Realizing the same PnL as Gamma Vs Vega

Question

Consider a delta hedged option postion. Futhermore assume that I can perfectly forecast realized volatility over the life of the option.

Vol I buy the option at = Implied Vol (IV)

Realized volatility over the life of the option = Realized Vol (RV) Futhermore, suppose RV > IV

Now, there are 2 ways in which I can monetize RV being greater than IV.

Method 1-> I remark the vol of the option to RV (realize the pnl as vega PnL today).

Then, given that I am heding the option using the correct realized volatility, my cumulative delta hedging pnl at expiry will be known, and should perfectly offset my theta.

In this case, PnL realized = Vega x (RV-IV)

This pnl will be a linear function of RV.

Method 2-> I do not remark my vol, and delta hedge the option using the IV as the marked vol.

In this case, of course, my pnl will be path dependent, but the expected pnl would be =

0.5 x $Gamma x (RV-IV)

The gamma PnL, of course, is a quadratic function of RV.

My questions are ->

a. Is the pnl in method 1 = expected PnL in method 2 ?

b. If yes, how is the PnL in method 1 a linear function of RV, while the PnL in method 2 a quadratic function of RV.

Elaborating on question b->

A common heuristic seems to be.

I pay 100cents for a swaption with IV=2bp/day

If realized vol = 2.1bp/day, total PnL = 10c

If realized vol = 2.2bp/day, total PnL = 30c

So it's not a linear function of realized vol. But if I remark to RV and realized the PnL as a vega pnl, the pnl will be a linear function of RV (since an atm straddle is a linear function of volatility).

Answer 1

It is well known that in the Black Scholes model with implied vol $s$ and realized vol $\sigma$ the PnL from delta hedging a long position in an option is

$$\tag{1} C(T,S_T)-\Pi_T=\frac{\sigma^2-s^2}{2}\int_0^TS_t^2\partial_x^2C(t,S_t)\,dt\,. $$ (see this post). The formula you use in Method 2 is similar but not exactly equal to this. The option price $C(t,S_t)$ in (1) uses implied vol $s$ throughout. If I understand Method 1 correctly you use the realized vol $\sigma$ to calculate $C(t,S_t)$ throughout and otherwise perform the same hedging strategy. This means that in the derivation that led to (1) we have $\sigma=s$ and therefore a zero hedging PnL. That's also intuitively clear because knowing the realized vol $\sigma$ in advance and using that to price and hedge the option will exactly replicate the final payoff $C(T,S_T)\,.$ I do not think that the expectation of (1) is zero. In fact it is known that if the realized vol $\sigma$ is larger than the implied vol $s$ there is almost always a profit from the delta hedging strategy of a long gamma position.

In the Black Scholes model with continuous dividend yield $q$ the following formulas hold for calls and puts: \begin{align} \text{ gamma }\quad&\partial_x^2C(t,S_t)=e^{-q(T-t)}\frac{\phi(d_1)}{S_t\sigma\sqrt{T-t}}\,,\\[3mm] \text{ vega }\quad&\partial_\sigma C(t,S_t)=S_te^{-q(T-t)}\phi(d_1)\sqrt{T-t}\,. \end{align}

• Because $d_1=\pm\frac{\ln(S_t/K)+\sigma^2(T-t)/2}{\sigma\sqrt{T-t}}$ the vega depends nonlinearly on vol. Hence, even the simplistic PnL $\text{vega}\cdot (\sigma-s)$ depends nonlinearly on $\sigma\,.$

• There is a simple relationship $S^2_t\partial_x^2C(t,S_t)\sigma(T-t)=\partial_\sigma C(t,S_t)$ between gamma and vega. We can therefore write (1) as $$\tag{2} C(T,S_T)-\Pi_T=\frac{\sigma^2-s^2}{2}\int_0^T \frac{\partial_\sigma C(t,S_t)}{\sigma(T-t)}\,dt\,. $$ which is similar but not exactly equal to your Method 1.