Integral Representations for the Hartman–Watson Density

Abstract

This paper concerns the density of the Hartman–Watson law. Yor (Z Wahrsch Verw Gebiete 53:71–95, 1980) obtained an integral formula that gives a closed-form expression of the Hartman–Watson density. In this paper, based on Yor’s formula, we provide alternative integral representations for the density. As an immediate application, we recover in part a result of Dufresne (Adv Appl Probab 33:223–241, 2001) that exhibits remarkably simple representations for the laws of exponential additive functionals of Brownian motion.

Appendix

We complete the proof of Lemma 2.1.

Proof of (ii) $\Rightarrow$ (iii) in Lemma 2.1 Given $x\in \mathbb {R}$, we integrate both sides of (2.3) multiplied by $e^{-r\cosh x}$ with respect to $r\ge 0$. Then by Fubini’s theorem, the left-hand side turns into that of (2.4). Therefore, it suffices to show that

$$\begin{aligned} \int _{0}^{\infty }\mathrm{d}r\,e^{-r\cosh x}\, \mathbb {E}\!\left[ G(B_{t})\cosh B_{t}\cos (r\sinh B_{t}) \right] =\mathbb {E}\!\left[ \frac{G(B_{t})}{\cosh (x+B_{t})} \right] . \end{aligned}$$

(A.1)

By the latter condition in (2.1), we may also use Fubini’s theorem to rewrite the left-hand side of the claimed identity (A.1) as

$$\begin{aligned}&\mathbb {E}\!\left[ G(B_{t})\cosh B_{t} \int _{0}^{\infty }\mathrm{d}r\,e^{-r\cosh x} \cos (r\sinh B_{t}) \right] \nonumber \\&\quad =\mathbb {E}\!\left[ G(B_{t}) \frac{\cosh x\cosh B_{t}}{\cosh ^{2}x+\sinh ^{2}B_{t}} \right] . \end{aligned}$$

(A.2)

On the other hand, by the symmetry of $G$, the right-hand side of (A.1) is equal to

$$\begin{aligned}&\frac{1}{2}\mathbb {E}\!\left[ G(B_{t})\left\{ \frac{1}{\cosh (x+B_{t})}+\frac{1}{\cosh (x-B_{t})} \right\} \right] \\&\quad =\mathbb {E}\!\left[ G(B_{t}) \frac{\cosh x\cosh B_{t}}{\cosh (x+B_{t})\cosh (x-B_{t})} \right] . \end{aligned}$$

Noting the fact that

$$\begin{aligned} \begin{aligned} \cosh (x+y)\cosh (x-y)&=\frac{1}{2}\left\{ \cosh (2x)+\cosh (2y) \right\} \\&=\cosh ^{2}x+\sinh ^{2}y \end{aligned} \end{aligned}$$

(A.3)

for any $x,y\in \mathbb {R}$, we compare the last expression with (A.2) to conclude identity (A.1). $\square$

We turn to the proof of the implication from (iii) to (i). To this end, we prepare the following lemma:

Lemma A.1

For every $x, b\in \mathbb {R}$, it holds that

$$\begin{aligned} \int _{\mathbb {R}} \frac{\mathrm{d}y}{\cosh (x+y)}\, \frac{1}{\cosh (2b)+\cosh (2y)} =\frac{\pi }{2\cosh b\left( \cosh b+\cosh x\right) }. \end{aligned}$$

Proof

We may assume $|x|\ne |b|$; validity in the case $|x|=|b|$ is verified by passing to the limit. By symmetrization and by the relation $\cosh (2b)+\cosh (2y)=2\left( \cosh ^{2}b+\sinh ^{2}y \right)$, the left-hand side of the claimed identity is equal to

$$\begin{aligned} \frac{1}{4}\int _{\mathbb {R}} \mathrm{d}y\left\{ \frac{1}{\cosh (x+y)}+\frac{1}{\cosh (x-y)} \right\} \frac{1}{\cosh ^{2}b+\sinh ^{2}y}, \end{aligned}$$

which is rewritten, due to relation (A.3), as

$$\begin{aligned}&\frac{\cosh x}{2}\int _{\mathbb {R}}\mathrm{d}y\, \frac{\cosh y}{ \left( \cosh ^{2}x+\sinh ^{2}y\right) \! \left( \cosh ^{2}b+\sinh ^{2}y\right) }\\&\quad =\frac{\cosh x}{2\left( \cosh ^{2}b-\cosh ^{2}x\right) } \int _{\mathbb {R}}\mathrm{d}z\left( \frac{1}{\cosh ^{2}x+z^{2}}-\frac{1}{\cosh ^{2}b+z^{2}} \right) \\&\quad =\frac{\cosh x}{2\left( \cosh ^{2}b-\cosh ^{2}x\right) } \left( \frac{\pi }{\cosh x}-\frac{\pi }{\cosh b} \right) , \end{aligned}$$

where we changed the variables with $\sinh y=z$ in the second line. Now the claimed identity follows. $\square$

