I am actually reading Lorenzo Bergomi's "Stochastic Volatility Modelling" book, and came across this bit :

enter image description here

I understand everything up to (5.5) included. But I don't see the point in mentioning the vanishing pricing drift. What is "pricing drift" ? After that he defines continuous (instantaneous in fact) VS forward variance and writes that they are driftless as well (same argument as for discrete VS forward variance), and write that in a diffusive setting they are equal to $\left(\ldots\right) dW_t^T$ where I guess $W^T$ is a standard Brownian motion under the forward $T$ measure.

Is pricing drift defined outside a diffusive setting or does everything here takes place in a diffusive setting ? (The "in a diffusive setting" is unsettling.)

What is general in Bergomi's remark about pricing drift ? I mean, is there a way to define being a martingle through linearity of a certain P&L or ?


1 Answer 1


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 $\Bbb{Q}$.

  • $\begingroup$ This makes total sense to me. Two questions though : 1) "a parallel between a forward variance trade and a futures trade" : you would use the same approach to show that a futures rate is martingale under the spot measure then ? I never understood the argument in the futures context because of margining. Could we show a futures is a "spot martingale" be saying it a rolled position of one-day forwards ? 2) How far does non arbitrage theory go without resorting to diffusions ? $\endgroup$
    – Olorin
    Mar 28, 2017 at 16:28
  • $\begingroup$ Assume that, over a given trading day, you enter a future position at $t$ and unwind it before the close at $t'$. At the end of the day, the clearing house credits/debits your margin account of the quantity $F(t',T)-F(t,T)$. Can you see the parallel now? $\endgroup$
    – Quantuple
    Mar 28, 2017 at 19:13
  • $\begingroup$ You get that $\Bbb{E}^\Bbb{Q}_{t}\left[ F(t',T) - F(t,T)\right] = 0$ as we entered at no cost, which gives the martingale relation as $F(t,T)$ is $\mathscr{F}_t$_measurable ? So basically the "linearity of the P&L in variation of the underlying" is just what allows you to do this cond. exp. calculation ? Why cannot I do the same for a forward ? I always tended to see that the fwd $T$ measure degenerates to the spot measure as $T-t\rightarrow 0$, but I may be wrong ? $\endgroup$
    – Olorin
    Mar 28, 2017 at 21:42
  • $\begingroup$ About futures, should I forget the following description seen in Bjork's Arbitrage Theory in continuous time : i.stack.imgur.com/NMYb5.jpg ? $\endgroup$
    – Olorin
    Mar 28, 2017 at 21:53
  • $\begingroup$ Yes. No. Yes, the same can be done for the forward. No the 2 measures coincide for deterministic rates. This is exactly the futures definition I used. Don't confuse the value of a future contract and the future price (same goes for forward). See here quant.stackexchange.com/questions/31162/…. $\endgroup$
    – Quantuple
    Mar 28, 2017 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.