Short answer

He's basically making a parallel between a forward variance trade and a futures trade. In both cases you should have that the underlying quotes are martingales in the absence of arbitrage.

Long(er) answer

Under the physical measure $\Bbb{P}$, an arbitrage is a (self-financing) trading strategy $V$ - or portfolio - for which there exists a time $T > 0$ such that $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{P}-\text{a.s. and } \Bbb{P}(V_T \ne 0) > 0$$

Suppose you define an equivalent probability measure $\Bbb{Q}\equiv\Bbb{P}$. Since by definition, both measures agree on null events, our arbitrage definition translates to $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{Q}-\text{a.s. and } \Bbb{Q}(V_T \ne 0) > 0 \tag{A}$$

Notice that if $\Bbb{Q}$ is further a martingale measure, that is if $(V_t)_{t\geq0}$ emerges as a $\Bbb{Q}$-martingale: $$ V_0 = \Bbb{E}_0^\Bbb{Q} [ V_T ] $$ then $(A)$ will never happen. This explains the central role of equivalent martingale measures in arbitrage pricing theory.

Putting that back into context, you've managed to identify a (self-financing) strategy $V$ (= buying and selling forward variance swaps), which starting from $V_t=0$ (no cost), allows you to earn a quantity $$V_{t'} = (T_2-T_1) \left( \hat{\sigma}_{VS,T_1T_2}^2(t') - \hat{\sigma}_{VS,T_1T_2}^2(t)\right)$$

Based on what we've said earlier, in the absence of arbitrage, there should exist a measure $\Bbb{Q} \equiv \Bbb{P}$ such that $$ \Bbb{E}^\Bbb{Q}_{t}[V_{t'}] = V_t$$ hence, using the definitions of $V_t$ and $V_{t'}$, $$ \Bbb{E}^\Bbb{Q}_{t}\left[ \hat{\sigma}_{VS,T_1T_2}^2(t') \right] = \hat{\sigma}_{VS,T_1T_2}^2(t) $$ hence forward variance swap quotes are martingales. Assuming a continuous paths process (= in a diffusive setting), by the martingale representation theorem we should then have $$ \hat{\sigma}_{VS,T_1T_2}^2(t) = ... dW_t^\Bbb{Q} $$ hence no pricing drift under $\Bbb{Q}$.

Short answer

He's basically making a parallel between a forward variance trade and a futures trade. In both cases you should have that the underlying quotes are martingales in the absence of arbitrage.

Long(er) answer

Under the physical measure $\Bbb{P}$, an arbitrage is a (self-financing) trading strategy $V$ - or rather the value of a portfolio implementing this strategy - for which there exists a time $T > 0$ such that $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{P}-\text{a.s. and } \Bbb{P}(V_T \ne 0) > 0$$

Suppose you define an equivalent probability measure $\Bbb{Q}\equiv\Bbb{P}$. Since by definition, both measures agree on null events, our arbitrage definition translates to $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{Q}-\text{a.s. and } \Bbb{Q}(V_T \ne 0) > 0 \tag{A}$$

Notice that if $\Bbb{Q}$ is further a martingale measure, that is if $(V_t)_{t\geq0}$ emerges as a $\Bbb{Q}$-martingale: $$ V_0 = \Bbb{E}_0^\Bbb{Q} [ V_T ] $$ then $(A)$ will never happen. This explains the central role of equivalent martingale measures in arbitrage pricing theory.

Putting that back into context, you've managed to identify a (self-financing) strategy (i.e. buying and selling forward variance swaps), which at no cost ($V_t=0$), allows you to earn a quantity $$V_{t'} = (T_2-T_1) \left( \hat{\sigma}_{VS,T_1T_2}^2(t') - \hat{\sigma}_{VS,T_1T_2}^2(t)\right)$$

Based on what we've said earlier, in the absence of arbitrage, there should exist a measure $\Bbb{Q} \equiv \Bbb{P}$ such that $$ \Bbb{E}^\Bbb{Q}_{t}[V_{t'}] = V_t$$ hence, using the definitions of $V_t$ and $V_{t'}$, $$ \Bbb{E}^\Bbb{Q}_{t}\left[ \hat{\sigma}_{VS,T_1T_2}^2(t') \right] = \hat{\sigma}_{VS,T_1T_2}^2(t) $$ hence forward variance swap quotes are martingales. Assuming a continuous paths process (= in a diffusive setting), by the martingale representation theorem we should then have $$ \hat{\sigma}_{VS,T_1T_2}^2(t) = ... dW_t^\Bbb{Q} $$ hence no pricing drift under $\Bbb{Q}$.

Short answer

He's basically making a parallel between a forward variance trade and a futures trade. In both cases you should have that the underlying quotes are martingales in the absence of arbitrage.

Long(er) answer

Under the physical measure $\Bbb{P}$, an arbitrage is a (self-financing) trading strategy $V$ - or portfolio - for which there exists a time $T > 0$ such that $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{P}-\text{a.s. and } \Bbb{P}(V_T \ne 0) > 0$$

Suppose you define an equivalent probability measure $\Bbb{Q}\equiv\Bbb{P}$. Since by definition, both measures agree on null events, our arbitrage definition translates to $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{Q}-\text{a.s. and } \Bbb{Q}(V_T \ne 0) > 0 \tag{A}$$

Notice that if $\Bbb{Q}$ is further a martingale measure, that is if $(V_t)_{t\geq0}$ emerges as a $\Bbb{Q}$-martingale: $$ V_0 = \Bbb{E}_0^\Bbb{Q} [ V_T ] $$ then $(A)$ will never happen. This explains the central role of equivalent martingale measures in arbitrage pricing theory.

Putting that back into context, you've managed to identify a (self-financing) strategy $V$ (= buying and selling forward variance swaps), which starting from $V_t=0$ (no cost), allows you to earn a quantity $$V_{t'} = (T_2-T_1) ( \hat{\sigma}_{VS,T_1T_2}^2(t') - \hat{\sigma}_{VS,T_1T_2}^2(t)$$

Based on what we've said earlier, in the absence of arbitrage, there should exist a measure $\Bbb{Q} \equiv \Bbb{P}$ such that $$ \Bbb{E}^\Bbb{Q}_{t}[V_{t'}] = V_t$$ hence, using the definitions of $V_t$ and $V_{t'}$, $$ \Bbb{E}^\Bbb{Q}_{t}\left[ \hat{\sigma}_{VS,T_1T_2}^2(t') \right] = \hat{\sigma}_{VS,T_1T_2}^2(t) $$ hence forward variance swap quotes are martingales. Assuming a continuous paths process (= in a diffusive setting), by the martingale representation theorem we should then have $$ \hat{\sigma}_{VS,T_1T_2}^2(t) = ... dW_t^\Bbb{Q} $$ hence no pricing drift under $\Bbb{Q}$.

Short answer

He's basically making a parallel between a forward variance trade and a futures trade. In both cases you should have that the underlying quotes are martingales in the absence of arbitrage.

Long(er) answer

Under the physical measure $\Bbb{P}$, an arbitrage is a (self-financing) trading strategy $V$ - or portfolio - for which there exists a time $T > 0$ such that $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{P}-\text{a.s. and } \Bbb{P}(V_T \ne 0) > 0$$

Suppose you define an equivalent probability measure $\Bbb{Q}\equiv\Bbb{P}$. Since by definition, both measures agree on null events, our arbitrage definition translates to $$ V_0=0,\,\, V_T \geq 0\,\, \Bbb{Q}-\text{a.s. and } \Bbb{Q}(V_T \ne 0) > 0 \tag{A}$$

Notice that if $\Bbb{Q}$ is further a martingale measure, that is if $(V_t)_{t\geq0}$ emerges as a $\Bbb{Q}$-martingale: $$ V_0 = \Bbb{E}_0^\Bbb{Q} [ V_T ] $$ then $(A)$ will never happen. This explains the central role of equivalent martingale measures in arbitrage pricing theory.

Putting that back into context, you've managed to identify a (self-financing) strategy $V$ (= buying and selling forward variance swaps), which starting from $V_t=0$ (no cost), allows you to earn a quantity $$V_{t'} = (T_2-T_1) \left( \hat{\sigma}_{VS,T_1T_2}^2(t') - \hat{\sigma}_{VS,T_1T_2}^2(t)\right)$$

Based on what we've said earlier, in the absence of arbitrage, there should exist a measure $\Bbb{Q} \equiv \Bbb{P}$ such that $$ \Bbb{E}^\Bbb{Q}_{t}[V_{t'}] = V_t$$ hence, using the definitions of $V_t$ and $V_{t'}$, $$ \Bbb{E}^\Bbb{Q}_{t}\left[ \hat{\sigma}_{VS,T_1T_2}^2(t') \right] = \hat{\sigma}_{VS,T_1T_2}^2(t) $$ hence forward variance swap quotes are martingales. Assuming a continuous paths process (= in a diffusive setting), by the martingale representation theorem we should then have $$ \hat{\sigma}_{VS,T_1T_2}^2(t) = ... dW_t^\Bbb{Q} $$ hence no pricing drift under $\Bbb{Q}$.