Discrete Taylor P&L Versus Itô Calculus in Delta Hedging

# Taylor expansion or Itô's formula


# Taylor expansion or Itô's formula












Consider a risky asset whose price at time $t$ is $S_t$, and an option whose price at time $t$ is $P(t,S_t)$.

I do not understand how to justify the following Taylor expansion without using Itô's formula: denoting $\delta S_t=S_{t+\delta t}-S_t$, $$ P(t+\delta t,S_t+\delta S_t)=P(t,S_t)+\frac{\partial P}{\partial t}(t,S_t)dt+\frac{\partial P}{\partial x}(t,S_t)\delta S_t+\frac12\frac{\partial^2 P}{\partial x^2}(t,S_t)(\delta S_t)^2. $$

For example, when computing the PnL of a delta-hedge option between $t$ and $t+\delta t$, using Black-Scholes formula we get the following: $$ PnL=\frac12S_t^2\frac{\partial^2 P}{\partial x^2}(t,S_t)\left(\frac{(\delta S_t)^2}{S_t^2}-\sigma^2\delta t\right), $$ where $\sigma$ is the implied volatility. This justifies statements such as "we make money if realised vol > implied vol".

Clearly if we had used Itô's formula above, then $(\delta S_t)^2$ should be understand as the quadratic variation $\delta\langle S\rangle_t$ which is $\sigma^2S_t^2\delta t$. Then the PnL above would be always 0, which is obviously wrong.

So it means the first formula above is indeed a Taylor expansion and not Itô's formula, but then:

- How do we justify the absence of $\delta t\delta S_t$, and of higher order terms? What justifies we consider $(\delta S_t)^2$? Why not $(\delta S_t)^3$ as well? I understand it in the context of Itô's formula with infinitesimal variation, which is not the case here.

- If the answer is just something like "we apply Itô, but just not with an infinitesimal increment". Then what is the justification that $\delta\langle S\rangle_t$ is replaced by $(\delta S_t)^2$? Why not write $\sigma^2S_t^2\delta t$ instead?

## Answer by Andrea (score 4)

https://quant.stackexchange.com/a/81283

I've re-written my answer as there is a more direct way to see it, and you don't need stochastic calculus.

Take a real (smooth) function $f(t)$, you might want to know

$\Delta_f = f(T) - f(0)$

There are 2 ways to do it



- Integrating its derivative $\Delta_f = \int_0^T df(t) = \int_0^T f'(t) \, dt$

Which to chose depends on the particular situation. But, one can immediately see some differences

- In the 2nd case, only the 1st derivative is used (unless you move to double - triple integrals)

- Which one is better? Not sure is a valid question. The first case only uses information at $t=0$, the second only the 1st derivative, but everywhere.

Now, if $P$ is a payoff function which depends on $S_t$, so $P(t, S_t)$. As a function of $t$ and $s$, $P(t, s)$ is super smooth, but as a function of $t$, $P(t, S_t)$ is not.

- no difference, but you need a 2D Taylor expansion (up to any order you like) and the increments are $\delta t$ and $\delta S_t$

- $\int_0^T dP(t, S_t)$ is not a Lebesgue integral ($P(t, S_t)$ is not of finite variation), so you need Ito, no alternative.

And a final remark, the Taylor expansion is often used in PnL explain, the integral in the PnL of the continuous delta hedge.