We are prepared to finish the proof of Lemma 2.1.

Proof of (iii) $\Rightarrow$ (i) in Lemma 2.1 We appeal to the injectivity of Fourier transform. For this purpose, we first observe that

$$\begin{aligned} \int _{\mathbb {R}}\mathrm{d}x\,\mathbb {E}\!\left[ \frac{|F(B_{t})|\cosh B_{t}}{\cosh (2B_{t})+\cosh (2x)} \right]<\infty ,&\int _{\mathbb {R}}\mathrm{d}x\,\mathbb {E}\!\left[ \frac{|G(B_{t})|}{\cosh (x+B_{t})} \right] <\infty . \end{aligned}$$

(A.4)

Indeed, the former observation is immediate from (2.6) and the former condition in (2.1), while the latter is clear by the latter condition in (2.1). For an arbitrarily fixed $\xi \in \mathbb {R}$, we integrate both sides of (2.4) multiplied by $\cos (\xi x)$ with respect to $x\in \mathbb {R}$. Then by the latter finiteness in (A.4) and Fubini’s theorem, the right-hand side turns into

$$\begin{aligned} \mathbb {E}\!\left[ G(B_{t})\int _{\mathbb {R}}\mathrm{d}x\,\frac{\cos (\xi x)}{\cosh (x+B_{t})} \right]&=\frac{\pi }{\cosh (\frac{\pi }{2}\xi )}\mathbb {E}\!\left[ G(B_{t})\cos (\xi B_{t}) \right] , \end{aligned}$$

(A.5)

where we used the fact that

$$\begin{aligned} \int _{\mathbb {R}}\mathrm{d}x\,\frac{\cos (\xi x)}{\cosh x} =\frac{\pi }{\cosh (\frac{\pi }{2}\xi )}, \end{aligned}$$

(A.6)

which is verified by standard residue calculus. On the other hand, as for the left-hand side of (2.4), we have

$$\begin{aligned}&\int _{\mathbb {R}}\mathrm{d}x\,\cos (\xi x)\mathbb {E}\!\left[ \frac{F(B_{t})}{\cosh B_{t}+\cosh x} \right] \\&\quad =\frac{2}{\pi }\int _{\mathbb {R}}\mathrm{d}x\,\cos (\xi x) \mathbb {E}\!\left[ F(B_{t})\cosh B_{t} \int _{\mathbb {R}}\frac{\mathrm{d}y}{\cosh (x+y)} \frac{1}{\cosh (2B_{t})+\cosh (2y)} \right] \\&\quad =\frac{2}{\pi }\int _{\mathbb {R}}\mathrm{d}y\, \mathbb {E}\!\left[ \frac{F(B_{t})\cosh B_{t}}{\cosh (2B_{t})+\cosh (2y)} \right] \int _{\mathbb {R}}\mathrm{d}x\,\frac{\cos (\xi x)}{\cosh (x+y)}\\&\quad =\frac{2}{\cosh (\frac{\pi }{2}\xi )}\int _{\mathbb {R}}\mathrm{d}y\,\cos (\xi y) \mathbb {E}\!\left[ \frac{F(B_{t})\cosh B_{t}}{\cosh (2B_{t})+\cosh (2y)} \right] , \end{aligned}$$

where we used Lemma A.1 for the second line, Fubini’s theorem for the third thanks to the former finiteness in (A.4), and fact (A.6) for the fourth. Since the last expression agrees with (A.5) for any $\xi \in \mathbb {R}$ and the function $G$ is assumed to be symmetric, the injectivity of Fourier transform entails relation (2.2). The proof completes. $\square$

Remark A.1

By using Lemma A.1, implication (i) $\Rightarrow$ (iii) may also be proven in the same manner as in the proof of (i) $\Rightarrow$ (ii) given in Sect. 2.