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Hi,Need some help with stochastic calculus...I'm currently investigating stochastic volatility models and am interested in the dynamics of the implied volatility curve that go with the model. In particular: I am looking at the Heston model and want to know the derivative of the instantaneous variance v(t) with respect to the underlying price S(t).. if such a thing exists. Following intuitive arguments I got the following expression:Exptected value [ dv(t)/dS(t) ] = rho * sigma / S(t), with rho the correlation of the two driving Wiener processes in the Heston model, sigma the volatility of volatility, and S(t) the price of the underlying at time t. Obviously, for time to infinity the derivative goes to zero, and for time to zero, the derivative goes to rho * sigma / S(0).I implemented the Milstein method, simulated the Heston model and calculated the mean of the derivative (with a minor adaptation that I capped and floored the derivative so that the mean could not be influenced too much by a spike in the derivative.. which happens when dS ~= 0). The results easily confirm the expression above.The problem is: how do I proof this mathematically..? If anyone has any ideas or thoughts about a mathematically tight proof.. leave a comment!Currently I think I may have found a foreward proof... 1. Rewrite the SDE of the underlying stock process to get the dWt - term on one side;2. Write the SDE of the instantaneous variance as a combination of two uncorrelated Wiener processes dWt and dZt;3. Plug in the dWt - term from the first step;4. Devide both sides of the subsequent expression by dS(t) .. (one side gives dv(t)/dS(t), other side gives two dt/dS(t) terms, one 1/S(t) term, and one dZt/dWt term);5. Assume that dt/dS(t) is zero;6. take expectation of both sides.. (the dZt/dWt term is zero in expectation due to no dependence / correlation=0);7. answer: Exptected value [ dv(t)/dS(t) ] = rho * sigma / S(t)...The thing I am not sure about is that in this proof I have assumed that dt/dS(t) is zero.. this would seem logical to me, given that dS(t)/dt is either plus or minus infinity. However, strictly speaking the time-derivative of a Wiener process is mathematically not defined, so I am not sure if I can make this assumption. If anyone knows how to proof that dt/dS(t) = 0, I think the above given proof is valid. Any thoughts?Thanks!!

I don't think what you have written exists.What exists that is vaguely similar is:Expected value [ (dv(t) dS(t))/ dt ] = rho * sigma * v(t) * S(t), using dW1(t) dW2(t) = rho dtor Expected value [ d (v(t)/S(t))/dt ], which you can work out using Ito's lemma.

Hi Alan,Thanks for your feedback. I agree with you on the existence of your expressions. However, they are not specifically useful for my needs.. What I would like to have is a difference function for the underlying S(t) and the variance v(t), such as delta_v(t)/delta_S(t)=... (or delta_S(t)/delta_v(t) =..). The reason I would like to have this, is that I can calculate how the implied volatility curve of the Heston model reacts to spot-changes (the vol-spot relation).Furthermore.. since simulations confirm my intuition.. wouldn't you agree that there should exist such an expression ?PS, I wrote LaTeX code for the proof in my first post; however, I cannot seem to attach pdf or dvi document... pasting the code in the equation editor doesn't work either.. how's this usually done @Wilmott?

zip-up the pdf; then upload that as an attachment.

Jeuj! Found a correct derivation of the expression as defined in the minimum-variance delta..minimum-variance delta = total delta = dP(t)/dS(t) = dP(t)/dS(t) + dP(t)/dv(t) * dv(t)/dS(t)A paper of Carel Alexander (see section three of http://papers.ssrn.com/sol3/papers.cfm? ... _id=763204) derives the minimum variance delta for the Heston model, and the dv(t)/dS(t) term can easily be extracted from this. Regards