Quantum algorithms for numerical differentiation of expected values with respect to parameters

Abstract

The quantum algorithms for Monte Carlo integration (QMCI), which are based on quantum amplitude estimation (QAE), speed up expected value calculation compared with classical counterparts and have been widely investigated along with their applications to industrial problems such as financial derivative pricing. In this paper, we consider an expected value of a function of a stochastic variable and a real-valued parameter and how to calculate derivatives of the expectation with respect to the parameter. This problem is related to calculating sensitivities of financial derivatives and so of industrial importance. Based on QMCI and the general-order central difference formula for numerical differentiation, we propose two quantum methods for this problem and evaluate their complexities. The first one, which we call the naive iteration method, simply calculates the formula by iterative computations and additions of the terms in it and then estimates its expected value by QAE. The second one, which we name the sum-in-QAE method, performs the summation of the terms at the same time as the sum over the possible values of the stochastic variable in a single QAE. We see that, depending on the smoothness of the function and the number of qubits available, either of two methods is better than the other. In particular, when the function is nonsmooth or we want to save the qubit number, the sum-in-QAE method can be advantageous.

Appendices

Appendix A: Proof of Lemma 1

Proof

First, we consider the case where $m\ge 2$. From the definition of $a^{(m)}_{n,j}$ in (6),

$$\begin{aligned} |a^{(m)}_{n,j}| \le \sum _{\begin{array}{c} \{l_1,\ldots \ldots ,l_{m-1}\}\in \qquad \quad \\ \mathcal {P}_{m-1}([-n:n]\setminus \{0,j\}) \end{array}} \frac{(n!)^2}{|jl_1\cdots l_{m-1}|} \end{aligned}$$

(A 1)

holds for any $j\in [-n:n]\setminus \{0\}$. Note that

$$\begin{aligned}&\sum _{\begin{array}{c} \{l_1,\ldots ,l_{m-1}\}\in \qquad \quad \\ \mathcal {P}_{m-1}([-n:n]\setminus \{0,j\}) \end{array}} \frac{1}{|l_1\cdots l_{m-1}|} \nonumber \\&\quad \le \frac{1}{(m-1)!} \left( \frac{1}{|-n|}+\cdots +\frac{1}{|-1|}+\frac{1}{1}+\cdots +\frac{1}{n}\right) ^{m-1} \nonumber \\&\quad \le \frac{\left( 2n\right) ^{m-1}}{(m-1)!} \nonumber \\&\quad \le \frac{m^{2n}}{(m-1)!}. \end{aligned}$$

(A 2)

Here, the third inequality follows from $\left( 2n\right) ^{m-1}\le m^{2n}$, which holds for any positive integers m and n such that $m\le 2n$. Then, we see that

$$\begin{aligned} |a^{(m)}_{n,j}| \le \frac{(n!)^2m^{2n}}{|j|(m-1)!}. \end{aligned}$$

(A 3)

This holds also when $m=1$, since

$$\begin{aligned} \left| a^{(1)}_{n,j}\right| =\left| \frac{\prod _{i\in [-n:n]\setminus \{0\}}i}{j}\right| =\frac{(n!)^2}{|j|}. \end{aligned}$$

(A 4)

Hence, combining this and (24) with the definition of $R_f(x,n,m,h)$ in (6), we have

$$\begin{aligned} |R_f(x,n,m,h)|\le & {} \frac{Mm!\left( n!\right) ^2m^{2n}}{(2n+1)!(m-1)!}\sum _{j=-n}^n\frac{|j|^{2n}}{(n+j)!(n-j)!} \nonumber \\\le & {} \frac{Mm\left( n!\right) ^2n^{2n}m^{2n}}{(2n+1)!(2n)!}\sum _{j=-n}^n\left( {\begin{array}{c}2n\\ n+j\end{array}}\right) \nonumber \\\le & {} \frac{Mm\left( n!\right) ^2(2n)^{2n}m^{2n}}{(2n+1)!(2n)!} \nonumber \\\le & {} \frac{Mm(2n)^{2n}m^{2n}}{(2n+1)}\left( \frac{n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}}{(2n)^{2n+\frac{1}{2}}e^{-2n}e^{\frac{1}{24n+1}}}\right) ^2\nonumber \\= & {} \frac{Mm}{2(2n+1)}\left( \frac{em}{2}\right) ^{2n}e^{\frac{1}{6n}-\frac{2}{24n+1}}\nonumber \\\le & {} Mm \left( \frac{em}{2}\right) ^{2n}. \end{aligned}$$

