Quasi-random numbers for copula models

Abstract

The present work addresses the question how sampling algorithms for commonly applied copula models can be adapted to account for quasi-random numbers. Besides sampling methods such as the conditional distribution method (based on a one-to-one transformation), it is also shown that typically faster sampling methods (based on stochastic representations) can be used to improve upon classical Monte Carlo methods when pseudo-random number generators are replaced by quasi-random number generators. This opens the door to quasi-random numbers for models well beyond independent margins or the multivariate normal distribution. Detailed examples (in the context of finance and insurance), illustrations and simulations are given and software has been developed and provided in the R packages copula and qrng.

Appendix

For all the randomization schemes mentioned in Sect. 2, in addition to having a simple method to estimate the variance of the corresponding RQMC estimator, results giving exact expressions for this variance are also known and typically rely on using a well-chosen series expansion of the function $\varPsi$ of interest. The following result recalls this variance expression in the case of randomly digitally shifted net; see Lemieux (2009) for a detailed proof. This result is used in the proof of Proposition 6 in Sect. 4.1.

Theorem 3

(Variance for randomly digitally shifted nets) Let $\tilde{P}_n = \{\tilde{\varvec{v}}_1,\ldots ,\tilde{\varvec{v}}_n\}$ be a randomly digitally shifted net in base b with corresponding RQMC estimator $\widehat{\mu }_n$ given by

$$\begin{aligned} \widehat{\mu }_n = \frac{1}{n} \sum _{i=1}^n \varPsi (\tilde{\varvec{v}}_i) \end{aligned}$$

and assume $\mathrm{Var}(\varPsi (\varvec{U})) < \infty$ for $\varvec{U} \sim U[0,1)^d$. Then we have that

$$\begin{aligned} \mathrm{Var}(\widehat{\mu }_n) = \sum _{\varvec{0} \ne \varvec{h} \in \mathscr {L}_d^*} |\widehat{\varPsi }(\varvec{h})|^2, \end{aligned}$$

where $\widehat{\varPsi }(\varvec{h})$ is the Walsh coefficient of $\varPsi$ at $\varvec{h}$, given by

$$\begin{aligned} \widehat{\varPsi }(\varvec{h}) = \int _{[0,1)^d} f(\varvec{u}) e^{-2\pi i \langle \varvec{h}, \varvec{u} \rangle _b}d\varvec{u} \end{aligned}$$

where $\langle \varvec{h}, \varvec{u} \rangle _b = \frac{1}{b} \sum _{j=1}^d \sum _{l=0}^{\infty } h_{j,l}u_{j,l+1}$ with $h_{j,l}$ and $u_{j,l}$ obtained from the base b expansion of $h_j$ and $u_j$, respectively, and $\mathscr {L}_d^*=\{\varvec{h} \in \mathbb {Z}^d: \langle \varvec{h}, \varvec{v}_i \rangle _b \in \mathbb {Z}, \forall i=1,\ldots ,n \}$ is the dual net of the deterministic net $P_n = \{\varvec{v}_i,i=1,\ldots ,n\}$ that has been shifted to get $\tilde{P}_n$.

1.1 Proofs

Proof (Proof of Proposition 4)

We start by providing more details on the expression (21), which is given by:

$$\begin{aligned}&\frac{\partial ^l \varPsi \circ \phi _C(v_{\alpha _1},\ldots ,v_{\alpha _l},\varvec{1})}{\partial v_{\alpha _1} \cdots \partial v_{\alpha _l}}\\&\quad = \sum _{1 \le |\varvec{\beta }| \le l} \frac{\partial ^{|\varvec{\beta }|} \varPsi }{\partial ^{\beta _1} u_{1} \ldots \partial ^{\beta _d} u_{d}}\\&\qquad \quad \sum _{s=1}^l \sum _{(\varvec{k},\varvec{\gamma }) \in p_s(\varvec{\beta },\varvec{\alpha })} c_{\gamma } \prod _{j=1}^s \frac{\partial ^{|\gamma _j|} \phi _{C,k_j}(v_{\alpha _1},\ldots ,v_{\alpha _l},\varvec{1})}{\partial ^{\gamma _{j,1}} v_{\alpha _1}\ldots \partial ^{\gamma _{j,l}} v_{\alpha _l}} \end{aligned}$$

