# Realizing the same PnL as Gamma Vs Vega

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).

• I would say that your first statement is correct. For the second, you will realise approximately the same on the delta hedging (you'll have slightly different deltas, but it will be close), the main difference will be that you pay less theta over the life of the trade. The amount less you pay will be thst same vega pnl in your first case.
– will
Jan 23 at 14:51
• Thank you, will. Any intuition on part b of the question would be very helpful! that's the bit that's tripping me up. Jan 23 at 16:08
• Intuition wise - if you take an out ofthe money option, then all the value it has will decay away as you head towards maturity. If you make that option more expensive by increasing the vol, then you're just going to decay away all that gain later.
– will
Jan 23 at 16:41
• How do I reconcile the gamma PnL being a non-linear function of realized vol, but the same pnl being realized as vega being a linear function of the remark ? Jan 23 at 17:24
• Heuristically, I think it is reconciled by the fact that there is less gamma on an option that is out of the money. So, gamma p/l would be quadratic if gamma were constant across all market prices , but it isn’t. Gamma is high when the option is atm and lower everywhere else. This results in the 2 methods agreeing.
– dm63
Feb 23 at 9:07

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 $$s$$ throughtout. If I understand Method 1 correctly you use the realized vol $$\sigma$$ to calculate $$C(t,S_t)$$ throughtout 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.