Japan Journal of Industrial and Applied Mathematics

, Volume 35, Issue 3, pp 1303–1308 | Cite as

Correction to: Towards the exact simulation using hyperbolic Brownian motion

  • Yuuki IdaEmail author
  • Yuri Imamura

1 Correction to: Japan J. Indust. Appl. Math. (2017) 34:833–843

The authors would like to correct the errors in the publication of the original article. The corrected details are given below for your reading.

Theorem 2 (i) claims that the derivative with respect to x of logarithm of the heat kernel of \({\mathbb {H}}^2 \) is bounded, but its proof presented in the paper has the following gap. In Line-7 of page 837, it is claimed that,
$$\begin{aligned} e^t(2\pi )p_4 (t,r) \le \frac{3}{2}\frac{1}{1+\cosh (r)} p_2 (t,r), \end{aligned}$$
which is implied by Gruet’s formula:
$$\begin{aligned} p_n (t,r) = \frac{e^{-(n-1)^2t/8}}{\pi (2\pi )^{n/2}t^{1/2}} \Gamma \left( \frac{n+1}{2} \right) \int _0^\infty \frac{e^{(\pi ^2-b^2)/2t}\sinh (b)\sin (\pi b/t)}{(\cosh (b)+\cosh (r))^{(n+1)/2}}db, \end{aligned}$$
where \( p_n (t, r(z,z')) \) is the heat kernel of \({\mathbb {H}}^n \), \( r (z,z') \) being the hyperbolic distance between z and \( z' \). However, since the integrand in the left-hand-side of (1) is not always positive, the inequality is not guaranteed.

Actually the claimed inequality is too sharp to prove, we should at this stage admit that Theorem 2 is an error. Instead here we present a corrected version.

$$\begin{aligned} \theta (t,(x,y),(x',y')):=\mu (x,y)\frac{\frac{\partial q_2}{\partial x}(t,(x,y),(x',y'))}{q_2(t,(x,y),(x',y'))} \end{aligned}$$

Theorem 2

  1. (i)
    For \(z=(x,y),z'=(x',y')\in {\mathbb {H}}^2\), we have that
    $$\begin{aligned} |\partial _x q_2(t,z, z' )| \le C' (yy')^{-\frac{1}{2}}t^{-\frac{1}{2}} q_2 (2t,z, z') \end{aligned}$$
    for some constant \( C' >0 \), and for each \(n\ge 2\), \(t>0\) and \((s_1,\ldots , s_{n-1})\in \Delta _{n-1}(t)\)
    $$\begin{aligned} {\mathbb {E}}\left[ \prod _{i=1}^n\left. \theta \left( s_i-s_{i-1},Z_{s_{i-1}}^0,Z_{s_i}^0\right) \right| Z^0_0=z, Z_t^0=z' \right] q(t,z,z') (y')^{-2} \in L^1(\Delta _{n-1}(t)) \end{aligned}$$
    where \(s_0=0,s_n=t\).
  2. (ii)
    $$\begin{aligned} h_1(t,z,z')=\mu (x,y)\frac{\partial }{\partial x}q_2(t,z,z') (y')^{-2}, \end{aligned}$$
    $$\begin{aligned} \begin{aligned} h_n (t,z,z')&:= \int _{\Delta _{n-1} (t)} {\mathbb {E}} \left[ \prod _{i=1}^n \theta (s_i-s_{i-1}, Z^0_{s_{i-1}}, Z^0_{s_i})|Z^0_0=z, Z^0_t=z' \right] \\&\qquad \times q_2 (t,z,z')/(y')^2 ds_1 \ldots ds_{n-1} \end{aligned} \end{aligned}$$
    for \(n\ge 2\). Then, the series \( \sum _{n=1}^N h_n (t, z,z')\) is absolutely convergent as \( N \rightarrow \infty \) uniformly in \( (t,z, z') \) on every compact set.
  3. (iii)
    The transition density of \( Z^{\mu } \) is given by
    $$\begin{aligned} s(t,z,z'):=\frac{q_2(t,z,z')}{(y')^2} +\int _{\mathbb {H}^2}\int ^t_0 \frac{q_2(t-s,z,z'')}{(y'')^2} \Phi (s,z'',z')dsdz'', \end{aligned}$$
    where \(\Phi (t,z,z')=\sum ^\infty _{n=1}h_n(t,z,z')\).

