identically distributed.

OK: I think I have said enough for now.