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.


2 Answers 2


I think the difference comes in that using returns and compounding is just a good approximation for lognormality, in the sense that it's the first term of the Taylor expansion

$\exp({\log(S_0) + r}) = S_0 + S_0 r + \frac{1}{2} S_0 r^2 + \dots $

The first order approximation $\exp({\log(S_0) + r)} \simeq S_0 (1 + r) $ only holds when $r$ is close to $0$, and this is why your intuition about $S$ becoming negative fails.


Juan's answer (which I have upvoted) is the important solution here, but there are a few other interesting points.

Mathematically, when we construct geometric brownian motion (GBM), we begin by constructing a brownian motion (BM) and then exponentiating it. So we never run into a product of $1+r$ terms.

Construction of brownian motion does not exclusively use forward-stepping increments. Instead, it starts with some coarse increments and then successively refines them using the brownian bridge.

Say we have generated standard brownian motion values $X_t$ for times $t_k < t_{k+1}$. Then setting $\tau=\frac12(t_k + t_{k+1})$ we choose a random value for $X_\tau$ as

$$ X_\tau \sim N\left(\frac12(X_{t_k} + X_{t_{k+1}}), \frac12( t_{k+1} -t_k ) \right) $$

Finally, we compute $S_\tau = \exp(X_\tau)$.

This process allows us to generate a BM or GBM approximation of arbitrary refinement, and converges in the limit to a true BM or GBM.

  • $\begingroup$ Just realized this myself recently. A GBM is actually solved by $Y=e^{X}$, where $e^X$ has $X~(N(\mu, \sigma^2))$ The question is as much about the bounds or math/numerical methods applied in X-space, where we have lots of instructions about working means, variances, etc. in X-space (hint: the product of two lognormals is also lognormal, since the sum of two normals is also normal. In Y-space). however, this looks like $e^{X_1 t}e^{X_2 t}=e^{(X_1+X_2)t}$, and since the X's are normal under addition, there is 1:1 with multiplication in lognormal space. $\endgroup$
    – eSurfsnake
    Jun 12, 2018 at 5:10

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