I was re-reading Chapter 6 of Stochastic Volatility Modeling by Lorenzo Bergomi. On page 203, he considers a forward variance of the following form: $$ d\xi_t^T=\lambda_t^T dZ_t^T, $$ where $Z_t^T$ is a standard Brownian motion for each $T>t$. Now he wants to prove on page 204 that: $$ dV_t=\frac{d \xi_t^T}{dT}\Biggr\rvert_{T=t} dt+\lambda_{t}^{t} dZ_t^t, $$ which means that the drift of a stochastic variance process $V_t^t=\xi_t^t$ is the slope at time $t$ of the short end of the forward variance curve. I am not entirely sure how he arrives at this formula. He says it is enough to differentiate the identity $V_t=\xi_t^t$, but I just don't see how one leads to other. Can you help me figure out?
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The instantaneous variance $V_t = \xi_t^t$. So $$ dV_t = \xi^{t + dt}_{t+dt} - \xi^t_t $$ But $$ \xi^{t + dt}_{t+dt} = \xi^{t + dt}_{t} + d \xi^{t + dt}_{t} $$ The second term to the right of the equal sign, $d \xi^{t + dt}_{t}$, is a martingale as it is simply the change in the value of a claim. So $$ dV_t = d \xi^{t + dt}_{t} + (\xi^{t + dt}_{t} - \xi^t_t) $$ But the second term to the right of the equality above is the term structure of the claims $\{\xi_t^T\}$ as observed at time $t$, for $T \in [t,\infty)$. In other words $$ (\xi^{t + dt}_{t} - \xi^t_t) = \left( \frac{ d\xi_t^T}{dT} dT \right)_{T=t} $$ Hope this helps.
EDIT As an historical aside, Bergomi was not the first to imply the dynamics of instantaneous variance from variance swaps. That, afaik, was Dupire in his unpublished paper "Arbitrage pricing with stochastic volatility, BNP, 1992."
