I'm self-studying and I encountered the following example. It seems to suggest that volatility is negative in this example. I was under the impression that volatility can never be negative, both from a mathematical perspective and in the "real world." Am I missing something?

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1 Answer 1


You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assumptions. While the first may be negative, the second may not.

[Interpretation 1]

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a standard Brownian motion $W_t^\mathbb{P}$. From a pure mathematical perspective, the dynamics: $$ dS_t = \dots + \vert \sigma \vert dW_t^\mathbb{P} $$ is strictly equivalent to \begin{align} dS_t &= \dots - \vert \sigma \vert dW_t^\mathbb{P}\\ &= \dots + \vert \sigma \vert d(-W_t^\mathbb{P}) \\ &= \dots + \vert \sigma \vert d \tilde{W}_t^\mathbb{P} \end{align} since $\tilde{W}_t^\mathbb{P} = -W_t^\mathbb{P}$ can be shown to be a $\mathbb{P}$-Brownian motion.

This means specifying a dynamics this way $$ dS_t = G S_t dt + 0.2 S_t dW_t $$ is the same as specifying it as $$ dS_t = G S_t dt - 0.2 S_t dW_t $$

[Interpretation 2]

Assuming a Geometric Brownian Motion as proposed in your exercise (Black-Scholes like dynamics), applying Itô's lemma and integrating from $0$ to $t$ will get you: $$ d\ln(S_t) = (\mu - \sigma^2/2)dt \pm \vert \sigma \vert dW_t^\mathbb{P} $$ $$ \ln(S_t) = \ln(S_0) + (\mu - \sigma^2/2)t \pm \vert \sigma \vert W_t^\mathbb{P} $$ where the $\pm$ sign is used for both the positive and negative diffusion coefficient convention;

Because of the properties of Brownian motion, in both cases, you'll end up with normally distributed log-returns: $$ \ln(S_t) \sim \mathcal{N}( \ln(S_0) + (\mu - \sigma^2/2)t, \sigma^2 t) $$ with a positive log-returns variance of $\sigma^2 t$ and the associated standard deviation $$\sqrt{\sigma^2 t } = \vert \sigma \vert \sqrt{t} > 0$$


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