Exact Sampling for the Maximum of Infinite Memory Gaussian Processes

Abstract

We develop an exact sampling algorithm for the all-time maximum of Gaussian processes with negative drift and general covariance structures. In particular, our algorithm can handle non-Markovian processes even with long-range dependence. Our development combines a milestone-event construction with rare-event simulation techniques. This allows us to find a random time beyond which the running time maximum will never be reached again. The complexity of the algorithm is random but has finite moments of all orders. We also test the performance of the algorithm numerically.

Appendix

1.1 Proof of Lemma 1

Proof

Note that $S_{k}$ conditional on $\mathcal {S}_{n}$ is still a Gaussian random variable with conditional mean

$$ \mu _{n}(k)=\mathbb {E}[S_{k}|\mathcal {S}_{n}]=-k\mu +\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\tilde{\mathcal {S}}_{n}, $$

and conditional variance

$$ \sigma _{n}(k)^{2}={\text {Var}}[S_{k}|\mathcal {S}_{n}]=\sigma ^{2}k^{2H}-\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\boldsymbol{U}_{nk}. $$

The proof of the lemma is divided into three steps. We first establish bounds for the conditional mean $\mu _{n}(k)$. Let $\tilde{\mu }_{n}(k)=\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\tilde{\mathcal {S}}_{n}$. As $\tilde{\mu }_{n}(k)$ is a linear combination of $\tilde{\mathcal {S}}_{n}$, it follows a Normal distribution with mean 0 and variance $\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\boldsymbol{U}_{nk}$. By the law of total variance, $\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\boldsymbol{U}_{nk}<\sigma ^{2}k^{2H}$. In this case, for any fixed $\delta \in (0,\mu )$,

$$\begin{aligned} \mathbb {P}(\tilde{\mu }_{n}(k)>\delta k) \le \mathbb {P}\left( \frac{\tilde{\mu }_{n}(k)}{\sqrt{\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\boldsymbol{U}_{nk}}}>\frac{\delta k}{\sigma k^{H}}\right) =\bar{\Phi }\left( \frac{\delta }{\sigma } k^{1-H}\right) . \end{aligned}$$

(11)

Then,

$$ \sum _{n=1}^{\infty }\sum _{k=n}^{\infty }\mathbb {P}(\tilde{\mu }_{n}(k)>\delta k) =\sum _{k=1}^{\infty }\sum _{n=1}^{k}\mathbb {P}(\tilde{\mu }_{n}(k)>\delta k) \le \sum _{k=1}^{\infty } k \bar{\Phi }\left( \frac{\delta }{\sigma }k^{1-H}\right) <\infty . $$

By Borel-Cantelli Lemma, there exits a random number $L_{0}\ge n$, which is finite almost surely, such that when $k>L_{0}$, $\tilde{\mu }_{n}(k)\le \delta k$, which further implies that $\mu _{n}(k)\le -(\mu -\delta )k$.

We next establish bounds for $\sum _{n=1}^{\infty }q(n)$. For $k>L_{0}$, we have $\mu _{n}(k)\le -(\mu -\delta )k$ and $\sigma _{n}(k)^2\le \sigma ^{2}k^{2H}$. Thus, for any $b\ge 0$,

$$\begin{aligned} \mathbb {P}_{n}(S_{k}>b)\le \mathbb {P}_{n}\left( \frac{S_{k}-\mu _{n}(k)}{\sigma _{n}(k)}>\frac{b+(\mu -\delta )k}{\sigma k^{H}}\right) \le \bar{\Phi }\left( \frac{\mu -\delta }{\sigma }k^{1-H}\right) . \end{aligned}$$

(12)

Based on the analysis above, let $b=\max _{1\le l\le n} S_{l}$. We decompose $\sum _{n=1}^{\infty }q(n)$ into three parts:

$$ \sum _{n=1}^{\infty }\sum _{k=n}^{\infty }\mathbb {P}_{n}(S_{k}>b) \le \underbrace{\sum _{n=1}^{L_{0}}\sum _{k=n}^{L_{0}}\mathbb {P}_{n}(S_{k}>b)}_{\text{(I) }} + \underbrace{\sum _{n=1}^{L_{0}}\sum _{k=L_{0}}^{\infty }\mathbb {P}_{n}(S_{k}>b)}_{\text{(II) }} + \underbrace{\sum _{n=L_{0}}^{\infty }\sum _{k=n}^{\infty }\mathbb {P}_{n}(S_{k}>b)}_{\text{(III) }}. $$

