# Smile Dynamics - forward variance

I was reading Smile Dynamics II by Lorenzo Bergomi. It is clear to me that on page 2

$$V_t^{T_1,T_2}=\frac{(T_2-t)V^{T_2}_{t}-(T_1-t)V^{T_1}_{t}}{T_2-T_1}$$ is the fair strike of a forward-starting variance swap that starts accumulating variance at $$T_1>t$$ and matures at $$T_2>T_1$$. However, I find it difficult to conceptualize quantity $$\xi_t^T=V_t^{T,T}$$. What does it really represent? And why is it a good idea to model a term structure of such quantities in $$T$$?

Another basic question I had is the following: if the dynamics of $$\xi_t^T$$ is postulated to be lognormal in the first formula of page 2, then how come we end up with a zero-mean Ornstein-Uhlenbeck process in eq. (2.1)?

• You agree that $\xi_t^{T,T'}$ is the expectation of future variance over the interval ${T,T'}$? So then $\xi_t^T := \xi_t^{T,T} = \lim_{T' \to T} \xi_t^{T,T'}$ which is the expectation of the future instantaneous variance. Just like forward rates in term structure theory. In regard to the OU process appearing, have you tried deriving it yourself? Jun 2 at 7:59
• @Frido You are right! Regarding the OU process, it is not clear to me what the starting point in that derivation would be? I mean $\xi_t^T$ is supposed to be a lognormal process and when I solve the SDE I should get $U_t$ instead of $X_t$ in the exponential, it seems to me. Am I wrong? (comment re-written to fix typo). Jun 3 at 17:40

Ok, so $$d\xi_t^T = \omega e^{-k(T-t)} \xi_t^T dW_t$$ where $$W$$ is standard Brownian. Then, just by applying Ito I hope you can see that $$\log \xi_t^T / \xi_0^T = \omega \int_0^t e^{-k(T-u)} dW_u - \frac12 \omega^2 \int_0^t e^{-2k(T-u)} du$$ Now just write $$e^{-k(T-u)} = e^{-k(T-t)}e^{-k(t-u)}$$ Then $$\log \xi_t^T / \xi_0^T = \omega e^{-k(T-t)} X_t - \frac12 \omega^2 e^{-2k(T-t)} E_0[X_t^2]$$ with $$X_t = \int_0^t e^{-k(t-u)} dW_u$$ which is an O-U process.