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Josip76
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Heston model probelm understanding Brownian motion correlation expression

March 5th, 2011, 6:42 am

Hi, I?m examining the Hesston model and whenever it is presented I see three equations two SDE for the price and volatility processes and one equation for the correlation between the Brownian motions. I?m having difficulty with the correlation equation understanding how the assumption that dWX_1(t) *dWX_2(t) = rho*dt was made. (sorry the equations are written poorly in text because I am not familiar with latex)To recap my understanding (too make sure it?s right): each of the coupled two-dimensional SDE must contain independent Brownian motions so to be able to apply the Girsanov theorem to change each equation to the risk neutral measure so that the discounted prices and volatility are both martingales. I am familiar with how to do this by expressing one of the Brownian motions in terms of a linear combination of Brownian motions. So if B and Z are correlated then Z could be expressed as a linear combination of B and another independent Brownian motion C (which serves only as an intermediary in this construction). As well as I am familiar with how to arrive at the result:Corr[WX_1 (t), WX_2 (t)] = E[WX_1 (t), WX_2 (t)] / t = rho Therefore, E[WX_1 (t), WX_2 (t)] = rho*tIf we call F the joint probability density function of WX_1 (t)*WX_2 (t) then we know that:E[WX_1 (t), WX_2 (t)] = (the double integral of) WX_1 (t) * WX_2 (t) * F * dWX_1(t) * dWX_2(t)It is from this point that I?m not sure how to get ? dWX_1(t) *dWX_2(t) = rho*dt ? and also why this assumption is important and where it will be used. Thanks for any help
 
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Alan
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Heston model probelm understanding Brownian motion correlation expression

March 5th, 2011, 6:46 pm

stick with the infinitesimals. dB1 dW = rho dt is just a convenience for dW = rho dB1 + sqrt(1-rho^2) dB2, (B1,B2) independent BM's.