Part (I) only involves a finite number of terms. For part (II), from (12), we have

$$ \text{(II) } \le L_{0} \sum _{k=L_{0}}^{\infty }\bar{\Phi }\left( \frac{\mu -\delta }{\sigma }k^{1-H}\right) < \infty . $$

Similarly, for part (III), from (12), we have

$$ \text{(III) }=\sum _{k=L_0}^{\infty }\sum _{n=L_0}^{k}\mathbb {P}_{n}(S_{k}>b) \le \sum _{k=L_0}^{\infty }(k-L_0)\bar{\Phi }\left( \frac{\mu -\delta }{\sigma }k^{1-H}\right) <\infty . $$

Putting parts (I)–(III) together, we have $\sum _{n=1}^{\infty }q(n)<\infty$. By Borell-Cantelli Lemma, there exits L, which is finite almost surely, such that for any $n>L$, $q(n)<a$.

Lastly, we show that $\mathbb {E}[L^{\eta }]<\infty$ for any $\eta >0$. Let $L_1$ denote a large enough constant, such that $\sum _{k=L_1}^{\infty }\bar{\Phi }\left( \frac{\mu -\delta }{\sigma }k^{1-H}\right) <a$. Then, $L\le \max \{L_{0},L_1\}$. Thus, to prove $\mathbb {E}[L^{\eta }]<\infty$, we only need to show that $\mathbb {E}[L_{0}^{\eta }]<\infty$. Define $\mathcal {A}_{n}=\bigcup _{k=n}^{\infty }\{\tilde{\mu }_{n}(k)>\delta k\}$. Then $L_{0}^{\eta }\le \sum _{n=1}^{\infty }1\{\mathcal {A}_{n}\}n^{\eta }$, and

$$\begin{aligned} \mathbb {E}[L_{0}^{\eta }]\le \mathbb {E}\left[ \sum _{n=1}^{\infty }1\{\mathcal {A}_{n}\}n^{\eta }\right]&=\sum _{n=1}^{\infty }\sum _{k=n}^{\infty }\mathbb {P}(\tilde{\mu }_{n}(k)>\delta k)n^{\eta }\\&=\sum _{k=1}^{\infty }\sum _{n=1}^{k}n^{\eta } \mathbb {P}(\tilde{\mu }_{n}(k)>\delta k) \le \sum _{k=1}^{\infty } k^{\eta } \bar{\Phi }\left( \frac{\delta }{\sigma }k^{1-H}\right) <\infty , \end{aligned}$$

where the last inequality follows from (11). $\square$

1.2 Proof of Lemma 2

Proof

With a little abuse of notation, we denote $\mathbb {Q}_n$ as the measure induced by the TBS procedure. First note that

$$\begin{aligned}&\mathbb {Q}_n((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k)= \sum _{m=n+1}^{\infty }f_{n}(m) \mathbb {P}_{n}((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k|S_{m}>b)\\ =&\sum _{m=n+1}^{\infty }f_{n}(m)\frac{\mathbb {P}_{n}((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k,S_{m}>b)}{\mathbb {P}_{n}(S_{m}>b)}\\ =&\sum _{m=n+1}^{\infty }\frac{\mathbb {P}_{n}((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k,S_{m}>b)}{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}\\ =&\sum _{m=n+1}^{\infty }\mathbb {P}_{n}((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k)\frac{\mathbb {P}_{n}(S_{m}>b|(S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k)}{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}\\ =&~\mathbb {P}_{n}((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k)\frac{\sum _{m=k}^{\infty }\mathbb {P}_{n}(S_{m}>b|(S_{n+1},...,S_{k})\in \cdot ,\tau (b)=k)}{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}. \end{aligned}$$

Thus, $\frac{\textrm{d}\mathbb {P}_{n}}{\textrm{d}\mathbb {Q}_n}(S_{n+1},...,S_{\kappa _n}, \kappa _n<\infty )=\frac{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}{\sum _{m=\kappa _n}^{\infty }\mathbb {P}_{\kappa _n}(S_{m}>b)}$. $\square$

