1
$\begingroup$

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 $T$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Thanks, this is a good way to look at the case of just diffusion. However, your answer only answers my question partially. $\endgroup$
    – Prb21245
    Commented Feb 14, 2023 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.