where $\varvec{\beta } \in \mathbb {N}_0^d, |\varvec{\beta }| = \sum _{j=1}^d \beta _j$, and the set $p_s(\varvec{\beta },\varvec{\alpha })$ includes pairs $(\varvec{k},\varvec{\gamma })$ such that $\varvec{k}$ is an s-dimensional vector $\varvec{k}=(k_1,\ldots ,k_s)$ where each $k_j\in \{1,\ldots ,d\}$, and $\varvec{\gamma }$ is an sl-dimensional vector $\varvec{\gamma }=(\varvec{\gamma }_1,\ldots ,\varvec{\gamma }_s)$ where each $\varvec{\gamma }_j$ is an l-dimensional vector whose entries are either 0 or 1, and $\sum _{j=1}^s {\gamma }_{j,i} = 1$ for $i\in \{1,\ldots ,l\}$. Finally, the $c_{\varvec{\gamma }}$ are constants, which are defined in detail in Constantine and Savits (1996), along with further information on the precise definition of $p_s(\varvec{k},\varvec{\gamma })$. As mentioned in Sect. 4.1, a sufficient condition to show that $\Vert \varPsi \circ \varPhi _C \Vert _{d,1} < \infty$ is to establish that all products of the form (22) are in $L_1$, which we recall is given by

$$\begin{aligned} \frac{\partial ^{|\varvec{\beta }|} \varPsi }{\partial ^{\beta _1} u_{1} \ldots \partial ^{\beta _d} u_{d}} \prod _{j=1}^s \frac{\partial ^{|\gamma _j|} \phi _{C,k_j}(v_{\alpha _1},\ldots ,v_{\alpha _l},\varvec{1})}{\partial ^{\gamma _{j,1}} v_{\alpha _1}\ldots \partial ^{\gamma _{j,l}} v_{\alpha _l}}, \end{aligned}$$

for $s\in \{1,\ldots ,l\}$ and $(\varvec{k},\varvec{\gamma }) \in p_s(\varvec{\beta },\varvec{\alpha })$.

Recall also that for the MO algorithm, $\phi _{C,l}$ is a function of $v_1$ and $v_{l+1}$ only, for $l=1,\ldots ,d$. Hence the only partial derivatives of $\phi _{C,l}$ that are nonzero are those with respect to variables in $\{v_1,v_{l+1}\}$.

Now, since we assume that (23) holds, then it means we just need to show that the product found in (22) is in $L_1$, under the conditions stated in the proposition. In turn, we first show that this holds if the following bounds hold for the mixed partial derivatives of $\phi _C$:

$$\begin{aligned}&\int _0^1 \left| \frac{\partial \phi _{C,l}(v_1=1,v_{l+1})}{\partial v_{l+1}}\right| dv_{l+1} < \infty , \end{aligned}$$

(26)

$$\begin{aligned}&\int _{[0,1)^{l}} \left| \frac{\partial ^2 \phi _{C,1}(v_1,v_2)}{\partial v_1 \partial v_2} \prod _{j=2}^{l-1} \frac{\partial \phi _{C,j}(v_1,v_{j+1})}{\partial v_{j+1}}\right| dv_1 dv_2 \ldots dv_{l} < \infty , \text{ and } \end{aligned}$$

(27)

$$\begin{aligned}&\int _{[0,1)^{l-1}} \left| \frac{\partial \phi _{C,r}(v_1,v_{r+1}=1)}{\partial v_1} \right| \nonumber \\&\quad \times \,\left( \prod _{j=1, j\ne r}^{l-1} \left| \frac{\partial \phi _{C,j}(v_1,v_{j+1})}{\partial v_{j+1}}\right| dv_j \right) dv_{l} < \infty \end{aligned}$$

(28)

for all $l \le d+1$,

We have three cases to consider.