1.3 Proof of Lemma 3

Proof

Let $\mathbb {E}_{\mathbb {Q}}$ denote the expectation under measure $\mathbb {Q}$. Suppose $M_n=b$. First note that by Lemma 2,

$$\begin{aligned} \begin{aligned} \mathbb {Q}_n(I=1) =&~\mathbb {E}_{\mathbb {Q}_n}\left[ \Bigr (\sum _{\ell =\kappa _n}^{\infty }\mathbb {P}_{\kappa _n}(S_{\ell }>b)\Bigr )^{-1}\right] \\ =&~\mathbb {E}_{\mathbb {P}_{n}}\left[ \frac{1}{\sum _{\ell =\kappa _n}^{\infty }\mathbb {P}_{\kappa _n}(S_{\ell }>b)}\frac{\sum _{\ell =\kappa _n}^{\infty }\mathbb {P}_{\kappa _n}(S_{\ell }>b)}{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}1\{\kappa _n<\infty \}\right] \\ =&~\mathbb {E}_{\mathbb {P}_{n}}\left[ \Bigr (\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)\Bigr )^{-1}1\{\kappa _n<\infty \}\right] = \frac{\mathbb {P}_{n}(\kappa _n<\infty )}{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)} \end{aligned} \end{aligned}$$

(13)

Next, by Bayes rule,

$$\begin{aligned} \mathbb {Q}_n((S_{n+1},...,S_{\kappa _n})\in \cdot ,\kappa _n\in \cdot |I=1) =\frac{\mathbb {Q}_n(I=1|\kappa _n,\mathcal {S}_{\kappa _n})\mathbb {Q}_n((S_{n+1},...,S_{\kappa _n})\in \cdot ,\kappa _n\in \cdot )}{\mathbb {Q}_n(I=1)}. \end{aligned}$$

(14)

As $\mathbb {Q}_n(I=1|\kappa _n,(S_{n+1},...,S_{\kappa _n}))=\frac{1}{\sum _{\ell =\tau (b)}^{\infty }\mathbb {P}_{\kappa _n}(S_{\ell }>b)}$, plugging (13) in (14), we have

$$\begin{aligned}&\mathbb {Q}_n((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k|I=1)\\ =&~ \frac{1}{\sum _{\ell =k}^{\infty }\mathbb {P}_{k}(S_{\ell }>b)}\mathbb {Q}_n((S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k) \frac{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}{\mathbb {P}_{n}(\kappa _n<\infty )}\\ =&~ \mathbb {E}_{\mathbb {Q}_n}\left[ 1\{(S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k\}\frac{\sum _{\ell =n+1}^{\infty }\mathbb {P}_{n}(S_{\ell }>b)}{\sum _{\ell =k}^{\infty }\mathbb {P}_{k}(S_{\ell }>b)}\right] \frac{1}{\mathbb {P}_{n}(\kappa _n<\infty )}\\ =&~ \frac{\mathbb {E}_{\mathbb {P}_{n}}\left[ 1\{(S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k\}\right] }{\mathbb {P}_{n}(\kappa _n<\infty )} \text{ by } \text{ Lemma } \text{2 }\\ =&~ \mathbb {P}_{n}(S_{n+1},...,S_{k})\in \cdot ,\kappa _n=k|\kappa _n<\infty ). \end{aligned}$$

$\square$

1.4 Proof of Lemma 4

Proof

Given $\mathcal {S}_n$, suppose $M_n=b$. We also define

$$ N_{1}=\left( \frac{2\sigma ^{2}n^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}\Vert \boldsymbol{\tilde{S}}_{n}\Vert _{1}}{\mu }\right) ^{\frac{1}{1-H}},~ N_{2}=\left( \frac{2\sigma ^{2}}{\pi \mu ^{2}}\right) ^{\frac{1}{2(1-H)}}, \text{ and } N_{3}=\left( \frac{2H-1}{1-H}\frac{16\sigma ^{2}}{\mu ^{2}}\right) ^{2}. $$

Note that for any $k>n$, $S_{k}$ conditional on $\mathcal {S}_{n}$ is still a Gaussian random variable with conditional mean $\mu _{n}(k)=\mathbb {E}[S_{k}|\mathcal {S}_{n}]=-k\mu +\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\tilde{\mathcal {S}}_{n}$, and conditional variance $\sigma _{n}(k)^{2}={\text {Var}}[S_{k}|\mathcal {S}_{n}]=\sigma ^{2}k^{2H}-\boldsymbol{U}_{nk}^{\top }\boldsymbol{\Sigma }_{n}^{-1}\boldsymbol{U}_{nk}$.

