I have a continuum of objects, all affected by Brownian motion with drift $\mu$ and variance $\sigma^2$. I want to find the mass of those that are moving upwards, which I will make more precise in the discrete-time approximation that follows.
Discretize a grid over a with step size $\Delta_a = \bar a - \underline a$. Time period is $\Delta_t = (\Delta a/\sigma)^2$. In the interior, $a$ increases with probability $1/2 (1 + \mu \Delta_a/\sigma^2)$, and decreases with probability $1/2 (1 - \mu \Delta_a/\sigma^2)$. This yields an expected value of $\mu \Delta_t$ with second moment $\sigma^2 \Delta_t$.
Now, what I am interested in, in discrete time, is separating the "upward-movers" from the "downward-movers", and compute the "speed" ($\Delta_a/\Delta_t$) for each of these.
For upward-movers, we have how much you move upwards X prob of moving upwards per unit of time:
$$\Delta_a \cdot \left[\frac{1}{2}(1 + \frac{\mu\Delta_a}{\sigma^2}) \right]/\Delta_t\\ = \frac{1}{2}\left[\sigma^2\frac{\Delta_t}{\Delta_a} + \frac{\mu\Delta_a^2}{\sigma^2})\right]/\Delta_t\\ = \frac{1}{2}\left[\sigma^2\frac{\Delta_t}{\Delta_a} + \mu\Delta_t\right]/\Delta_t\\ = \frac{1}{2}\left[ \sigma^2\frac{1}{\Delta_a} + \mu\right] $$
Now, when $\Delta_t \to 0$, we have $\Delta_a \to 0$, so the first term explodes. What am I missing?