How is variance derived in BS?

Question

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 1

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).