Quantitative Analysis.
Trading Platform.
Python for Excel.
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I.Basic math.
II.Pricing and Hedging.
III.Explicit techniques.
1.Black-Scholes formula.
2.Change of variables for Kolmogorov equation.
3.Mean reverting equation.
4.Affine SDE.
5.Heston equations.
A.Affine equation approach to integration of the Heston equations.
B.PDE approach to integration of the Heston equations.
6.Displaced Heston equations.
7.Stochastic volatility.
8.Markovian projection.
9.Hamilton-Jacobi Equations.
IV.Data Analysis.
V.Implementation tools.
VI.Basic Math II.
VII.Implementation tools II.
Bibliography.
Forum Notation Index Contents

Affine equation approach to integration of the Heston equations.


he Heston equations are of the affine type. Hence, we proceed as discussed in ( section about Affine equation).

We transform the equations to the $a,b$ -form ( Affine equation). Let $Y_{t}=\ln S_{t}$, MATHMATH henceMATH In particular,MATH According to the summary ( Affine characteristic function 1)-( Affine boundary conditions) the functionMATH is given by the expressionMATH where the functions MATH and MATH must satisfy the following system of ODEs:MATHMATHMATH We transform the above relationships:MATH The MATH are to satisfy the final conditions:MATH Hence, the next task is to solve the equationMATH We introduce the convenience notationMATH and writeMATH We perform the change of the unknown functionMATH as followsMATH We further perform the transformationMATH as below MATH but alsoMATH Hence,MATH and consequentlyMATH We arrived to a linear equation. Therefore, we look for solutions of the formMATH Hence, it is enough to haveMATH We mean to integrate over $z$ because this is the argument of the characteristic function and we want to take the inverse Fourier or Laplace transform. Note that when $z=0$ the expression under the square root is positive. For large $z$ it is negative. Hence, we would rather change $z=i\zeta$ for real $\zeta$. Then the square root is never zero. Indeed, the real part would be MATH, where the correlation $\rho$ is not greater then 1.

We perform the backward substitutionsMATH We introduce the quantity $k$ according to the relationship MATH and transform the expression for $\beta_{2}$MATH to emphasize that the requirement MATH uniquely identifies the $\beta_{2}$.

We now consider the equation for $\alpha$:MATH where the $\beta_{1}$ is a constant and the $\beta_{2}$ has just been calculated:MATH We haveMATH henceMATH It remains to evaluate the integralMATH Since $\alpha$ satisfies the final conditionMATH we haveMATH





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