To prove Theorem 2, we use the following


[1].  Let \(p_n\) be the heat kernel of the hyperbolic space \({\mathbb {H}}^n\), and
$$\begin{aligned} k_{n+1}(t,r):=(2\pi t)^{-\frac{n+1}{2}} e^{-\frac{n^2}{8}t-\frac{r^2}{2t}-\frac{nr}{2}} (1+r+t)^{\frac{n}{2}-1}(1+r), \end{aligned}$$
for \(n\ge 1\), \( t > 0 \), and \( r > 0 \). For any integer \(n>1\), then
$$\begin{aligned} p_n(t,r)\sim k_n(t,r), \end{aligned}$$
uniformly in \(r \ge 0\) and \(t> 0\).

Proof of Theorem 2

By (4) and (5), we have in particular,
$$\begin{aligned} p_2(t,r)\sim \frac{e^{-\frac{t}{8} -\frac{r^2}{2t}-\frac{r}{2}}}{2\pi t} (1+r+t)^{-\frac{1}{2}} (1+r), \end{aligned}$$
$$\begin{aligned} p_4(t,r)\sim \frac{e^{-\frac{9t}{8}-\frac{r^2}{2t}-\frac{3r}{2}}}{(2\pi t)^2} (1+r+t)^{\frac{1}{2}}(1+r). \end{aligned}$$
To start with, we calculate the derivative with respect to x of \(q_2\). Using Milson’s formula, we have
$$\begin{aligned} \begin{aligned} \left| \frac{\partial q_2}{\partial x}(t,(x,y),(x',y')) \right| = \frac{|x-x'|}{yy'} 2\pi e^{t}p_4(t,r). \end{aligned} \end{aligned}$$
$$\begin{aligned} \frac{|x-x'|}{yy'}2\pi e^{t}p_4(t,r)&\le K \frac{|x-x'|}{yy'} 2\pi e^{t} \frac{e^{-\frac{9t}{8}-\frac{r^2}{2t}-\frac{3r}{2}}}{(2\pi t)^2} (1+r+t)^{\frac{1}{2}}(1+r)\nonumber \\&=K \frac{|x-x'|}{yy'} \frac{e^{-\frac{t}{4}-\frac{r^2}{4t}-\frac{r}{2}-\frac{r^2}{4t}+\frac{t}{8}-r}}{2\pi t^2} (1+r+t)^{\frac{1}{2}}(1+r)\nonumber \\&\le K' \frac{|x-x'|}{yy'} (1+r+t)^{\frac{1}{2}} (1+r+2t)^{\frac{1}{2}} \frac{e^{-\frac{r^2}{4t}+\frac{t}{8}-r}}{t} p_2(2t,r)\nonumber \\&\le K' \frac{|x-x'|}{yy'} (1+r+2t) \frac{e^{-\frac{r^2}{4t}+\frac{t}{8}-r}}{t} p_2(2t,r)\nonumber \\&= \frac{K'|x-x'|}{yy'} \frac{e^{-\frac{r^2}{8t}}}{r} (1+r+2t) \frac{re^{-\frac{r^2}{8t}+\frac{t}{8}-r}}{\sqrt{t}} \frac{p_2(2t,r)}{\sqrt{t}}, \end{aligned}$$
where we have used (5) twice. We shall estimate the last term of (6). First, we observe that
$$\begin{aligned} \begin{aligned} \frac{|x-x'|}{yy'}&=\sqrt{\frac{2}{yy'}} \sqrt{\frac{|x-x'|^2}{2yy'}} \\&\le \sqrt{\frac{2}{yy'}} \sqrt{\frac{|x-x'|^2+(y-y')^2}{2yy'}} \\ {}&=\sqrt{\frac{2}{yy'}} \sqrt{\cosh (r)-1}\\&\le \sqrt{\frac{2}{yy'}} \sqrt{\cosh ^2(r)-1} =\sqrt{\frac{2}{yy'}} \sinh (r). \end{aligned} \end{aligned}$$
Then we see that
$$\begin{aligned} \begin{aligned} \frac{|x-x'| }{r\sqrt{yy'}} e^{-\frac{r^2}{8t}} \end{aligned} \end{aligned}$$
is bounded since \( \lim _{r \rightarrow 0} \sinh (r)/r = 1 \) and \( \lim _{r \rightarrow \infty } \sinh (r)e^{-\frac{r^2}{8t}}/r = 0 \). Noting that \((1+r+2t)e^{\frac{t}{8}-r}\), and \( \frac{r}{\sqrt{t}} e^{-\frac{r^2}{8t}} \) with \( (t,r) \in [0,T] \times [0,\infty ) \), are bounded, we obtain (2).
Subsequently, we see that \(h_1\) is bounded by \(Cq_2/\sqrt{yy't}\) with some constant C. Therefore,
