Why a Negative SDE Diffusion Coefficient Does Not Mean Negative Volatility
Summary
The document distinguishes the sign of a diffusion coefficient in a stochastic differential equation from the standard deviation of modeled returns. A negative coefficient can be represented by reversing the sign of the Brownian motion, which remains a Brownian motion under the same probability measure. Thus, in the stated model, choosing a positive or negative coefficient of equal magnitude describes equivalent dynamics.
For geometric Brownian motion, applying Itô’s lemma gives normally distributed log returns with variance equal to the squared coefficient times elapsed time. The corresponding standard deviation is the absolute value of the coefficient multiplied by the square root of time, so it cannot be negative. This resolves the apparent contradiction by separating a sign convention in the equation from a nonnegative statistical measure. The explanation relies on the geometric Brownian motion assumptions in the example; it does not address other meanings of volatility or broader model calibration issues.
Key ideas
- A diffusion coefficient in an SDE may be written with either sign because Brownian motion can be sign-reversed.
- The sign of the coefficient does not change the distribution of modeled returns in this setup.
- Under geometric Brownian motion, log-return variance is the squared diffusion magnitude times time.
- Standard deviation is nonnegative and uses the absolute coefficient magnitude.
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# Is this a poorly written example, or could volatility in fact be negative?
# Is this a poorly written example, or could volatility in fact be negative?
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?
## Answer by Quantuple (score 5, accepted)
https://quant.stackexchange.com/a/28127
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 $$\sqrt{\sigma^2 t } = \vert \sigma \vert \sqrt{t} > 0$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.