When a 0 a similar result holds with cosh at replaced

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When A < 0 a similar result holds, with cosh( At ) replaced by cos( p | A | t ) etc. When A = 0, the generalized Laplace transform takes on a simpler form. The case where γ = 1 , B = 0 case can be found in [9]. In addition, if g = 0, then taking u 0 = 1 yields a fundamental solution p ( x, y, t ) with R 0 p ( x, y, t ) dy = 1 .
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FUNDAMENTAL SOLUTIONS 13 These Laplace transforms are less tractable than the previous two cases and so a detailed analysis of the inversion process will be given elsewhere. The drifts covered by this case include some very important examples, such as the so called double square root model of Longstaff, [16] which is an alternative to the Cox-Ingersoll-Ross model for interest rates dynamics. Thus this result is potentially very useful in financial modelling. 2.1. When is a Fundamental Solution a Transition Density? We have already seen that a fundamental solution of the Kolmogorov forward equation is not necessarily a transition density. We also require that the fundamental solution integrate to one. However, even this is not enough for the fundamental solution to be the desired density, as the following new result for squared Bessel processes shows. Proposition 2.10. Consider a squared Bessel process X = { X t : t 0 } of dimension 2 n . The Kolmogorov forward equation is u t = 2 xu xx + 2 nu x . (2.17) There are two linearly independent stationary solutions u 0 ( x ) = 1 and u 1 ( x ) = x 1 - n . For the stationary solution u 0 , the inverse Laplace trans- form of U λ ( t, x ) coming from Theorem 2.3 yields the transition density for a squared Bessel process of dimension 2 n. If n = 2 , 3 , 4 , .... and we use the second stationary solution u 1 to construct U λ ( t, x ) , then the inverse Laplace transform of the symmetry solution produces a second fundamental solution q ( t, x, y ) with R 0 q ( t, x, y ) dy = 1 . Moreover, if E q denotes expectation taken with respect to this fundamental solution, then E q £ ( X t ) 1 - n fl fl X 0 = x / = x 1 - n . Proof. Taking u 0 ( x ) = 1 gives U λ ( x, t ) = 1 (1+2 λt ) n exp ( - λx 1+2 λt ) , and inversion of this Laplace transform gives the well known transition density for a squared Bessel process of dimension 2 n . To prove the second claim, we use the fact that if u 1 ( x ) = x 1 - n then U λ ( x, t ) = x 1 - n (1 + 2 t λ ) n - 2 e - λx 1+2 λt . The fundamental solution will be q ( x, y, t ) = (2 t ) n - 2 y x · n - 1 e - x + y 2 t L - 1 £ λ n - 2 e k/λ / . (2.18) Here k = x (2 t ) 2 . Since n - 2 is a nonnegative integer, we have L - 1 £ λ n - 2 e k/λ / = n - 2 X l =0 k l l ! δ ( n - 2 - l ) ( y ) + k y n - 1 2 I n - 1 2 p ky · . (2.19) From which q ( x, y, t ) = 1 2 t y x · n - 1 2 e - x + y 2 t I n - 1 2 xy t + (2 t ) n - 2 y x · n - 1 n - 2 X l =0 x l δ ( n - 2 - l ) ( y ) (2 t ) 2 l l ! .
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14 MARK CRADDOCK Now Z 0 (2 t ) n - 2 y x · n - 1 n - 2 X l =0 x l (2 t ) 2 l l ! δ ( n - 2 - l ) ( y ) dy = 0 , since the Dirac delta functions and their derivatives select the value of the test function y n - 1 and its derivatives at zero. Also Z 0 1 2 t y x · n - 1 2 e - x + y 2 t I n - 1 2 xy t dy = 1 , as the integrand is simply the transition density of a squared Bessel process of dimension 2 n. To complete the proof we note that U 0 ( x, t ) = u 1 ( x ) and U λ ( x, t ) = Z 0 e - λy u 1 ( y ) q ( t, x, y ) dy. (2.20) Which implies that R
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