Case 1 $1 \notin I$. Then the product in (22) is given by

$$\begin{aligned} \prod _{j=1}^l\left| \frac{\partial \phi _{C,j}(v_1=1,v_{j+1})}{\partial v_{j+1}}\right| , \end{aligned}$$

where we assumed w.l.o.g. that $I = \{2,\ldots ,l+1\}, s=l$ and $k_j=j+1$ for $j\in \{1,\ldots ,s\}$. Since each term in the product depends on a distinct variable, the product is in $L_1$ if (26) holds.

Case 2 $1\in I$ and j such that $\gamma _{j,1}=1$ has $k_j+1 \notin I$. This case can be analyzed w.l.o.g. by assuming I is of the form $I=\{1,\ldots ,r,r+2,\ldots ,l+1\}$ for some $r\ge 1$. In that case, the products in (22) are of the form

$$\begin{aligned} \left| \frac{\partial \phi _{C,r}(v_1,v_{r+1}=1)}{\partial v_1} \prod _{j=1, j\ne r}^{l-1} \frac{\partial \phi _{C,j}(v_1,v_{j+1})}{\partial v_{j+1}}\right| \end{aligned}$$

and is thus in $L_1$ as long as (28) holds.

Case 3 $1\in I$ and j such that $\gamma _{j,1}=1$ has $k_j+1 \in I$. In this case, we can assume w.l.o.g. that $I=\{1,\ldots ,l\}$ and therefore the products in (22) are of the form

$$\begin{aligned} \left| \frac{\partial ^2 \phi _{C,r}(v_1,v_{r+1})}{\partial v_1\partial v_{r+1}} \prod _{j=1, j\ne r}^{l-1} \frac{\partial \phi _{C,j}(v_1,v_{j+1})}{\partial v_{j+1}}\right| \end{aligned}$$

and is thus in $L_1$ as long as (27) holds.

The last part of the proof is to show that (26), (27), and (28) hold. First we study the partial derivatives involved in these expressions and find they are given by:

$$\begin{aligned} \frac{\partial \phi _{C,1}(v_1,v_2)}{\partial v_1}&= \psi '\left( -\frac{\log (v_2)}{x_1}\right) \frac{\log v_2}{x_1^2} \frac{\partial x_1}{\partial v_1}, \\ \frac{\partial \phi _{C,1}(v_1,v_2)}{\partial v_2}&= -\psi '\left( -\frac{\log (v_2)}{x_1}\right) \frac{1}{x_1 v_2},\\ \frac{\partial ^2 \phi _{C,1}(v_1,v_2)}{\partial v_1 \partial v_2}&= \frac{\partial x_1}{\partial v_1} \frac{1}{v_2x_1^2}\left[ \psi '\left( -\frac{\log v_2}{x_1}\right) \right. \\&\left. \quad -\,\frac{\log v_2}{x_1} \psi ''\left( -\frac{\log v_2}{x_1}\right) \right] , \end{aligned}$$

where $x_1 = F^{-1}(v_1)$ and $\frac{\partial x_1}{\partial v_1}= 1/f(x_1)$, where f is the pdf corresponding to F, which exists since we assumed F was continuous. Now, the partial derivatives with respect to either $v_1$ or $v_2$ are clearly non-negative for all $v_1$ and $v_2$. Hence it is easy to see that (26) and (28) hold, because we can remove the absolute value inside the integrals and therefore, these integrals amount to take differences/sums of $\phi _{C,r}(\cdot ,\cdot )$ at different values over its domain, which obviously yields a finite value since $\phi _{C,r}(\cdot ,\cdot )$ always takes values in [0, 1].

