Deriving Diffusion Variance in Black–Scholes from Independent Increments
Summary
The document asks why the variance accumulated by a geometric Brownian motion over a time horizon is represented by the integral of instantaneous variance, and how time averaging enters the stated realized-variance expression. The answer builds the result from independent increments: over an interval with constant volatility, variance equals volatility squared multiplied by the interval length. Variances from successive nonoverlapping intervals add, so a sequence of intervals with different volatility levels contributes the sum of each squared volatility times its duration.
Taking the continuum limit turns that sum into the integral of instantaneous variance over the horizon. Dividing by the horizon length gives the time average. The explanation assumes a diffusion without jumps and, in its opening example, flat volatility; it does not derive the jump contribution in the question’s Merton model. It therefore clarifies the diffusion component but leaves the full jump-diffusion variance discussion incomplete.
Key ideas
- For a constant-volatility diffusion, variance accumulated over an interval is volatility squared times its duration.
- Independent increments allow variances across nonoverlapping intervals to be added.
- With changing volatility, the accumulated variance is the sum of interval variances.
- The continuum limit gives an integral of instantaneous variance, and dividing by the horizon yields its time average.
- The explanation covers diffusion variance and does not derive the jump term.
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# How is variance derived in BS?
# How is variance derived in BS?
The realized variance under classical Black Scholes where the stock price process follows a GBM is given as $$V_T = \frac1T\int_0^T\sigma_s^2ds\qquad (1)$$ however, the texts I have been reading do not give a derivation of this fact. Further, it is stated, that in the case of the merton jump diffusion model, term $(1)$ must be added to $$\frac1T\sum_{i=1}^{N(T)}\ln(Y_i)^2\,\,\,\,\,\,\,\,\,\,\,\qquad (2)$$
where $N(T)\sim\text{Poisson}(\lambda)$ and $Y_i$ denotes the relative jump size in the stock price. My naive approach to derive this was to find the variance on the $\log$ process (dynamic). Doing so the log price becomes a sum of normal random variables and is therefore a normal variable as well (we can add variances -- no correlation). In order to find the variance of the 2 random processes (diffusion and jump), my idea was to apply the Ito Isometry. However, by doing so I cannot recover the $1/T$ term in both $(1)$ and $(2)$.
What I am wondering is if this procedure is correct. How can I incorporate the averaging over $T$?
## Answer by KT8 (score 2)
https://quant.stackexchange.com/a/74638
Assume a flat (both in strike and time) volatility input, $\sigma$. Then, the variance a GBM accumulates from $t_0$ up to time $t_1$ is $$ \text{Var}(t_0, t_1) = \sigma_{t_0}^2 (t_1 - t_0). $$ Now consider the volatility from time $t_1$ to $t_2$ changes to $\sigma_{t_1}$, as a GBM has independent increments, the variance from $t_1$ to $t_2$ is $$ \text{Var}(t_1, t_2) = \sigma_{t_1}^2 (t_2 - t_1), $$ and $$ \text{Var}(t_0, t_2) = \sigma_{t_0}^2 (t_1 - t_0) + \sigma_{t_1}^2 (t_2 - t_1). $$ Taking the continuum limit leads to that integral, for a process with just a diffusion (no jumps).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.