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?

  • 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 $\endgroup$ 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



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

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

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


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