# What are $\mu$ and $a$ in $\mu = a + \frac{\sigma^2}{2}$

Considering GBM:

$$$$S(t_i) = S_0 \exp(a \cdot t_i + \sigma \cdot W(t_i)) = S_0 \exp\left((\mu - \frac{\sigma^2}{2}) \cdot t_i + \sigma \cdot W(t_i)\right)$$$$

I am interested in this part: $$$$a = \mu - \frac{\sigma^2}{2}$$$$ or reformulated:

$$\mu = a + \frac{\sigma^2}{2}$$

I experimented a bit with simulating price paths in Python. From there I gained the intuition that when we simulate say 100 stocks (all have same inital stock price $$S_0$$) with GBM over a period of 1 year and take the mean of the simulated prices (after 1 year) we end up with approximately $$S_0 \cdot exp(a + \frac{\sigma^2}{2})$$ rather than just $$S_0 \cdot exp(a)$$.

From this experiment, $$a$$ would denote whatever we assumed to be the rate of growth (e.g. the riskless rate). And consequently $$\mu$$ would denote the corresponding expected rate of change in the price after 1 year, i.e. the drift (e.g. risk-neutral drift). See also Pricing European Options with Monte Carlo.

But after having read "The drift of stock price becomes the risk-free interest rate" under RNP and looking through the blog-post of Oxymoron, I am not sure anymore whether my understanding is decent.

The same topic occured also in this post: Drift rate vs. Riskless rate in the Black-Scholes model. (see Chris Taylor's answer and the comment of Antoine Conze to it)

Any input is welcome! Would be great if you could explain in simple terms (as I am not too familiar with differential equations and stochastic calculus in general).

Edit: More generally spoken, can we consider $$\mu$$ to be related to percentage change, whereas $$a$$ denotes log returns?

• I would encourage you to learn the fundamentals of Stochastic Calculus if you are seriously interested in this type of question. Intutitively if a stock goes up one day 1% and goes down 1% the next day, over the long term is goes down by -0.5 bps per day (try to do the calculation yourself). This -0.5 bps or $-\frac{1}{2} (0.01)^2$ corresponds to the $-\frac{1}{2} \sigma^2$ term for the difference between short term expectation and long term geometric growth rate. Commented May 7 at 12:18
• Thanks! Any books you can recommend where this is covered? And what is the broader topic, that this question is related to? Commented May 7 at 12:49

The above stochastic exponential is merely a solution of

$$dS/S = \mu dt + \sigma dW_t$$

which is equivalent to, by Ito

$$d \log{S} = ( {\mu - \sigma^2/2 }) dt + \sigma dW_t$$,

which can be immediately integrated.

If $$\sigma = 0$$, then you get the elementary result from calculus

$$dS/S = \mu dt \equiv d \log S = \mu dt$$,

but if $$\sigma \neq 0$$, hence $$S$$ is stochastic, then transformation from $$dS/S$$ to $$d \log S$$ requires addition of the term to $$dt$$. This is the consequence of the definition of the Ito stochastic integral. In Stratonovich, for example, the non-stochastic formula holds.

• Great, thanks a lot! What about the story of the volatility drag (i.e. volatility is lowering actual portfolio growth below the average return)? Is that just the "intuition" behind what you have written in more formal terms? Commented May 9 at 12:49
• The way to look at it is that if $X_t$ is stochastic then the full differential of a function $f(X)$ will be $df(X) = f_x(X_t) dX_t + 1/2 f_{xx}(X_t) {\sigma_t }dt$ This can be seen as the 1st order Taylor expansion, so the additional drift term emerges due to contribution of the second order Taylor expansion term due to $(dW)^2$. This is of the order of $dt$, since variance of $(dW)^2$ is $dt$. Terms with $dt^2$ and $dt dW$ tend to zero faster than $dt$ and can be ignored. So it is the quadratic variation of the Brownian motion that brings this linear in time yet second order term. Commented May 10 at 10:21