As for the mixed partial derivative with respect to $v_1$ and $v_2$, our assumption on $\psi '(t)+t\psi ''(t)$ implies we have at most one sign change over the domain of the integral. If there is no sign change, the argument used in the previous paragraph to handle (26) and (28) can be used to show (27) is bounded. If there is one sign change, then we let $t^*$ be such that

$$\begin{aligned}&\psi '(t)+t \psi ''(t) \le 0\quad \text {for } 0 \le t \le t^*\text { and }\\&\quad \psi '(t) +\,t \psi ''(t) \ge 0\quad \text {for } t^* \ge t. \end{aligned}$$

Then let q(v) be a function such that $-\log q(v)/F^{-1}(v) = t^*$. For instance, one can verify that for the Clayton copula, $q(v)=e^{-\theta F^{-1}(v)}$. When integrating the absolute value of the mixed partial derivative $\partial ^2 \phi _{C,1}(v_1,v_2)/\partial v_1 \partial v_2$, we get

$$\begin{aligned}&\int _0^1 \frac{\partial x_1}{\partial v_1} \frac{1}{x_1^2} \left[ \int _0^{q(v_1)} \frac{1}{v_2}\left( \psi ' \left( -\frac{\log v_2}{x_1}\right) \right. \right. \nonumber \\&\quad \qquad \left. \left. -\,\frac{\log v_2}{x_1}\psi ''\left( -\frac{\log v_2}{x_1}\right) \right) dv_2 \right. \nonumber \\&\quad \qquad \left. +\, \int _{q(v_1)}^1\frac{1}{v_2}\left( -\psi '\left( -\frac{\log v_2}{x_1}\right) \right. \right. \nonumber \\&\quad \qquad \left. \left. +\,\frac{\log v_2}{x_1}\psi ''\left( -\frac{\log v_2}{x_1}\right) \right) dv_2\right] dv_1 \nonumber \\&\qquad =\,2\int _0^1 \frac{\partial x_1}{\partial v_1} \frac{1}{x_1^2} \left[ \psi '(-\log q(v_1)/x_1) \log q(v_1)\right] dv_1 \nonumber \\&\qquad =-2t^* \psi '(t^*) \int _0^1 \frac{1}{F^{-1}(v_1)} \frac{\partial F^{-1}(v_1)'}{\partial v_1} dv_1 \nonumber \\&\qquad =-2t^* \psi '(t^*) \log F^{-1} (v_1)\bigl |_0^1. \end{aligned}$$

(29)

Now, in most cases $F^{-1}(1)$ is not bounded, and thus we cannot prove that $\varPsi \circ \phi _C$ has bounded variation. However, from there we can still get the upper bound on the error given in the result, by using a technique initially developed by Sobol’ (1973) to handle improper integrals, and later by Hartinger et al. (2004) to deal with unbounded integration problems taken w.r.t. to a measure that is not necessarily uniform (as studied in Sect. 4.2). Note that to apply their approach more easily, we need to make a small change and assume that rather than generating V as $F^{-1}(v_1)$, we use $F^{-1}(1-v_1)$, so that in our study of the variation above (via the integral of the absolute value of the mixed partial derivatives), the boundedness condition fails at $v_1=0$ instead of $v_1=1$. Following the approach in Hartinger et al. (2004) (see their Equation (24)) and taking $\varvec{c}=(1/pn,0,\ldots ,0)$, the integration error satisfies

$$\begin{aligned}&\biggl |\frac{1}{n} \sum _{i=1}^n \varPsi (\varvec{u}_i) -\mathbb {E}[\varPsi (\varvec{U})]\biggr |\\&\quad \le \frac{1}{pn} \varPsi (1,\ldots ,1) +D^*(P_n) V_{[\varvec{c},\varvec{1}]}(\varPsi \circ \phi _C) + I_{rest} \end{aligned}$$

where $V_{[\varvec{c},\varvec{1}]}(\varPsi \circ \phi _C)$ denotes the variation of $\varPsi \circ \phi _C$ over $[\varvec{c},\varvec{1}]$ and

$$\begin{aligned} I_{rest}&= \left| \int _{\varvec{0}}^{\varvec{1}} \varPsi \circ \phi _C(\varvec{v}) d\varvec{v} - \int _{\varvec{c}}^{\varvec{1}} \varPsi \circ \phi _C(\varvec{v}) d\varvec{v} \right| \\&\le \frac{M}{pn} \text{ for } \text{ some } M>0, \end{aligned}$$

since we assumed $|\psi (\varvec{u})|$ was bounded. As for $V_{[\varvec{c},\varvec{1}]}(\varPsi \circ \phi _C)$, we can infer from the steps that led to (29) that it is bounded by a constant times $\log F^{-1}(1-1/pn) \le a \log n + \log c$ by assumption. Therefore there exists a constant $K^{(d)}$ such that $V_{[\varvec{c},\varvec{1}]}(\varPsi \circ \phi _C) \le K^{(d)} \log n$. $\square$

