# Drift of stochastic variance as slope of the short end of the forward variance curve

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?

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.