(A 5)

Here, we used $\sum _{j=0}^k\left( {\begin{array}{c}k\\ j\end{array}}\right) =2^k$, which holds for any $k\in \mathbb {N}$, at the third inequality, and

$$\begin{aligned} \sqrt{2\pi }n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n+1}}<n!<\sqrt{2\pi }n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}, \end{aligned}$$

(A 6)

which is given in [58], at the fourth inequality. $\square$

Appendix B: Proof of Lemma 2

Proof

When $m\ge 2$, because of (A 1), we have

$$\begin{aligned} \left| d^{(m)}_{n,j}\right|\le & {} \frac{m!(n!)^2}{(n+j)!(n-j)!|j|}\sum _{\begin{array}{c} \{l_1,\ldots ,l_{m-1}\}\in \qquad \quad \\ \mathcal {P}_{m-1}([-n:n]\setminus \{0,j\}) \end{array}} \frac{1}{|l_1\cdots l_{m-1}|} \nonumber \\\le & {} \frac{m!}{|j|}\sum _{\begin{array}{c} \{l_1,\ldots ,l_{m-1}\}\in \qquad \quad \\ \mathcal {P}_{m-1}([-n:n]\setminus \{0,j\}) \end{array}} \frac{1}{|l_1\cdots l_{m-1}|} \end{aligned}$$

(B1)

for any $j\in [-n:n]\setminus \{0\}$, where we used

$$\begin{aligned} \frac{(n!)^2}{(n+j)!(n-j)!}=\frac{n(n-1)\cdots (n-j+1)}{(n+j)(n+j-1)\cdots (n+1)}\le 1. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} D^{(m)}_n= & {} \sum _{j=-n}^n\left| d^{(m)}_{n,j}\right| \nonumber \\\le & {} 2\sum _{j\in [-n:n]\setminus \{0\}}\left| d^{(m)}_{n,j}\right| \nonumber \\\le & {} \sum _{j\in [-n:n]\setminus \{0\}}\frac{2m!}{|j|}\sum _{\begin{array}{c} \{l_1,\ldots ,l_{m-1}\}\in \qquad \quad \\ \mathcal {P}_{m-1}([-n:n]\setminus \{0,j\}) \end{array}} \frac{1}{|l_1\cdots l_{m-1}|} \nonumber \\\le & {} 2m!m \sum _{\begin{array}{c} \{l_1,\ldots ,l_{m}\}\in \qquad \quad \\ \mathcal {P}_{m}([-n:n]\setminus \{0\}) \end{array}} \frac{1}{|l_1\cdots l_{m}|}\nonumber \\\le & {} 2m!m\times \frac{1}{m!} \left( \frac{1}{|-n|}+\cdots +\frac{1}{|-1|}+\frac{1}{1}+\cdots +\frac{1}{n}\right) ^m\nonumber \\\le & {} 2m\left[ 2\left( 1+\log n\right) \right] ^m, \end{aligned}$$

(B2)

where the first equality follows from the definition of $d^{(m)}_{n,0}$ in (6).

When $m=1$, $|a^{(1)}_{n,j}| = (n!)^2/|j|$ for $j\in [-n:n]\setminus \{0\}$, and therefore,

$$\begin{aligned} \left| d^{(1)}_{n,j}\right| =\frac{(n!)^2}{(n+j)!(n-j)!|j|} \le \frac{1}{|j|}. \end{aligned}$$

(B3)

This leads to

$$\begin{aligned} D^{(m)}_n\le & {} 2 \sum _{j\in [-n:n]\setminus \{0\}}\left| d^{(m)}_{n,j}\right| \nonumber \\\le & {} 4\sum _{j=1}^n\frac{1}{j}\nonumber \\\le & {} 4(1+\log n), \end{aligned}$$

(B4)

which indicates that (27) holds also for $m=1$. $\square$

Appendix C: Proof of Lemma 3

To prove Lemma 3, let us introduce some additional lemmas.

