Comparing the Volatility of a Price and Its Reciprocal
Summary
The document compares a geometric Brownian motion for an asset price with an SDE proposed for its reciprocal. It distinguishes instantaneous diffusion coefficients from the variance of the process at a fixed time: the coefficient on the Brownian increment is state-dependent, so the two equations do not have the same absolute diffusion coefficient along a path.
The replies use Itô’s lemma to derive the reciprocal-price dynamics from the original price process, then give lognormal solutions and compare their fixed-time variances. A key caveat is that the displayed reciprocal SDE in the question uses a positive Brownian term, whereas applying Itô’s lemma yields a negative one. The answers therefore mix a comparison of the stated equation with a derivation for the corrected reciprocal equation; sign conventions and the meaning of “volatility” must be specified before comparing them.
Key ideas
- Applying Itô’s lemma to the reciprocal of a geometric Brownian motion produces a negative diffusion term.
- The diffusion coefficient for the reciprocal is proportional to the reciprocal price, while the price coefficient is proportional to the price.
- Instantaneous diffusion coefficients and fixed-time variances are distinct measures of variability.
- The sign of the Brownian term in the question differs from the sign in the derived reciprocal process.
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# Is the volatility for these two SDEs the same
# Is the volatility for these two SDEs the same
$$ (1) \ \ d\left(\frac{1}{S_t}\right) =\frac{1}{S_t}\left(\sigma^2-r\right)dt +\frac{1}{S_t}\sigma dW_t $$ and $$ (2) \ \ dS_t = S_t rdt + \sigma S_t dW_t $$
How can you prove that?
## Answer by Richi Wa (score 1)
https://quant.stackexchange.com/a/15602
Let's first rewrite the tow processes and let $X_t = 1/S_t$ Then we have $$ dX_t/X_t = (\sigma^2-r)dt + \sigma dW_t, $$ with the solution (apply Ito) $$ X_t = X_0 \exp((\sigma^2/2-r) t + \sigma W_t), $$ and $$ dS_t/S_t = r dt + \sigma dW_t, $$ with the solution (apply Ito) $$ S_t = S_0 \exp((r-\sigma^2/2) t + \sigma W_t). $$ If we look at the two processes at point $t$ then their volatilities defined as the coefficient of $dW_t$ in the SDEs is different. It is $X_t \sigma = \sigma/S_t$ for the first and $\sigma S_t$ for the second - these are very different numbers for any path of $S_t$.
If you look at the variance of the processes then we know from results about log-normal distributions (http://en.wikipedia.org/wiki/Log-normal_distribution) that $$ Var[X_t] = X_0^2 \exp(-2rt+2\sigma^2t)(\exp(\sigma^2 t)-1), $$ and $$ Var[S_t] = S_0^2 \exp(2rt)(\exp(\sigma^2t)-1). $$ Also very different: so the answer is no.
## Answer by SolitonK (score 0)
https://quant.stackexchange.com/a/15596
We need to start from $(2)$. We start from Ito's Lemma which stipulates for the single variable case that: $$ \\ df(S_t) = f'(S_t)dS_t + \frac{1}{2}df''(S_t)Var[dS_t] $$ Setting $f=1/S_t$ yields: $$ \\ d\bigg(\frac{1}{S_t}\bigg) = -\frac{1}{(S_t)^2}dS_t+\frac{1}{2}\frac{2}{(S_t)^3}\sigma^2(S_t)^2Var[dW_t] \Rightarrow $$ $$\\d\bigg(\frac{1}{S_t}\bigg) = -\frac{1}{(S_t)^2}(S_trdt+\sigma S_tsW_t)+\frac{1}{2}\frac{2}{(S_t)^3}\sigma^2(S_t)^2Var[dW_t] $$ As $Var[dW_t] = dt$, after some algebra and common factoring we end up with: $$ \\ d\bigg(\frac{1}{S_t}\bigg) = \frac{1}{S_t}(\sigma^2 - r)dt - \frac{1}{S_t}\sigma dW_t$$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.