Estimation in Financial Models - RiskLab

If we do not specialize at this point, we are not able to calculate much more
further and therefore we restrict ourselves to two cases.
First, we consider the case a 1 (t) b 1 (t) 0 and denote a 2 (t) =: (t)
and b 2 (t) =: (t). In order to obtain conditions for a and b it is useful
to dierentiate equation (5.19) with respect to x and equation (5.20) with
respect to t:
!
@ 2 U @
(t; x)+ a(t; x)@U
@t@x @x @x (t; x)+1 2 b2 (t; x) @2 U
@x (t; x) =0 (5.21)
2
and
b(t; x) @2 U
(t; x)+@b(t; x)@U
@t@x @t @x (t; x) =0 (t): (5.22)
By means of (5.21) and (5.22) we obtain
"
0 1 @b(t; x)
(t) =(t)
(t; x) , b(t; x) @ a(t; x)
b(t; x) @t
@x b(t; x) , 1 !#
@b
2 @x (t; x) :
(5.23)
In (5.21) and hence in (5.23) we assumed that the left hand side of (5.19) is
independent of x, in other words, we assumed that (t) can be determined.
Since is independent ofx we now follow the condition for the determination
of from equation (5.23)
@f
(t; x) =0; (5.24)
@x
where
f(t; x) = 1
b(t; x)
@b(t; x)
(t; x) , b(t; x) @
@t
@x
a(t; x)
b(t; x) , 1 !
@b
2 @x (t; x) : (5.25)
If condition (5.24) is satised we are able to determine and . Thus condition
(5.24) is sucient for the reducibility of equation (5.16) to the linear
equation
dX t = (t)dt + (t)dW t (5.26)
by means of a transformation U. Integrating equations (5.20) and (5.23) we
obtain the explicit expression
U(t; x) =C exp
with an arbitrary constant C.
Z t
0
Z x
f(s; x)ds
0
1
dz; (5.27)
b(t; z)
64