This is of course the same result we obtained from

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This is of course the same result we obtained from Theorem 2.3 with u 0 = 1 . 3. Some Applications and Examples Now that we are able to determine when a fundamental solution is a transition density, we are in a position to give some examples. Of course the transition densities will always arise from taking u 0 = 1 as the stationary solution. This is obvious in practice, though rather more
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16 MARK CRADDOCK difficult to prove in general. The virtue of our results is that they are extremely easy to use. Despite the apparently complicated expressions for U ² in Theorem 2.5, their Laplace transforms can easily be inverted. Example 3.1. We will now obtain the transition density for the Cox- Ingersoll-Ross process of interest rate modelling from Theorem 2.5. That is, we will obtain the transition density for the process X = { X t : t 0 } where dX t = ( a - bX t ) dt + p 2 σX t dW t , X 0 = x. (3.1) We assume that a is positive and real for convenience. The drift is a solution of σxf 0 - σf + 1 2 f 2 = 1 2 Ax 2 + Bx + C with A = b 2 , B = - ab, C = 1 2 a 2 - aσ. We require a fundamental solution of u t = σxu xx + ( a - bx ) u x , (3.2) which is positive and integrates to one. Using Theorem 2.5, with u 0 ( x ) = 1 , F ( x ) = a ln x - bx and after some cancellations, we arrive at Z 0 e - λy p ( x, y, t ) dy = b a σ e ab σ t ( λσ ( e bt - 1) + be bt ) a σ exp - λbx λσ ( e bt - 1) + be bt . (3.3) We easily invert this Laplace transform to obtain p ( x, y, t ) = be b ( a σ +1) t σ ( e bt - 1) y x · ν 2 exp - b ( x + e bt y ) σ ( e bt - 1) I ν ˆ b xy σ sinh ( bt 2 ) ! , (3.4) with ν = a σ - 1 . This is the transition density for the CIR process. We can check this using Proposition 2.11. Example 3.2. Consider now the diffusion X = { X t : t 0 } satisfying the SDE dX t = 2 X t tanh( X t ) + p 2 X t dW t , X 0 = x > 0 . (3.5) Then the transition density is a fundamental solution of u t = xu xx + 2 tanh( x ) u x . The drift satisfies xf 0 - f + 1 2 f 2 = 2 x 2 . So A = 4 , B = C = 0 . An application of Theorem 2.5 with u 0 = 1 reveals the existence of a fundamental solution p ( x, y, t ) such that Z 0 e - λy p ( x, y, t ) dy = cosh( x 1+ λ sinh( t )(2 cosh( t )+ λ sinh( t )) ) cosh( x ) × exp - λx (cosh(2 t ) + λ cosh( t ) sinh( t )) 1 + λ sinh( t ) (2 cosh( t ) + λ sinh( t )) .
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FUNDAMENTAL SOLUTIONS 17 Inversion of this Laplace transform requires a certain amount of work, which we omit for the sake of brevity. The result is that p ( x, y, t ) = exp n - x + y tanh( t ) o sinh( t ) cosh( y ) cosh( x ) r x y I 1 2 xy sinh t + sinh( t ) δ ( y ) . This is the transition density for X satisfying (3.5). An important application of our results is to the calculation of Laplace transforms of joint densities. We present some examples. Example 3.3. We will find a fundamental solution of the PDE u t = xu xx + ( 1 2 + x ) u x - μ x u, x 0 , 0 μ < 15 16 , which at μ = 0 reduces to the transition density of the process X = { X t : t 0 } , where dX t = ( 1 2 + X t ) dt + 2 X t dW t . To obtain such a fundamental solution, we require a stationary solution u μ 0 ( x ) with the property that u 0 0 ( x ) = 1. Such a stationary solution is not difficult to find. It is u μ 0 ( x ) = x 1 4 e - x ( I α ( x ) + I - α ( x ), where α = 1 2 1 + 16 μ.
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