How Volatility Drag Lowers Geometric Brownian Motion Growth

# Intuition behind prices modeled by Geometric Brownian Motion


# Intuition behind prices modeled by Geometric Brownian Motion












Suppose that we model a price $P_t$ to evolve per

$$\frac{dP_t}{P_t}=\mu dt+\sigma dW_t$$

for $\mu\in\mathbb{R}$ and $\sigma>0$. The unique strong solution to this diffusion is

$$P_t=P_0e^{(\mu-\sigma^2/2)t+\sigma W_t}$$

My question is the following: by the law of iterated logarithm, one can show that as $t\to\infty$, the drift term $(\mu-\sigma^2/2)t$ dominates the stochatic part $\sigma W_t$, and $P_t$ goes to $\pm \infty$ depending on the sign of the drift. I am interested on intuition behind the following fact: if the volatility increases to $\sigma'>\sigma$, then $$(\mu-(\sigma')^2/2)<(\mu-\sigma^2/2),$$ So for $t$ big we have $$P_t(\sigma')\leq P_t(\sigma)$$ systematically. I understand this is due to Itô's correction, but I'm wondering at an intuitive level why, if the volatility is bigger, then the prices/value of a project tend to be smaller.

For reference of what I'm talking about, you can see this picture where I show two geometric Brownian motions, with the same draw of $W_t$, with the black one having a bigger volatility:

## Answer by Daneel Olivaw (score 8, accepted)

https://quant.stackexchange.com/a/53217

Because of volatility drag.

In very simple terms, assume three periods, $t=t_0, t_1, t_2$, and a process which starts with value $100$ at $t_0$ and which can either go up or go down with same probability when transitioning from one period to the next. Let us assume the same expected change for each period, for example $1\%$, but with different up and down moves: in the first case, the process either decreases by $2\%$ or increases by $4\%$; in the second case, it either decreases by $3\%$ or increases by $5\%$. Now, suppose the process goes down one period and up the other one (the order obviously does not matter). In the first case, the final value of the process will be $$0.98\times1.04=1.0192,$$ whereas in the second case it would be $$0.97\times1.05=1.0185.$$ Hence the process with higher volatility (the 2nd one) ends up with a lower value that the one with lower volatility. Said otherwise, volatility has a cost.

Another way to see this mathematically is through logarightmic return, which is the appropriate return to look at for log-normally distributed processes. The logarithmic return for an asset $S$ and times $s<t$ is: $$\ln\frac{S_t}{S_s}=\ln\left(1+\frac{S_t-S_s}{S_s}\right)=\ln(1+r)$$ The return $r$ over $[s,t]$ of asset $S$ will be in $[-1;\infty)$, yet the derivative of the logarithm over that interval is: $$\frac{\partial}{\partial r}\ln (1+r)=\frac{1}{1+r}$$ which shows that the marginal contribution of negative returns ($r<0$) is higher than the contribution of positive returns ($r>0$): $$\ln(1+r)+\ln(1-r)\leq0$$ Alternatively, this can be seen by the properties of logarithms: $$\begin{align} \ln(1+r)+\ln(1-r)&=\ln((1+r)(1-r)) \\&=\ln(1-r^2) \\&\leq\ln(1) \\&=0 \end{align}$$