The classic treatment of GBM for asset pricing leads to a point where eventually one gets a solution that is the same as assuming an underlying arithmetic Brownian motion, $X_t$, which has (over unit time) drift equal to $\mu-\frac{\sigma^2}{2}$and a random term $\sigma B_t$, then getting $S_t=S_0e^{X_t}$, and prices are log-normally distributed.
When viewed this way, it seems any path $X_t$ is the "total return" from $t=0$ until time $T$. It is formed by taking the drift/return for one period of time, Adjusted for the "offset" that arises from the random term after taking $e^x$ (Due to Jensen's inequality) and dividing time into more and more tiny increments where each one has a deterministic size $\frac{\mu -\frac{\sigma^2}{2}}{n}$, added to infinitessimal and independent normally distributed random increments to create the path from $X_0$ to $X_t$.
I am trying to get this to reconcile with a discrete intuition about returns and compounding. If we define $r=\mu -\frac{\sigma^2}{2}$for one period, one is adding an increment each time that is distributed as $N(\frac{r}{n}, \frac{\sigma^2}{n})$. In price space, this is like multiplying the price times $(1+\frac{r}{n})$ (plus the random term, of course).
Now, as $n\to \infty$, is it true that (1) these additive micro-returns in $X$-space drive $X_t$ such that $e^{X_t}$ gives the same result as applying the multiplications in price-space; and (2) one gets log-normally distributed prices?
The sort of key question I am trying to get at is this: with varying returns, where does the log-normality come into prices, and how is that squared with compounding in the limit with returns in price space?
We know that $e^{X_t}$ is always non-negative. Yet using an arithmetic BM and then taking the exponent seems like the same actions as taking infinitessimally-sized, normally distributed returns $r_t$ and calculating, as $n\to \infty$,
$$S_t=S_0\prod (1+r_t)$$ (plus, of course, the randomness).
If that is the case, then the log-normality comes not from $\mu$ when we take $e^\mu...$, since that is just continuous compounding if $\sigma=0$ - it comes from the random term (hence why we need to subtract $\frac{\sigma^2}{2}$ to maintain alignment with discrete/deterministic compounding calculations.)
The problem I see with the equivalence is that the approach I am trying to use in price-space,chaining the multiplication of tiny amounts of $(1+r_t)$ where each of the $r_t$ is normally distributed with a tiny r and a tiny $\sigma$ in the limit, still has the chance of having a negative return large enough that it makes prices go negative. The odds of this are tiny: as $n$ grows, $(1+N(\cdot ,\cdot))$ ends up with a mean of$ (1+\epsilon _1)$ and variance of $\epsilon_2^2$, where both $\epsilon _1$ and $\epsilon_2 <<1$ - so the chance of a negative return over 100% becomes vanishingly small.
My suspicion is one could argue that $P(1+N(\frac{r}{n}, \frac{\sigma^2}{n})<0)=0$ as $n\to \infty$ rapidly enough that it is OK.