$$\begin{aligned} \begin{aligned}&|h_n(t,z,z')|\\&\quad \le \int _{\Delta _{n-1}(t)} {\mathbb {E}} \left[ \prod _{i=1}^n \left| \theta (s_i-s_{i-1}, Z^0_{s_{i-1}}, Z^0_{s_i}) \right| \left| Z^0_0=z, Z^0_t=z'\right. \right] \\&\qquad \times q_2 (t,z,z')/(y')^2 ds_1 \ldots ds_{n-1}\\&\quad \le \int _{\Delta _{n-1}(t)} \int _{\mathbb {H}}\cdots \int _{\mathbb {H}} |\mu (z)|\left| \frac{\partial _xq_2(s_1,z,z_1)}{(y_1)^2} \right| \times \cdots \times |\mu (z_{n-1})|\left| \frac{\partial _xq_2(t-s_{n-1},z_{n-1},z')}{(y')^2} \right| \\&\qquad \times dz_1\ldots dz_{n-1} ds_1 \ldots ds_{n-1}\\&\quad \le \int _{\Delta _{n-1}(t)}\int _{\mathbb {H}}\cdots \int _{\mathbb {H}} |\mu (z)\times \cdots \times \mu (z_{n-1})|\\&\qquad \times \frac{q_2(2s_1,z,z_1)/(y_1)^2 \times \cdots \times q_2(2(t-s_{n-1}),z_{n-1},z')/(y')^2}{\sqrt{y}y_1\ldots y_n\sqrt{y'}\sqrt{s_1\times \cdots \times (t-s_{n-1})}} \\&\qquad \times dz_1\ldots dz_{n-1} ds_1 \ldots ds_{n-1}\\&\quad \le \int _{\Delta _{n-1}(t)}\int _{\mathbb {H}}\cdots \int _{\mathbb {H}} K_0^nC^n\sqrt{\frac{y}{y'}} \frac{q_2(2s_1,z,z_1)/(y_1)^2 \times \cdots \times q_2(2(t-s_{n-1}),z_{n-1},z')/(y')^2}{\sqrt{s_1\times \cdots \times (t-s_{n-1})}} \\&\qquad \times dz_1\ldots dz_{n-1} ds_1 \ldots ds_{n-1}. \end{aligned} \end{aligned}$$
Using the Chapman–Kolmogorov equation, we then obtain that
$$\begin{aligned} \begin{aligned} |h_n(t,z,z')|&\le K_0^nC^n\sqrt{\frac{y}{y'}}\frac{q_2(2t,z,z')}{(y')^2}\int _{\Delta _{n-1}(t)} \frac{1}{\sqrt{s_1\times \cdots \times (t-s_{n-1})}} ds_1 \ldots ds_{n-1} \\&=K_0^nC^n \sqrt{\frac{y}{y'}}\frac{q_2(2t,z,z')}{(y')^2}t^{\frac{n-1}{2}} \prod ^n_{i=1}\beta \left( \frac{i}{2},\frac{1}{2}\right) . \end{aligned} \end{aligned}$$
Here we have in particular shown (3).
If \(n=2m\) with some natural number m, we have
$$\begin{aligned} \prod ^{2m}_{i=1}\beta \left( \frac{i}{2},\frac{1}{2}\right)&=\prod ^{2m}_{i=1} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) }\\&=\prod _{i=2k,k\in \{1\cdots m\}} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) } \prod _{i=2k-1,k\in \{1\cdots m\}} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) }\\&=\prod ^m_{k=1} \frac{\Gamma \left( k\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{2k+1}{2}\right) } \prod ^m_{k=1} \frac{\Gamma \left( \frac{2k-1}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( k\right) }\\&=\prod ^m_{k=1}\left( \frac{\Gamma \left( k\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{2k+1}{2}\right) } \frac{\Gamma \left( \frac{2k-1}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( k\right) } \right) \\&= \pi ^{m} \prod ^m_{k=1}\left( \frac{ \Gamma \left( \frac{2k-1}{2}\right) }{\frac{2k-1}{2}\Gamma \left( \frac{2k-1}{2}\right) } \right) = \pi ^{m} \prod ^m_{k=1}\left( \frac{1 }{\frac{2k-1}{2}} \right) =\frac{\pi ^{m-1/2}}{\Gamma \left( \frac{2m+1}{2} \right) }. \end{aligned}$$