Proof (Proof of Proposition 5)

Let $p_l$ be such that $P(V=l)=p_l$, for $l \ge 1$. Let $P_l = \sum _{k=1}^l p_k$ for $l \ge 1$ and $P_0=0$. We also let $\phi _C^l(v_2,\ldots ,v_{d+1})$ $= \phi _C(P_{l-1},v_2,\ldots ,v_{d+1})$ for $l \ge 1$ (transformation $\phi _C$ when $v_1$ generates the value l for V). Consider a given value of n and low-discrepancy point set $P_n$. If we use inversion to generate V, then we have that the subset $P_n^l = \{\varvec{v}_i: P_{l-1} < v_{i,1} \le P_l\}$ will be used to produce copula samples with $V=l$. Let $\tilde{n}_l = |P_n^l|$ and $n_l = np_l$. It is clear that if l becomes too large, then $\tilde{n}_l$ will eventually be 0. Let L(n) be the largest value of l such that $\tilde{n}_{l} >0$, and let $\tilde{p}_l = \tilde{n}_l/n$. Then we can write

$$\begin{aligned}&\biggl | \int _{[0,1)^{d+1}} \varPsi \circ \phi _C(\varvec{v})d\varvec{v} - \frac{1}{n} \sum _{i=1}^n \varPsi \circ \phi _C(\varvec{v}_i)\biggr | \\&\,\quad \le \biggl | \sum _{l=1}^{L(n)} p_l \left( \int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} - \frac{1}{n_l}\sum _{P_n^l}\varPsi \circ \phi _C(\varvec{v}_i) \right) \biggr | \\&\,\qquad + \sum _{l=L(n)+1}^{\infty } p_l \biggl |\int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} \biggr | \\&\,\quad \le \sum _{l=1}^{L(n)} \tilde{p}_l \biggl | \int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} - \frac{1}{\tilde{n}_l}\sum _{P_n^l}\varPsi \circ \phi _C(\varvec{v}_i) \biggr |\\&\,\qquad + \sum _{l=L(n)+1}^{\infty } p_l \biggl |\int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} \biggr | \\&\,\qquad + \sum _{l=1}^{L(n)} \biggl |(p_l-\tilde{p}_l) \int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} \biggr | \\&\,\quad \le \sum _{l=1}^{L(n)}( \tilde{p}_l A(n,d)) + B(n,d)+C(n,d), \end{aligned}$$

where A(n, d), B(n, d), and C(n, d) are bounds such that

$$\begin{aligned}&\biggl | \int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} - \frac{1}{\tilde{n}_l}\sum _{P_n^l}\varPsi \circ \phi _C(\varvec{v}_i) \biggr | \le A(n,d) \\&\qquad \times \,\sum _{l=L(n)+1}^{\infty } p_l \biggl |\int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1} \biggr | \le B(n,d) \\&\qquad \times \,\sum _{l=1}^{L(n)} \biggl |(p_l-\tilde{p}_l) \int _{[0,1)^d} \varPsi \circ \phi _C^l (\varvec{v}) dv_2 \ldots dv_{d+1}\biggr | \le C(n,d). \end{aligned}$$

