Confusion about replicating a call option

Question

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?

Answer 1

• 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

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• 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

Answer 2

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.