# 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 $$T$$?

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).