We first establish the sequence of bounds. The lower bound is straightforward. For the upper bound, note that for $k\ge N_{1}$,

$$ \mu _{n}(k) \le -k\mu +\sigma ^{2}(nk)^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\tilde{\mathcal {S}}_{n}\Vert _{1} \le -k\mu +\sigma ^{2}(nk)^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}\Vert \tilde{\mathcal {S}}_{n}\Vert _{1}\le -\frac{k\mu }{2}. $$

Next, note that for $k\ge \max \{N_{1}, N_{2}\}$,

$$\begin{aligned} \mathbb {P}_{n}(S_{k}>b)\le \frac{1}{\sqrt{2\pi }}\frac{\sigma _{n}(k)}{b-\mu _{n}(k)}\exp \left( -\frac{(b-\mu _{n}(k))^{2}}{\sigma _{n}(k)^{2}}\right) \le \exp \left( -\frac{\mu ^{2}}{8\sigma ^{2}}k^{2-2H}\right) . \end{aligned}$$

(15)

To see the second inequality, note that when $k\ge N_{1}$, $\mu _{n}(k)\le -k\mu /2$ and $\sigma _{n}(k)\le \sigma k^{H}$. Thus, $\frac{b-\mu _{n}(k)}{\sigma _{n}(k)} \ge \frac{b+k\mu }{2\sigma k^{H}}\ge \frac{\mu }{2\sigma }k^{1-H}$. And for $k\ge N_{2}$, $\frac{1}{\sqrt{2\pi }}\left( \frac{\mu }{2\sigma }k^{1-H}\right) ^{-1} \le 1$.

Lastly, we have for $\ell \ge \max \{N_{1},N_{2},N_{3}\}$,

$$\begin{aligned} \sum _{k=\ell +1}^{\infty }\mathbb {P}_{n}(S_{k}>b)&\le \sum _{k=\ell +1}^{\infty } \exp \left( -\frac{\mu ^{2}}{8\sigma ^{2}}k^{2-2H}\right) ~~~ \text{ from } \text{15 } \text{ as }\, \ell \ge \max \{N_1, N_2\}\\&\le \int _{\ell }^{\infty } \exp \left( -\frac{\mu ^{2}}{8\sigma ^{2}}k^{2-2H}\right) \textrm{d}k\\&=\frac{1}{2-2H} \int _{\ell ^{2-2H}}^{\infty } y^{(2H-1)/(2-2H)}\exp \left( -\frac{\mu ^{2}}{8\sigma ^{2}}y\right) \textrm{d}y\\&\le \frac{1}{2-2H} \int _{\ell ^{2-2H}}^{\infty } \exp \left( -\frac{\mu ^{2}}{16\sigma ^{2}}y\right) \textrm{d}y ~~~ \text{ as }\, \ell \ge N_3\\&\le \frac{8\sigma ^{2}}{(1-H)\mu ^{2}} \exp \left( -\frac{\mu ^{2}}{16\sigma ^{2}}\ell ^{2-2H}\right) =h(\ell ). \end{aligned}$$

For $\mathbb {E}[B(n)^{\eta }]$, we first note that $N_{2}$ and $N_{3}$ are finite constants. Thus, we only need to show that $\mathbb {E}[N_{1}^{\eta }]<\infty$. For any fixed n,

