Log Return Volatility from a Time-Varying Diffusion
Summary
The document considers a stock price driven by a stochastic differential equation with time-dependent drift and a two-dimensional Brownian motion. It corrects the proposed solution by expressing the log price as integrals of the drift adjustment and the volatility vector against Brownian motion. For the conditional variance of a log return over an interval, the drift contribution is deterministic given the coefficient functions and does not affect variance.
Applying the Itô isometry to the stochastic integral gives the log return variance as the integral over the interval of the squared norm of the volatility vector; its square root is the standard deviation. This result assumes the volatility coefficients are deterministic functions of time, as posed in the question. The answer’s displayed proposed solution appears to contain a notation error in the drift correction, so the reliable takeaway is the integral form and the variance result under the stated assumptions.
Key ideas
- The solution to the price process with time-varying coefficients is expressed using integrals over time.
- The drift terms do not contribute to the conditional variance of the log return under the stated deterministic-coefficient setup.
- Itô isometry converts the variance of the Brownian integral into an integral of squared volatility magnitude.
- The log return standard deviation is the square root of integrated variance across the observation interval.
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# Deriving an Analytical Expression for Standard Deviation of Log Returns
# Deriving an Analytical Expression for Standard Deviation of Log Returns
I am looking to find an expression for the standard deviation log returns of a stock price process.
I have a stock price which follows the following dynamics:
- $dY(t) = Y(t)(r(t)dt + η(t)dW(t))$
Here, W (t) is a two–dimensional vector–valued Brownian motion with independent components and η(t) is a two–dimensional (time-dependent) vectors. With:
- $\eta(t)$ = $ξ(t)$ $ \begin{bmatrix}η_{1}\\η_{2}\end{bmatrix} $
- $\eta_1^2 + \eta_2^2 = 1$
How would I find an expression for log returns, which is defined:
$\sqrt{(Var[\log{}Y(t + ∆t) − \log{} Y (t)|F_t])}$
I believe you can use Ito's lemma we can get the solution to the above SDE:
- $Y(t) = Y(0)\exp(r(t) - \frac{1}{2} η(t)^2)t + η(t)W(t))$
Is the above done correctly? Considering the drift and volatility both depend on a time component
If so, or otherwise, what is the next steps to find an expression for the standard deviation of log returns
Any helps/tips is highly appreciated
## Answer by NN2 (score 2, accepted)
https://quant.stackexchange.com/a/75431
You solution formula in the question is not correct, it should be $$Y(t) = Y(0)\cdot \exp\left(\int_0^t\left(r(s)-\frac{1}{2}\eta^T(s)\cdot \mathbf{W}(s) \right)dt +\int_0^t\ \eta^T(s)d\mathbf{W}(s) \right) $$ where $^T$ is the transpose of a vector.
For the log return formula, you have $$\begin{align} R&:=\sqrt{V(\ln Y(t + ∆t) − \ln Y (t)|\mathcal{F}_t)}\\ &=\sqrt{V\left(\underbrace{\int_t^{t+\Delta t}\left(r(s)-\frac{1}{2}\eta^T(s)\cdot \mathbf{W}(s) \right)dt}_{\text{constant term, don't affect the variance, so ignore}} +\int_t^{t+\Delta t}\ \eta^T(s)d\mathbf{W}(s)\right)}\\ &=\sqrt{V\left(\underbrace{\int_t^{t+\Delta t}\ \eta^T(s)dW(s)}_{\text{expectation is equal to }0}\right)}\\ &=\sqrt{\underbrace{\mathbb{E}\left(\left(\int_t^{t+\Delta t}\ \eta^T(s)d\mathbf{W}(s)\right)^2\right)}_{\text{Ito isometry }}}\\ R&=\color{red}{\sqrt{\int_t^{t+\Delta t}\ \left(\eta^T(s)\cdot \eta(s) \right)ds}}\\ \end{align}$$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.