What is the differential probability notation used in this paper by Einstein? Earlier today, I saw a discussion on Einstein's 1917 paper On The Quantum Theory of Radiation. Section 2 of the paper introduces three equations that I'm neither sure how they are derived nor what they're supposed to represent. 
For instance, section 2 defines the following equation (the other two are similar, but introduce $\rho$, a radiation density):
$$dW = A_m^ndt$$
$A_m^n$ is a constant, and $dt$ is a "very small" unit of time. On the other hand, $dW$ is said to be a probability that emission of radiation occurs during $dt$.
My first confusion is that the probability is represented using a differential $dW$. Earlier in the paper the same variable $W$ is also used to represent a density function. However, setting aside that confusion for a moment, since $A_m^n$ is a constant, and $t$ is presumably unbounded:
$$\int_{-\infty}^{\infty}A_m^ndt = \infty$$
The above should equal 1 for $dW$ to represent a distribution.
This makes me think I'm either missing some implicit assumptions, and/or Einstein is not calculating a traditional integral using Riemann sums.
According to this question, taking the differential of a probability density function occurs in stochastic calculus. Considering we are dealing with molecules in a gas, this makes sense. 
So is the above equation an example of describing a stochastic process? If so, is there a less-terse way in writing the equation that would also show more clearly that this is not an integral calculated using Riemann sums? If so, I do not have the background to understand this paper.
 A: Radiation from a material (neglecting the overall decrease in radioactive atoms) is a stochastic process in which in a given time interval every atom has a certain chance of decaying. If the time interval $\Delta t$ is short enough then there is a very small probability of having 1 decay and an extremely small of having two or more decays. Einsteins $dW$ denotes this probability of having 1 decay (and thus neglects the possibility of two or more). 
From temporal invariance it must be proportional to the length of
the time interval: $dW = a \; dt$. This only makes sense when $dW$ is small compared to 1. 
Another way to think of it is as the average number of decays in the interval $dt$. Chopping a (not so small) time interval $[0,T]$ into $n$ tiny intervals where Einsteins argument works allow you to show that e.g. $X$, the total number of decays is a random variable following a Poisson distribution: If  $\Delta t=T/n$ is small enough so that we can neglect multiple decays in each interval, then $p=a \Delta t=aT/n$ is the (approx) probability of one decay in each interval, and these decays may be assumed independent. Having $k$ decays in total is (approx) a binomial distribution (let $n\rightarrow \infty$ for fixed $k$):
$$ P(X=k)=\left( \begin{matrix} n\\k \end{matrix} \right) \left(\frac{aT}{n}\right)^k \left(1- \frac{aT}{n}\right)^{n-k} \rightarrow \frac{1}{k!} (aT)^k e^{-aT}$$
You may also derive an ODE for the probability of having no decays up to a certain time. Your integral (over some fixed 'long' period of time) represents the average total number of decays in that period. It may  not be interpreted as a probability anymore.