First, by definition of $D^*(P_n)$ we have $|\tilde{n}_l - n_l| \le 2nD^*(P_n)$ and thus $|\tilde{p}_l-p_l| \le 2D^*(P_n)$. Hence we can take $C(n,d) = 2{\text {E}}(|\varPsi (\varvec{U})|)D^*(P_n)$. Similarly, we can show that $\sum _{l=L(n)+1}^{\infty } p_l \le D^*(P_n)$ and can therefore take $B(n,d) = {\text {E}}(|\varPsi (\varvec{U})|) D^*(P_n)$. The analysis of the expression to be bounded by A(n, d) is more complicated. First, we note that under the assumption we have on $\varPsi$ and its partial derivatives, we need to show that the product in (22) is in $L_1$, but where each $\phi _{C,k_j}$ is replaced by $\phi _{C,k_j}^l$ for a given l. Since $\phi _{C,k_j}^l$ is solely a function of $v_{k_j+1}$, then it means that the only relevant products to consider are of the form

$$\begin{aligned} \prod _{j=1}^r \frac{\partial \phi _{C,k_j}^l(v_{k_j+1})}{\partial v_{k_j+1}} \end{aligned}$$

(30)

in which each term is of the form $-\psi ^{'} \left( \frac{-\log v_{k_j+1}}{l}\right) \frac{1}{l v_{k_j+1}}$ which is non-negative for any $v_{k_j+1}$. Using a similar reasoning to the one used in the proof of Proposition 4 (to conclude that (26) and (28) hold), it is easy to see that (30) is in $L_1$.

What remains to be done is to analyze the discrepancy of $P_n^l$. That is, here we are looking for a bound on $\sup _{\varvec{z} \in \mathscr {J}^*} |E(\varvec{z};P_n^l)|$, where we recall that $\mathscr {J}^*$ is the set of intervals of $[0,1)^d$ of the form $\varvec{z} = \prod _{j=1}^d [0,z_j)$, where $0 < z_j \le 1$. So consider a given $\varvec{z} \in [0,1)^d$. Then $E(\varvec{z};P_n^l) = A(\varvec{z};P_n^l)/\tilde{n}_l - \lambda (\varvec{z})$. Let $\varvec{z}_1 = (P_l,\varvec{z})$ and $\varvec{z}_2 = (P_{l-1},\varvec{z})$, which are both in $[0,1)^{d+1}$. Note that $A(\varvec{z}_1;P_n) - A(\varvec{z}_2;P_n) = A(\varvec{z};P_n^l)$. By definition of $D^*(P_n)$, it is not hard to see that

$$\begin{aligned} \biggl |\frac{(A(\varvec{z}_1;P_n)-A(\varvec{z}_2;P_n))}{n}-p_l \lambda (\varvec{z})\biggr | \le 2D^*(P_n) \end{aligned}$$

and therefore

$$\begin{aligned} \biggl |\frac{A(\varvec{z};P_n^l)}{\tilde{n}_l} -\frac{n_l}{\tilde{n}_l}\lambda (\varvec{z})\biggr | \le 2D^*(P_n) \frac{n}{\tilde{n}_l}. \end{aligned}$$

Using the fact that $|\tilde{n}_l-n_l| \le 2nD^*(P_n)$, after some further simplifications we get that

$$\begin{aligned} \biggl |\frac{A(\varvec{z};P_n^l)}{\tilde{n}_l} -\lambda (\varvec{z})\biggr | \le 4D^*(P_n) \frac{n}{\tilde{n}_l}. \end{aligned}$$

Hence we can take $A(n,d) = 4D^*(P_n)\frac{n}{\tilde{n}_l}$ and then get $\sum _{l=1}^{L(n)} \tilde{p}_l A(n,d) \le 4 L(n)D^*(P_n)$. To show that the overall bound for the integration error is of the form $(\log n)D^*(P_n)$ times a constant, we simply need to show that $L(n) \in O(\log n)$. But this follows from our assumptions on $P_n$ and F, since by definition, L(n) is the largest integer such that $1-F(L(n)) > 1/pn$ but we also have $1-F(L(n)) \le cq^{L(n)}$, hence

$$\begin{aligned} 1/pn< cq^{L(n)} \Leftrightarrow L(n) \log (1/q) - \log c <\log p + \log n \end{aligned}$$

and thus $L(n) \le (\log n + \log p + \log c)/\log (1/q)$, as required. $\square$