Quantitative Analysis
Parallel Processing
Numerical Analysis
C++ Multithreading
Python for Excel
Python Utilities
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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 Heston equations.
B. PDE approach to integration of 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.
VIII. Bibliography
Notation. Index. Contents.

Affine equation approach to integration of Heston equations.

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

We transform the equations to the $a,b$ -form ( Affine equation ). Let $Y_{t}=\ln S_{t}$ , MATH MATH hence MATH In particular, MATH According to the summary ( Affine characteristic function 1 )-( Affine boundary conditions ) the function MATH is given by the expression MATH where the functions MATH and MATH must satisfy the following system of ODEs: MATH MATH MATH We transform the above relationships: MATH The MATH are to satisfy the final conditions: MATH The next task is to solve the equation MATH We introduce the convenience notation MATH and write MATH We perform the change of the unknown function MATH as follows MATH We perform the transformation MATH as follows: MATH In addition, MATH Hence, MATH and, consequently, MATH We arrived to a linear equation. We look for solutions of the form MATH It suffices to have MATH 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 substitutions MATH 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 have MATH hence MATH It remains to evaluate the integral MATH Since $\alpha$ satisfies the final condition MATH we have MATH

Notation. Index. Contents.

Copyright 2007