How does a Poisson process generate the exponential decay?

The exponential decay is derived by

\displaystyle \frac{dN}{dt} = - k N \rightarrow N(t) = N_0 \exp\left(-kt\right) = N_0 2^{- t / \tau}, ~ k = \frac{\log(2)}{\tau}

which means the rate of the loss of the number of nuclei is proportional to the number of nuclei. and \tau is the half-life.

This is a phenomenological and macroscopic equation that tells us nothing about an individual decay.

Can we have a microscopic derivation that, assumes the decay of an individual nucleus following a random distribution?

From this post, we derived the distribution for the differences in a list of sequential random numbers. The key idea is the probability of having a decay at exactly time t - \delta t to t, denote as f(t, \delta t) , is

\displaystyle f(t, \delta t) = P(t-\delta t, 0)P(\delta t, 1)

where P(t,k) is the probability of having k decay within (time) length t, which is a Poisson distribution.

\displaystyle P(t, k) = \frac{t^k}{k!} \exp(-t)

\displaystyle f(t, \delta t) = \frac{(t-\delta t)^0}{0!} \exp(-(t-\delta t)) \frac{\delta t}{1!} \exp(-\delta t) = \exp(-t) \delta t

Now, the total probability of all possible decay within time T, assume not more than 1 decay can happen in \delta t ,

\displaystyle F(T) = \sum_{\delta t \rightarrow 0}^{t < T} f(t, \delta t) = \int_0^{T} \exp(-t) dt = 1- \exp(-T)

Thus, the number of nucleus not decay or survived within time T is

\displaystyle N(t) = N_0 (1 - F(t)) = N_0 \exp(-t) = N_0 2^{-t / \tau}