# Why do we adjust the drift in the geometric brownian motion

I am building a monte carlo based on the GMB, and I am having a hard time understanding why we subtract 1/2 variance from the drift. If I have a drift of 12% and a volatility of 50%, that would give me "mean" drift of -.24% based on this. That doesn't make sense to me. If I am expecting the asset to drift higher then doesn't my mean drift need to be positive? I am also assuming that drift and expected return are synonomous in this context. Please correct me if drift and expected/required return are not the same, and if so then what exactly is "drift". Not a mathematician, so factor that into your answers and thank you for the help!

$$dS_t = \mu S_t dt + \sigma S_t dW_t$$

with the solution to the SDE above given as:

$$S_t = S_0 exp((\mu - \sigma^2/2)t + \sigma W_t)$$

The drift is the $$\mu$$ term above and I guess when you say "mean drift" you are referring to $$\mu - \sigma^2/2$$? The "mean drift" is actually 0.5% from my calculations but that is besides the point.

I originally wanted to write up my own answer but I think Geometric Brownian motion - Volatility Interpretation (in the drift term) does it much better.

• Thanks, that was helpful! Also you are correct, the example I gave is negative .5%. I had rounded up the inputs in my model. Apologies for the confusion and thanks again.
– John
Commented Oct 11, 2023 at 19:12

Following the notation of Lars Tyge Nielsen, stock prices $$S_t$$ at time t follow a lognormal distribution in the Black Scholes world. At time 0, $$log(S_t) \sim \mathcal{N}(log(S_0) +(\mu -\sigma^2/2)t, \sigma^2t).$$

The continuously compounded rate of return over an interval $$[0,t]$$ is $$\frac{log(S_t)-log(S_0)}{t}.$$ Given the current stock price $$S$$, this rate follows the normal distribution $$\mathcal{N}((\mu -\sigma^2/2),\sigma^2/t).$$ In plain English, its logarithm is normally distributed with mean $$(\mu -\sigma^2/2)$$ and variance $$\sigma^2/t$$. As $$t$$ grows, variance decreases towards zero, whereas the mean of the rate of return does not depend on time $$t$$. However, the mean depends on volatility. The chart below shows this relationship for high vol ($$t=1$$).

The reason you get a negative value (for high IV) all pins down to this answer.

All else equal, increasing vol results in the distribution trying to extend itself on both sides of the definition domain but hits a boundary at zero, where probability accumulates (probability mass).

Intuitively, this indeed sounds a bit odd. In my opinion, it is best answered by looking at the probability density function (PDF) and cumulative density function (CDF). I ran some code in Julia to demo this:

• The higher σ, the more the global maximum of the probability density function (the mode) shifts towards the lower bound of the lognormal distribution.

• The cumulated distribution function (CDF) shows the increase in the probability of $$S_T$$ being very small.

• Last but not least, the effect on the mean return can also be demonstrated by plotting the PDF of the normal distribution.

The drift in Black Scholes is in the risk-neutral measure $$\mathbb Q$$. Details can be found in this answer, or here.