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Doe dC/dS (delta) = dC/dX ?

Posted: April 8th, 2016, 4:52 pm
by yeahmoon
In general, what the relationship between delta ([$]\frac{\partial C}{\partial S}[$]) and [$]\frac{\partial C}{\partial X}[$], where [$]X[$] is the strike price?Do we have [$]\frac{\partial C}{\partial S}[$] [$] =- \frac{\partial C}{\partial X}[$]?In the Black-Schole world, [$]\frac{\partial C}{\partial S} = N(d_1)[$] and [$]\frac{\partial C}{\partial X}=-e^{-rT}N(d_2)[$], it is not obvious that [$]\frac{\partial C}{\partial S}[$] [$] = -\frac{\partial C}{\partial X}[$].What is behind here? Any books/papers?Thanks!

Doe dC/dS (delta) = dC/dX ?

Posted: April 8th, 2016, 5:46 pm
by Cuchulainn
QuoteOriginally posted by: yeahmoonIn general, what the relationship between delta ([$]\frac{\partial C}{\partial S}[$]) and [$]\frac{\partial C}{\partial X}[$], where [$]X[$] is the strike price?Do we have [$]\frac{\partial C}{\partial S}[$] [$] =- \frac{\partial C}{\partial X}[$]?In the Black-Schole world, [$]\frac{\partial C}{\partial S} = N(d_1)[$] and [$]\frac{\partial C}{\partial X}=-e^{-rT}N(d_2)[$], it is not obvious that [$]\frac{\partial C}{\partial S}[$] [$] = -\frac{\partial C}{\partial X}[$].What is behind here? Any books/papers?Thanks!Espen Haug Options book pages 22-25 has a bunch of greeks.

Doe dC/dS (delta) = dC/dX ?

Posted: April 8th, 2016, 5:48 pm
by Cuchulainn
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: yeahmoonIn general, what the relationship between delta ([$]\frac{\partial C}{\partial S}[$]) and [$]\frac{\partial C}{\partial X}[$], where [$]X[$] is the strike price?Do we have [$]\frac{\partial C}{\partial S}[$] [$] =- \frac{\partial C}{\partial X}[$]?In the Black-Schole world, [$]\frac{\partial C}{\partial S} = N(d_1)[$] and [$]\frac{\partial C}{\partial X}=-e^{-rT}N(d_2)[$], it is not obvious that [$]\frac{\partial C}{\partial S}[$] [$] = -\frac{\partial C}{\partial X}[$].What is behind here? Any books/papers?Thanks!Espen Gaarder Haug Options book pages 22-25 has a bunch of greeks.EGH has this for stock options[$]\frac{\partial C}{\partial S} = N(d_1)[$] and [$]\frac{\partial C}{\partial X}=-e^{-rT}N(d_2)[$],

Doe dC/dS (delta) = dC/dX ?

Posted: April 8th, 2016, 6:20 pm
by LocalVolatility
They are in general not equal. Merton (1973), Theorem 9, showed however that if the underlying asset exhibits constant returns to scale, then the European call price is homogeneous of degree one in the underlying asset price and the strike price, i.e.[$]C = S \frac{\partial C}{\partial S} + K \frac{\partial C}{\partial K}[$]Most common models exhibit constant returns to scale. One exception is the local vol. model (and its stochastic local vol. relatives).

Doe dC/dS (delta) = dC/dX ?

Posted: April 9th, 2016, 7:25 am
by Cuchulainn
QuoteOriginally posted by: LocalVolatilityThey are in general not equal. Merton (1973), Theorem 9, showed however that if the underlying asset exhibits constant returns to scale, then the European call price is homogeneous of degree one in the underlying asset price and the strike price, i.e.[$]C = S \frac{\partial C}{\partial S} + K \frac{\partial C}{\partial K}[$]Most common models exhibit constant returns to scale. One exception is the local vol. model (and its stochastic local vol. relatives).Maybe a stupid question: is there no dependence on t? What the above 1st order PDE useful for/used for? Can it be solved for C?

Doe dC/dS (delta) = dC/dX ?

Posted: April 9th, 2016, 11:30 am
by LocalVolatility
The time dependence is in the derivatives, I just didn't make that explicit. As you noted before in the Black Scholes setting you have[$]\frac{\partial C}{\partial S} = \mathcal{N} \left( d_1 \right)[$],where [$]d_1[$] depends on time.This relationship is quite useful to directly infer the delta form a pricing formula for European call options without having to carefully take the partial derivatives. I.e. if you look at the Heston paper you find a valuation function of the form[$]C(S, v, t) = S P_1 - K P(t, T) P_2[$],where [$]P_1[$] and [$]P_2[$] are some integrals that both depend on [$]S[$]. From the homogeneity result we can directly conclude that[$]\frac{\partial C}{\partial S} = P_1[$].