Assume standard Black-Scholes model, $$dS(t)=S(t)(rdt+\sigma dW(t))$$ where $\sigma$ is a constant and $W(t)$ is a Brownian motion under the risk neutral measure.
A call option is replicable, so if we are long a call and continuously (in theory) trade according to the negative of the delta of the option, we should in theory end up with 0 at the end since the two positions cancel out, and this is how we determine the price of the call option. There is one thing I do not understand here. Among the input parameters in the Black-Scholes model, $\sigma$ is treated as a constant, so there is no Pnl associated with $\sigma$; we are delta neutral so there is no Pnl associated with $\delta$ as well, and since we hedge continuously, there is no gamma Pnl (I guess?); but why do we not have a Pnl associated with theta in this case since theta is not hedged?


2 Answers 2

  • In Black Scholes $$\frac{dS}{S}=rdt+\sigma dW$$

  • $dC_{BS}(S,t)=\underbrace{\frac{\partial C_{BS}}{\partial t}dt}_{Theta PnL}+\underbrace{\frac{\partial C_{BS}}{\partial S}dS}_{DeltaPnL}+\underbrace{\frac{1}{2}\frac{\partial^2 C_{BS}}{\partial S^2}dS^2}_{GammaPnL}$

  • $dC_{BS}(S,t)=\frac{\partial C_{BS}}{\partial t}dt+\frac{\partial C_{BS}}{\partial S}dS+\frac{1}{2}\frac{\partial^2 C_{BS}}{\partial S^2}\sigma^2S^2dt$

  • Note that $dC_{BS}(S,t)$ is only the PnL of option that exists in BS world, since the spot $S$ follow BS dynamics

  • Assuming zero rates dividends, $\theta_{BS} = -\frac{1}{2}\Gamma_{BS} S^2 \sigma^2$

  • Delta hedged option PnL in BS world = $\frac{1}{2}\Gamma_{BS} S^2 [(\frac{dS}{S})^2-\sigma^2dt]= \frac{1}{2}\Gamma_{BS} S^2 [\sigma^2dt-\sigma^2dt]=0$

  • It makes sense since $S$ follows BS dynamics, if you hedge acc to BS delta, your PnL is indeed zero, since theta PnL is offset by gamma PnL

  • However, this spot $S$ follows BS dynamics which is not true in real world

  • In real world, spot $S$ follows unknown dynamics

  • Denote $C_{mkt}(S,t)$ as market price of option at spot $S_1$ and time $t$

  • $dC_{mkt}=C_{mkt}(S_1,t_1)-C_{mkt}(S_0,t_0)$

  • $𝑑𝐶_{mkt}=\underbrace{\frac{\partial C_{BS}(S,t|\hat\sigma)}{\partial t}dt}_{ThetaPnL}+\underbrace{\frac{\partial C_{BS}(S,t|\hat\sigma)}{\partial S}dS}_{Delta PnL}+\underbrace{\frac{1}{2}\frac{\partial^2 C_{BS}(S,t|\hat\sigma)}{\partial S^2}dS^2}_{GammaPnL}+\underbrace{\frac{\partial C_{BS}(S,t|\hat\sigma)}{\partial \sigma}d\sigma}_{VegaPnL}+\underbrace{\frac{\partial^2 C_{BS}(S,t|\hat\sigma)}{\partial \sigma\partial S}dSd\sigma}_{VannaPnL}+\underbrace{\frac{1}{2}\frac{\partial^2 C_{BS}(S,t|\hat\sigma)}{\partial \sigma^2}(d\sigma)^2}_{VolgaPnL}+...$

  • Spot/vol correlation would generate vanna P&L, e.g. plot VIX log return against SPX log return would get a -70% correlation