On the other hands, when \(n=2m-1\) for any \(m\in {\mathbb {N}}\), we have
$$\begin{aligned} \begin{aligned} \prod ^{2m-1}_{i=1}\beta \left( \frac{i}{2},\frac{1}{2}\right)&=\prod ^{2m-1}_{i=1} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) }\\&=\prod _{i=2k,k\in \{1\cdots m-1\}} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) } \prod _{i=2k-1,k\in \{1\cdots m\}} \frac{\Gamma \left( \frac{i}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{i+1}{2}\right) }\\&=\prod ^{m-1}_{k=1} \frac{\Gamma \left( k\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{2k+1}{2}\right) } \prod ^m_{k=1} \frac{\Gamma \left( \frac{2k-1}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( k\right) }\\&=\frac{\Gamma \left( \frac{2m-1}{2}\right) }{\Gamma (m)} \prod ^{m-1}_{k=1}\left( \frac{\Gamma \left( k\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( \frac{2k+1}{2}\right) } \frac{\Gamma \left( \frac{2k-1}{2}\right) \Gamma \left( \frac{1}{2}\right) }{\Gamma \left( k\right) } \right) \\&=\frac{\pi ^{m-1}\Gamma \left( \frac{2m-1}{2}\right) }{\Gamma (m)} \prod ^{m-1}_{k=1}\frac{ \Gamma \left( \frac{2k-1}{2}\right) }{\frac{2k-1}{2}\Gamma \left( \frac{2k-1}{2}\right) }\\&=\frac{\pi ^{m-1}\Gamma \left( \frac{2m-1}{2}\right) }{\Gamma (m)} \prod ^{m-1}_{k=1}\frac{1}{\frac{2k-1}{2}} =\frac{\pi ^{m-3/2}\Gamma \left( \frac{2m-1}{2}\right) }{\Gamma (m)\Gamma \left( \frac{2m-1}{2}\right) } =\frac{\pi ^{m-3/2}}{(m-1)!}. \end{aligned} \end{aligned}$$
$$\begin{aligned} \begin{aligned} \sum _{n=1}^\infty |h_n(t,z,z')|&\le \sum _{n=1}^\infty K_0^nC^n \sqrt{\frac{y}{y'}}\frac{q_2(2t,z,z')}{(y')^2}t^{\frac{n-1}{2}} \prod ^n_{i=1}\beta \left( \frac{i}{2},\frac{1}{2}\right) \\&=\sqrt{\frac{y}{y'}}\frac{q_2(2t,z,z')}{(y')^2} \sum _{n=1}^\infty \left( K_0^{2n}C^{2n}t^{\frac{2n-1}{2}} \prod _{i=1}^{2n} \beta \left( \frac{i}{2},\frac{1}{2}\right) \right. \\&\left. \quad +K_0^{2n-1}C^{2n-1}t^{n-1} \prod ^{2n-1}_{i=1} \beta \left( \frac{i}{2},\frac{1}{2}\right) \right) \\&=\sqrt{\frac{y}{y'}}\frac{q_2(2t,z,z')}{(y')^2} \sum _{n=1}^\infty \left( K_0^{2n}C^{2n}t^{\frac{2n-1}{2}} \frac{\pi ^{n-1/2}}{\Gamma \left( \frac{2n+1}{2} \right) }\right. \\&\left. \quad + K_0^{2n-1}C^{2n-1}t^{n-1} \frac{\pi ^{n-3/2}}{(n-1)!} \right) , \end{aligned} \end{aligned}$$
which shows that the series is convergent.

The proof of (iii) in the paper need not be corrected. \(\square \)


  1. 1.
    Davies, E.B., Mandouvalos, N.: Heat Kernel bounds on hyperbolic space and Kleinian groups. Proc. Lond. Math. Soc. (3) 52(1), 182–208 (1988)MathSciNetCrossRefGoogle Scholar

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© The JJIAM Publishing Committee and Springer Japan KK, part of Springer Nature 2018

Authors and Affiliations

  1. 1.Ritsumeikan UniversityKusatsuJapan
  2. 2.Tokyo University of ScienceTokyoJapan

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