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
form (
Affine equation
). Let
,
hence
In
particular,
According to the summary (
Affine
characteristic function 1
)(
Affine
boundary conditions
) the
function
is given by the
expression
where the functions
and
must satisfy the following system of
ODEs:
We transform the above
relationships:
The
are to satisfy the final
conditions:
The next task is to solve the
equation
We introduce the convenience
notation
and
write
We perform the change of the unknown
function
as
follows
We perform the
transformation
as follows:
In
addition,
Hence,
and,
consequently,
We arrived to a linear equation. We look for solutions of the
form
It suffices to
have
We mean to integrate over
because this is the argument of the characteristic function and we want to
take the inverse Fourier or Laplace transform. Note that when
the expression under the square root is positive. For large
it is negative. Hence, we would rather change
for real
.
Then the square root is never zero. Indeed, the real part would be
,
where the correlation
is not greater then 1.
We perform the backward
substitutions
We introduce the quantity
according to the relationship
and transform the expression for
to emphasize that the requirement
uniquely identifies the
.
We now consider the equation for
:
where the
is a constant and the
has just been
calculated:
We
have
hence
It remains to evaluate the
integral
Since
satisfies the final
condition
we
have
