I am currently reading Volatility Trading, I have only just started, but I am trying to understand a "derivation from first principles" of the BSM pricing model.

I understand how the value of a long call ($C$) and delta-hedged short position ($\Delta S$) in the underlying is given by:

$$ C - \Delta S_t $$


  • $C$ is the value of the long call option
  • $S_t$ is the spot price of the underlying at time $t$
  • $\Delta$ is the hedge ratio.

On page 9, I also understand that the change in the value of said portfolio, as the underlying moves from $S_{t}$ to $S_{t+1}$ is given by:


$$ C(S_{t+1}) - C(S_t) -\Delta(S_{t+1} - S_t) + r(C-\Delta S_t) $$

where the last term is money earned from reinvesting net received funds obtained in establishing the position at a rate $r$.

The change in the option value is then obtained via a second-order Taylor series approximation:


$$ \Delta(S_{t+1} - S_t) + \frac{1}{2}(S_{t+1} - S_t)^2\frac{\partial^2C}{\partial{S}^2} + \theta - \Delta(S_{t+1}-S_t) + r(C- \Delta S_t) $$

where $\theta$ is time decay.

I don't see how the author moves from equation 1.2 to equation 1.3, as it is not clear (at least to me) what function $f(x)$ he is approximating in 1.3

I would be grateful if someone could explain how the author makes the leap from equation 1.2 to the Taylor approximation (1.3) given on page 9.


1 Answer 1


He is approximating $C(S_{t+1})$ around $t$:

$$C(S_{t+1})=C(S_{t}) + \frac{\partial C(S_{t})}{\partial S_{t}}(S_{t+1}-S_{t})+\frac{(S_{t+1}-S_t)²}{2}\frac{\partial^{2}C(S_{t})}{\partial S_{t}^{2}} + ...$$

In addition, he takes the time value of $C(S_t)$ into account (and I look only at the time contribution here):

$$C(S_{t+1})-C(S_t)=\Delta t\frac{\partial C}{\partial t}+...=\Delta t\cdot\theta+..$$

There, the first equation is just the derivative of the option with regard to t. Usually, $\theta$ is the loss of the option value in a day, so it is just a question of normalization here. If you put everything together, you get the step you are looking at.

  • $\begingroup$ Thanks. +1 for the effort. I now get the first part of the derivation. I am trying to understand the theta derivation (i.e. derivation wrt time), and how that leads to the final equation presented in 1.3 $\endgroup$ Feb 23, 2013 at 20:01

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