$$\begin{aligned} \mathbb {E}[N_{1}^{\eta }]&=\mathbb {E}\left[ \left( \frac{2\sigma ^{2}n^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}\Vert \boldsymbol{\tilde{S}}_{n}\Vert _{1}}{\mu }\right) ^{\tfrac{\eta }{1-H}}\right] \\&=\left( \frac{2\sigma ^{2}n^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}}{\mu }\right) ^{\tfrac{\eta }{1-H}}\mathbb {E}\left[ \left( \sum _{k=1}^{n}|S_{k}+k\mu |\right) ^{\tfrac{\eta }{1-H}}\right] \\&\le \left( \frac{2\sigma ^{2}n^{H}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}}{\mu }\right) ^{\tfrac{\eta }{1-H}} n^{\tfrac{\eta }{1-H}-1}\sum _{k=1}^{n}\mathbb {E}\left[ |S_{k}+k\mu |^{\tfrac{\eta }{1-H}}\right] \\&=\left( \frac{2\sigma ^{2}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}}{\mu }\right) ^{\tfrac{\eta }{1-H}} n^{\tfrac{\eta H+\eta +H-1}{1-H}} \frac{\Gamma (\tfrac{\eta /(1-H)+1}{2})}{\sqrt{\pi }}(2\sigma ^{2})^{\tfrac{\eta }{2(1-H)}}\sum _{k=1}^{n}k^{\tfrac{\eta H}{1-H}} \\&\le \left( \frac{2^{3/2}\sigma ^{3}\Vert \boldsymbol{\Sigma }_{n}^{-1}\Vert _{1}}{\mu }\right) ^{\tfrac{\eta }{1-H}}\frac{\Gamma (\tfrac{\eta /(1-H)+1}{2})}{\sqrt{\pi }}n^{\tfrac{2\eta H+\eta }{1-H}}. \end{aligned}$$

$\square$

1.5 Proof of Lemma 5

Proof

Given $\kappa _n$ and $\mathcal {S}_{\kappa _n}$, suppose $M_n=b$. First note that

$$ \tilde{q}(n,\ell ) \le \tilde{q}(n,\ell ) \le \cdots \le \sum _{i=\kappa _n+1}^{\infty } \mathbb {P}_{\kappa _n}(S_{i}>b). $$

Next, following the proof of Lemma 4, we have for $\ell \ge B(\kappa _n)$,

$$ \tilde{q}(n,\ell )+h(\ell )\ge \tilde{q}(n,\ell +1)+h(\ell +1)\ge \cdots \ge \sum _{i=\kappa _n+1}^{\infty } \mathbb {P}_{\kappa _n}(S_{i}>b). $$

Since $\mathbb {P}_{k}(S_{k}>b)=1$, $p(k)=(1+\sum _{i=k+1}^{\infty } \mathbb {P}_{k}(S_{i}>b))^{-1}$, and for $\ell \ge B(\kappa _n)$, $(1+\tilde{q}(n,\ell )+h(\ell ))^{-1} \le p(\kappa _n)\le (1+\tilde{q}(n,\ell ))^{-1}$. The rest of the results follow similarly. $\square$

1.6 Proof of Lemma 6

Proof

We first note that in Step 2.1 in Algorithm 2, $\mathbb {P}_{n}(N(n)=\ell , J=1)= \mathbb {P}_{n}(S_{\ell }>M_{n})$. Next, following the same lines of analysis as the proof of Lemma 1, we have for any $\delta >0$, there exists $L_{0}>0$ such that for $\ell >L_{0}$, $\mathbb {P}_{n}(S_{\ell }>M_{n})\le \bar{\Phi }\left( \frac{\mu -\delta }{\sigma } \ell ^{1-H}\right)$, and for any $\eta >0$, $\mathbb {E}[L_{0}^{\eta }]<\infty$. Then for any $\eta >0$,

$$\begin{aligned} \mathbb {E}[N(n)^{\eta }|\mathcal {S}_{n}]= & {} \sum _{\ell =n+1}^{\infty }\ell ^{\eta }\mathbb {P}_{n}(N(n)=\ell ) \\\le & {} \mathbb {E}[N_{0}^{\eta }|\mathcal {S}_{n}] + \mathbb {E}\left[ \left. \sum _{\ell =N_{0}+1}^{\infty } \ell ^{\eta }\bar{\Phi }\left( \frac{\mu -\delta }{\sigma } \ell ^{1-H}\right) \right| \mathcal {S}_{n}\right] . \end{aligned}$$

Thus,

$$\mathbb {E}[N(n)^{\eta }]= \mathbb {E}[\mathbb {E}[N(n)^{\eta }|\mathcal {S}_{n}]]\le \mathbb {E}[N_{0}^{\eta }] + \mathbb {E}\left[ \sum _{\ell =N_{0}+1}^{\infty } \ell ^{\eta }\bar{\Phi }\left( \frac{\mu -\delta }{\sigma } \ell ^{1-H}\right) \right] <\infty .$$

$\square$