Please help me figure out what is the mathematical relationship between $\frac{dV}{dS}$ (Delta) and $\frac{dV}{dK}$ ($K$=strike), taking into account vol skew.

I ask this because I want to figure out the value of a digital at a certain strike, given that I know the vanilla option delta there, and also I have the volatility smile.

I know that there is the equation: $$ \frac{d V}{d S} = \frac{\partial V}{\partial S} + \frac{\partial V}{\partial \sigma} \frac{\partial \sigma}{\partial S} , $$ so I suppose there is something similar for what I need.

As I recall , $dV/dK = N(d2)$ and $dV/dS = N(d1)$ (ignoring dividends and risk-free rates), so I just need some simple approximation linking $\frac{dV}{dS}$ and $\frac{dV}{dK}$ (approximation if an exact relationship is too complicated).

  • $\begingroup$ If your modelling assumptions are such that the dynamics of $ln(S_t)$ is homogeneous in space, then you can derive a relationship tying these two quantities using Euler theroem on the call price function (which is homogeneous in spot/strike of a degree 1 under the above circumstances). $\endgroup$ – Quantuple Jul 27 '17 at 7:09
  • $\begingroup$ thanks for fixing the algebra! please give me an approx if its too complicated to do in precise formal maths. i am just using black formula , with vol smile $\endgroup$ – Randor Jul 27 '17 at 7:19
  • $\begingroup$ No it is pretty straightforward. Just note that $$C(xS,xK,\tau,\sigma,r,q) = x C(S,K,\tau,\sigma,r,q)$$ and derive both sides of this equality with respect to $x$ to get what you need: $S \partial_S C + K \partial_K C = C$ $\endgroup$ – Quantuple Jul 27 '17 at 7:22
  • $\begingroup$ your equation ends with C- what is your final result - i have dv/dS and i want to calc dv/dK $\endgroup$ – Randor Jul 27 '17 at 7:27
  • $\begingroup$ My C is your V (option price). Sorry cannot edit. $\endgroup$ – Quantuple Jul 27 '17 at 7:46

If your working modelling assumptions are such that the dynamics of the log price process $\ln(S_t)$ is space homogeneous, you have that the price of a European vanilla option is itself a space-homogeneous function of degree one. You can then appeal to Euler theorem to get the relationship you need.

More specifically, define the price at time $t$ of the option expiring at $T$ and struck at $K$ as

$$ V = DF(t,T)\, \Bbb{E}_t^\Bbb{Q} \left[ (w(S_T - K))^+ \right] := V(S_t, K, T-t, \theta) $$ where $\theta$ figures the relevant model parameters and $w=\pm1$ the call/put factor. Now under the space homogeneity assumption we've just mentioned, you can write that $$ V(xS_t,xK,T-t,\theta) = x V(S_t,K,T-t,\theta), \forall x \geq 0$$

Taking the derivative with respect to $x$ on both sides and then setting $x=1$ gives:

$$ \frac{\partial V}{\partial S} S + \frac{\partial V}{\partial K} K = V $$ hence $$ \frac{\partial V}{\partial K} = \frac{1}{K} \left( V - \frac{\partial V}{\partial S} S \right) $$

which is what you are looking for.

And indeed if you are pricing a digital call ($D$ below) for instance, using the notation $C$ to denote the European call price \begin{align} D &= -\frac{dC}{dK} \\ &= -\left[ \frac{\partial C}{\partial K} + \frac{\partial C}{\partial \Sigma} \frac{\partial \Sigma}{\partial K} \right] \\ &= -\left[ \frac{1}{K}\left( C - \Delta S\right) + \nu \frac{\partial \Sigma}{\partial K} \right] \end{align} where for a maturity $T$ and strike level $K$, $C$ is the corresponding European call price, $\Delta$ its BS Delta, $\nu$ its BS Vega and $\partial \Sigma/\partial K$ the IV skew. We have moved from the second line to the third using the result which we just derived.

| improve this answer | |
  • $\begingroup$ many thanks quantuple! also i now realise that this result exactly is shown by Uwe Wystup , and he calls dv/dk dual delta $\endgroup$ – Randor Jul 27 '17 at 11:09
  • $\begingroup$ Does this mean we can find N(d2) under standard Black-Scholes without Gaussian functions, provided we know N(d1)? If so, I think this would be useful. $\endgroup$ – David Addison Jul 28 '18 at 17:15
  • $\begingroup$ For a call in BS, $C = DF(0,T) ( F(0,T) N(d_1) - K N(d_2) )$ so indeed if you know $C$ and $N(d_1)$ (and the discount factor + forward price) you can find $N(d_2)$. But yes the formula I derive above can be used anyway. $\endgroup$ – Quantuple Jul 30 '18 at 8:20

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.