  • Vol-of-vol would generate volga PnL

  • It actually means you pay theta for gamma, vanna and volga

  • More sophisticated models like LV/SV tries to address these mkt phenomenon

  • $\begingroup$ Empirically do you find SV and SVJ model accurate? Like increase in vol of vol actually leads to steeper smile and changes in spot/vol correlation leading to changes in skew? I've always been curious about this but don't have access to data. $\endgroup$
    – confused
    Jul 21, 2020 at 8:06
  • $\begingroup$ Actually you don’t need too many data, you can start with SPX options quoted on CBOE [cboe.com/delayedquote/quote-table-download] Start with a simple SV model like Heston’s model, calculate vanilla prices for various strikes and expiries with the SV model, and translate prices to a IV surface fitted by the model. Play around with the parameters (spot/vol corr, vol-of-vol, mean reversion rate, long run average of variance) and see how well SV can fit the market’s IV surface. Skew in SV is essentially a product of spot/vol correlation and vol-of-vol, Smile is generated by vol-of-vol $\endgroup$
    – ryc
    Jul 21, 2020 at 10:38
  • $\begingroup$ You will see SV models don’t have enough skew for short expiries. Also since you only have 4 parameters available, you won’t be able to fit the market’s iv surface. Jump parameters are hard to estimate. Unlike parameters like spot/vol correlation, vol-of-vol, which can be traded with risk reversals and butterflies, jump parameter is not really tradable. So jump models are not so popular among practitioners $\endgroup$
    – ryc
    Jul 21, 2020 at 10:39
  • $\begingroup$ Hello, I know in the model when you change parameters it shifts the skew. I also have read that Heston doesn't fit market that well. My question was empirically, when vol of vol changes or more importantly when spot/vol corr changes do actually option skew change accordingly. For example, usually they're seems to be a relationship between implied vol and realized vol with implied vol a bit lagged. $\endgroup$
    – confused
    Jul 21, 2020 at 12:36
  • $\begingroup$ My question is, when we see realized vol/vol increase do we also see the wings get bid at the same time in actual options. I guess what I am curious about is how does the implied model parameters correlate with empirical data. I don't think IV can predict FV but I haven't done the research but I think HV can predict IV to some extent. Wondering if there was anything to be said with vol/vol and wings and spot/vol corr and skew $\endgroup$
    – confused
    Jul 21, 2020 at 12:38

When you replicate the option, you negatively scalp yourself when hedging deltas (if you are short the option). That negative scalp should be offset by theta you make by being short the option, and thus on net your option + hedge has 0 pnl. This obviously assumes realized volatility = implied volatility.

If your option has high IV but underlying doesn't move, then obviously you will lose/gain money on theta (depending on long/short option) but you would have 0 PnL change from underlying hedging. In this scenario, realized volatility < implied volatility. Also in this scenario, option is "mispriced" and hence there is non-zero PnL.

Obviously this all assumes option prices follow BSM model, so just take everything with a grain of salt when you enter the real world. And if you approach options from a P or Q perspective.

TLDR: Options make lose money from theta, underlying make lose money from gamma. Under BSM, if IV = RV, then they cancel out and net PnL is 0.

  • $\begingroup$ could you explain what you mean when you say: "you negatively scalp yourself when hedging deltas". or maybe a simple example would help. thanks a lot. $\endgroup$
    – mark leeds
    Jul 22, 2020 at 2:38
  • $\begingroup$ You are short a call option, let's say ATM so you are short -50 deltas. To hedge it, you have to buy 50 underlying. In the BSM world, if the underlying moves up in price, your call gets even shorter deltas, so now let's say the call is now -60 deltas. To hedge that you have to buy 10 more underlying, but you will now have to buy it at a higher price than before. Now let's say the underlying moves back down in price to where it started and the call option once again has -50 deltas. You are long 60 underlying, so you have to sell 10 more to hedge your position. $\endgroup$
    – confused
    Jul 22, 2020 at 3:34
  • $\begingroup$ But now you sold at a lower price. And thus, as you hedge your option by buying/selling the underlying, you are negatively scalping yourself because you are buying high, selling low. Being short the call, you hope that this negative scalp is offset by the theta you collect from being short the option. $\endgroup$
    – confused
    Jul 22, 2020 at 3:35
  • $\begingroup$ Obviously everything in BSM model is in continuous time, but the effect is still the same. The gamma in the call option creates delta mismatches (with your replicating portfolio) and if you are short an option, you negatively scalp against your self. If you are long the option, you positively scalp against your option. If you look at BSM greeks, theta and gamma are linked hand in hand. High gamma options must have high theta because theta must offset trading the underlying. $\endgroup$
    – confused
    Jul 22, 2020 at 3:45
  • $\begingroup$ thanks a lot. I never worked with options (only read books) so I will print out and study it carefully. thanks for through and clear explanation. this is definitely related to a different question that I asked a few weeks ago in a different thread so it will help with that one also. $\endgroup$
    – mark leeds
    Jul 23, 2020 at 2:18

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