0
$\begingroup$

I'm working my way through the following paper:

Malz. A. M. (2014). A Simple and Reliable Way to Compute Option-Based Risk-Neutral Distributions

I am completely stuck on the following derivation. The author expresses the price of a call option at time-$t$ (with strike $K$ and on underlier $S_t$) as \begin{equation} c(t; K, T) = v[S_t , K, T, σ_\text{imp}(t, K), r]. \end{equation}

where $v(.)$ denotes the Black-scholes pricing formula for a European call, and $\sigma_\text{imp}(t, K)$ is the B-S implied vol.

I understand this, but the following step is not clear to me. The author differentiates both sides of the above equation with respect to the strike, $K$. This gives:

\begin{equation} \frac{\partial c}{\partial K} = \frac{\partial v}{\partial K} + \frac{\partial v}{\partial \sigma_\text{imp}}\frac{\partial \sigma_\text{imp}}{\partial K} \end{equation}

But how can this be true?

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ It is a slight modification of Black Scholes- it assumes implied volatility varies with Strike (which is what one sees in practice: smile). Then the derivative is just what is called total derivative: when k changes, option price changes, but the volatility also changes with k, which has an additional impact on the price. Pls see here for the definition: mathworld.wolfram.com/TotalDerivative.html $\endgroup$
    – Magic is in the chain
    Jul 6, 2020 at 21:59
  • $\begingroup$ precisely, the volatility varies with the strike hence you must perform a partial derivative there too. you can think of it as dc/dK = dv/dK (straight derivative) $\endgroup$
    – John
    Jul 7, 2020 at 1:21

1 Answer 1

Reset to default
2
$\begingroup$

Hint:

$$f(x) = g(h(x),k(x)) $$

$$ f'(x) = \partial_1 g (h(x),k(x)) h'(x) + \partial_2 g (h(x),k(x)) k'(x)$$

(ignore red herrings: $t$, $T$, $S_t$ and $r$; focus on $K$ only)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.