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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).

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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