Deriving nuclear decay law from exponential waiting time. I was trying to derive the decay law $N(t)=Ne^{-\lambda t}$ starting from exponential waiting time for a nucleus to decay.
The probability that a nucleus doesn't decay in time $t$ is $e^{-\lambda t}$.
So for a sample of $N$ nuclei, the probability for $n$ decays in time $t$ is,
$$P_t(n)=\binom{N}{n}(1-e^{-\lambda t})^n(e^{-\lambda t})^{N-n}$$
I was expecting this to be strongly peaked at $N-Ne^{-\lambda t}$ as $N$ becomes large.
Since this is just  a binomial distribution with mean $N(1-e^{-\lambda t})$ and variance $Ne^{-\lambda t}(1-e^{-\lambda t})$, the mean looks correct to me but the variance does not as I was expecting it to decrease with $N$.
 A: The law $$N(t) = N_0 e^{-\lambda t}$$ represents the mean amount of nuclei remaining at time $t$, given the initial amount $N_0$ at time $t = 0$ and decay intensity $\lambda$ events per unit of time.
Therefore, the fact that the decay process is well-approximated by a binomial model with $N_0$ nuclei, each of which has "independent" probability $p = 1 - e^{-\lambda t}$ of decaying by time $t$, immediately leads to the desired relationship:  specifically, if $X(t)$ is a random variable that counts the number of decays by time $t$, then $$X(t) \sim \operatorname{Binomial}(n = N_0, p = 1 - e^{-\lambda t}),$$ and while it is true that $$\Pr[X(t) = x] = \binom{N_0}{x} (1 - e^{-\lambda t})^x (e^{-\lambda t})^{N_0 - x},$$ what we want is the expected value of the random number of decays $\operatorname{E}[X(t)]$ observed by time $t$, which is  $$\operatorname{E}[X(t)] = N_0 p = N_0 (1 - e^{-\lambda t}).$$  This gives us the mean number of nuclei remaining at time $t$
$$\operatorname{E}[N_0 - X(t)] = N_0 e^{-\lambda t}.$$
As for the variance, there is no reason to expect that it will decrease as $N_0$ increases, since the more nuclei you have to begin with, the more variability you can expect from the decay process.  For instance, if you only had $1$ nucleus, then your outcomes can only ever be $1$ remaining, or $0$ remaining.  That's not much variation, compared to say, $N_0 = 10^{10}$ nuclei, and at any given time $t$, you can have many different possible values for the number of decays $X(t)$.
But if we were talking about the proportion of nuclei remaining at time $t$, which is
$$Y(t) = 1 - \frac{X(t)}{N_0},$$ then this indeed has decreasing variance with increasing $N_0$, since we must always have $$0 \le Y(t) \le 1$$ and in particular,
$$\operatorname{Var}[Y(t)] = \frac{\operatorname{Var}[X(t)]}{N_0^2} = \frac{N_0 p (1-p)}{N_0^2} = \frac{e^{-\lambda t}(1-e^{-\lambda t})}{N_0},$$ which for a fixed $\lambda$ and $t$, is a strictly decreasing function of $N_0$.
