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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)?

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    $\begingroup$ 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? $\endgroup$
    – Frido
    Commented Jun 2, 2023 at 7:59
  • $\begingroup$ @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). $\endgroup$
    – fwd_T
    Commented Jun 3, 2023 at 17:40

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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.

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  • $\begingroup$ Do you have any insight on why Bergomi chose that particular form of lognormal process with exponentially decaying volatility? $\endgroup$
    – fwd_T
    Commented Jun 3, 2023 at 17:39
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    $\begingroup$ @fwd_T Probably because it's the simplest reasonable form and it ensures that at t=T there is no vol of vol, which makes sense because then the forward start varswap expires. $\endgroup$
    – Frido
    Commented Jun 3, 2023 at 18:55
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    $\begingroup$ @fwd_T Sorry, not entirely true what I wrote: there is still vol of vol at t=T, and it will be equal to omega which is the vol of instantaneous variance at T. $\endgroup$
    – Frido
    Commented Jun 3, 2023 at 19:22

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