Λx 1 t 2 σ u ye λy px y tdy exp 4 1 λt cosh x 1

Info icon This preview shows pages 4–6. Sign up to view the full content.

View Full Document Right Arrow Icon
λ(x + 1 t 2 ) Σ u 0 (y)e λy p(x, y, t)dy = √ exp 4 . 0 (1 + λt) cosh( x) 1 + λt Inversion of the Laplace transform gives the fundamental solution 1 . Σ t 4 t cosh( x) exp y t (x, y, t) = . t I α . t I α t 4 t cosh( x) exp y t (x, y, t) = This reduces to the transition density of the process at µ = 0. (See [9] for the derivation of the transition density). This fundamental so- lution is integrable near zero for 0 ≤ µ < 1 . We may calculate by the Feynman-Kac formula E x Σ e λX t µ t ds 0 X s = e λy p(x, y, t)dy but 0 this integral again does not seem to be easy to evaluate analytically. Of course it is possible to evaluate it numerically. We leave it to the reader to show that for X = {X t : t ≥ 0} where dX 1 X coth( X )) dt + 2 X dW , we may solve the equation u t = xu xx + ( 1 + x coth( x))u x µ u to obtain the fundamental solution 1 y t (x, y, t) = 2 x . Σ , t I t 4 t 0 X s 2
Image of page 4

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
at µ = 0. y = 0 for all µ > 1 . 2 x t 0 X s . Σ 0 X s 2 4 y d p(x, y, t)e λy dy = p(x, y, t) = 2t 2 2 I 2 d + n 1 = e 2 t +4 t 2 λ Γ(β)(1 + t 0 X s x , t I t 4 t sinh( x) exp y t (x, y, t) = where α = 1+16 µ . This reduces to the transition density of the process It is integrable at 16 The range of µ values in these examples for which the fundamental solution is integrable at y = 0 depends on the properties of the process. For µ outside the given ranges we may find other fundamental solutions, and investigate their properties. Some will contain distributional terms. A full study of this is beyond the scope of the current paper. Example 5. Let X = { X t : t 0 } b e a squared Bessel pro- cess, where dX t = ndt + 2 X t dW t , X 0 = x. The joint density of (X , t ds ) arises in the pricing of Asian options and other prob- lems, see [6]. To obtain its Laplace transform we require a funda- mental solution for the PDE u t = 2xu xx + nu x µ u. From Theorem 2.3 the reader may check that the stationary solution u 0 (x) = x d , where d = 1 (2 n + ( n 2) 2 + 8 µ
Image of page 5
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern