Quantum speedup of Monte Carlo integration with respect to the number of dimensions and its application to finance

Abstract

Monte Carlo integration using quantum computers has been widely investigated, including applications to concrete problems. It is known that quantum algorithms based on quantum amplitude estimation (QAE) can compute an integral with a smaller number of iterative calls of the quantum circuit which calculates the integrand, than classical methods call the integrand subroutine. However, the issues about the iterative operations in the integrand circuit have not been discussed so much. That is, in the high-dimensional integration, many random numbers are used for calculation of the integrand and in some cases similar calculations are repeated to obtain one sample value of the integrand. In this paper, we point out that we can reduce the number of such repeated operations by a combination of the nested QAE and the use of pseudorandom numbers (PRNs), if the integrand has the separable form with respect to contributions from distinct random numbers. The use of PRNs, which the authors originally proposed in the context of the quantum algorithm for Monte Carlo, is the key factor also in this paper, since it enables parallel computation of the separable terms in the integrand. Furthermore, we pick up one use case of this method in finance, the credit portfolio risk measurement and estimate to what extent the complexity is reduced.

Appendix A: Proof of (26)

We can transform $H(\theta _j,M)$ to

$$\begin{aligned}&H(\theta _j,M) \nonumber \\&\quad = \frac{\sin ^2(M\theta _j \pi )}{M^2}\left[ M\cos (2\theta _j\pi ) + \sin (2\theta _j\pi )\sum _{{\tilde{\theta }}\in I_M} \cot (({\tilde{\theta }}-\theta _j)\pi )\right] . \end{aligned}$$

(A1)

Here, we used (9), some formulae on trigonometric functions and the relation $\sin ^2(M({\tilde{\theta }}-\theta _j)\pi )=\sin ^2(M\theta _j \pi )$, which follows from ${\tilde{\theta }}\in I_M$. We can show that the second term in the parenthesis in (A1) is of subleading order with respect to 1/M, since the terms in the sum of cotangents over ${\tilde{\theta }}\in I_M$ nearly cancel out. In the limit that $\theta _j \rightarrow {\tilde{\theta }}$ where ${\tilde{\theta }}\in I_M$, one term in the cotangent sum diverges but $H(\theta _j,M)$ itself goes to 0 due to the overall factor $\sin ^2(M\theta _j\pi )$. As a result, we get

$$\begin{aligned} H(\theta _j,M)=\frac{\sin ^2(M\theta _j\pi )\cos (2\theta _j\pi )}{M} + O\left( \frac{1}{M^2}\right) . \end{aligned}$$

(A2)

Since $|\sin ^2(M\theta _j \pi )\cos (2\theta _j\pi )/M|<1/M$, we finally obtain (26).