Distribution of half-life from radioactive decay Suppose that $X$ measures the half-life of a radioactive element, with decay rate $\lambda$ (per unit of time).
Starting from a population of $N$ particle, I believe you can model the number of particles that decay over a unit of time as a $Poisson(N\lambda)$, and that the time until the first decay is detected is modeled using an $Exponential(N\lambda)$.
My question is: How can we model the distribution of the half-life $X$, i.e., starting from a population of $N$ particles, the amount of time it takes for $N/2$ of the particles to have decayed ?
Assuming the decays are independent from one another, it seems an Erlang distribution could be appropriate https://en.wikipedia.org/wiki/Erlang_distribution? Should I then use $f(x,N/2,N\lambda)$ (where the notation refers to the Wikipedia article)? If the half-life intuition is correct, it would seem $f(x,N/2,N\lambda)$ should then be invariant to the choice of $N$, but that's not obvious to me looking at the algebraic for of the Erlang's pdf. Is that correct? If so, there should be a single distribution $g = f(x,N/2,N\lambda)$ for all $N$ that models half-lifes?

You ask about the distribution of an order statistic $X_{(k)}$ of $N$ independent and identically distributed random variables $X_1, X_2, \ldots, X_N$ and $k=N/2$ when $N$ is even and $k=(N+1)/2$ when $N$ is odd.  Because this order statistic is the median when $N$ is odd and otherwise is a median, I will refer to it henceforth as the "median."
When the common distribution function $F$ has a density $F^\prime = f,$ the order statistic has a density, too, given by
$$f_{k;X}(x) = \binom{N}{k-1;1;N-k}\, F^{k-1}(x) f(x) (1-F(x))^{N-k}.$$
By choosing time units in which the half-life is $1$, the exponential distribution function is $F(x) = 1 - e^{-x},$ whence $f(x) = e^{-x}$ and therefore
$$f_{k;X}(x) = \frac{N!}{(k-1)!(N-k)!} e^{-(N-k+1)x}\left(1 - e^{-x}\right)^{k-1}.$$
One way to understand this is to express it in terms of $U = e^{-X}$ (which, incidentally, has a Uniform$[0,1]$ distribution).  The density of $U$ is
$$f_{k;U}(u) \ \propto\ u^{N-k}(1-u)^{k-1},$$
immediately recognizable as that of a Beta$(N+1-k, k)$ distribution.  Thus, $X_{(k)}$ is seen as the (negative) logarithm of a Beta variable.
Here, for example, are histograms of the median of $N=12$ Exponential variables (from 40,000 simulations) and its negative exponential on which, in red, graphs of $f_{k;X}$ and $f_{k;U}$ have been overlaid to demonstrate the correctness of this analysis.

At this point you can apply what is known about the Beta distribution to determine in great detail anything you like about the distribution of $X_{(k)},$ so I won't belabor the point.  Instead, consult the references, including the following CV threads:

*

*For more about the distributions of medians generally, see https://stats.stackexchange.com/a/160148/919.


*For a detailed analysis of how such distributions approach a Normal distribution as $N$ increases, see https://stats.stackexchange.com/a/86804/919.  Use this for accurate approximations when $N \gg 10$ or so.


*See https://stats.stackexchange.com/a/225990/919 for a sketch of how the initial formula for $f_k$ can be derived.