Lemma 4

For any $a,x\in \mathbb {R}_+$ satisfying $a^2\le x$,

$$\begin{aligned} x\ge a\log x \end{aligned}$$

(C1)

holds.

Proof

Consider a function $f_a(x)=x-a\log x$ on ${\mathbb {R}}_+$. $f_a^\prime (x)=1-\frac{a}{x}$, and therefore, it takes the minimum $a(1-\log a)$ at $x=a$. Then, we consider the following cases.

(i) $a\le e$

Since the minimum of $f_a(x)$ is larger than 0,

$$\begin{aligned} \forall x\in \mathbb {R}_+, f_a(x)\ge 0, \end{aligned}$$

(C2)

which means (C1) holds for any $x\in \mathbb {R}_+$.

(i) $a> e$

As a special case of (C2), $f_2(y)=y-2\log y\ge 0$ holds for any $y\in {\mathbb {R}}_+$, which indicates that $f_a(a^2)=a(a-2\log a)\ge 0$. Besides, $f_a^\prime (x)>0$ for any x larger than $a^2$. Combining these, we see that, for any x larger than $a^2$, $f_a(x)\ge 0$ holds, and so does (C1).

Lemma 5

For any $a\in {\mathbb {R}}_{\ge 0}$, $x\in {\mathbb {R}}_+$, and $\epsilon \in {\mathbb {R}}_+$ such that

$$\begin{aligned} \epsilon \le {\left\{ \begin{array}{ll} 2^{a-\left( \frac{a}{\log 2}\right) ^2} &{}; \ \mathrm{for} \ a \ge \log 2 \\ 2^{a-1} &{}; \ \mathrm{for} \ a < \log 2 \end{array}\right. } \end{aligned}$$

(C3)

and

$$\begin{aligned} x\ge \tilde{x}_{a,\epsilon }&:= \log _2\left( \frac{2^a}{\epsilon }\right) +a\log _2\left( \log _2\left( \frac{2^a}{\epsilon }\right) \right) , \end{aligned}$$

(C4)

$$\begin{aligned}&\qquad \qquad \qquad \qquad \frac{x^a}{2^x} \le \epsilon \end{aligned}$$

(C5)

holds.

Proof

When $a=0$, (C5) becomes $2^{-x}\le \epsilon$ and holds trivially, since (C4) becomes $x\ge \log _2\left( \frac{1}{\epsilon }\right)$. Therefore, we consider the case where $a>0$ below. First, note that, under the condition (C3),

$$\begin{aligned} \log _2 \left( \frac{2^a}{\epsilon }\right) \ge \left( \frac{a}{\log 2}\right) ^2 \end{aligned}$$

(C6)

holds, and this means

$$\begin{aligned} \log _2 \left( \frac{2^a}{\epsilon }\right) \ge \frac{a}{\log 2} \log \left( \log _2 \left( \frac{2^a}{\epsilon }\right) \right) = a \log _2\left( \log _2 \left( \frac{2^a}{\epsilon }\right) \right) , \end{aligned}$$

(C7)

because of Lemma 4. Then, we see that

$$\begin{aligned} \frac{(\tilde{x}_{a,\epsilon })^a}{2^{\tilde{x}_{a,\epsilon }}}= & {} \frac{\left[ \log _2 \left( \frac{2^a}{\epsilon }\right) + a \log _2\left( \log _2 \left( \frac{2^a}{\epsilon }\right) \right) \right] ^a}{\frac{2^a}{\epsilon } \left( \log _2\left( \frac{2^a}{\epsilon }\right) \right) ^a}\nonumber \\\le & {} \frac{2^a \left( \log _2 \left( \frac{2^a}{\epsilon }\right) \right) ^a}{\frac{2^a}{\epsilon } \left( \log _2\left( \frac{2^a}{\epsilon }\right) \right) ^a}\nonumber \\= & {} \epsilon , \end{aligned}$$

(C8)

where we use the definition (C4) of $\tilde{x}_{a,\epsilon }$ at the first equality and (C7) at the inequality.

On the other hand, defining $g(x):=\frac{x^a}{2^x}$, we obtain

$$\begin{aligned} \frac{d}{dx}\log g(x) = \frac{a}{x}-\log 2, \end{aligned}$$

(C9)

which implies that, when $x\ge \frac{a}{\log 2}$, $\log g(x)$ is monotonically deceasing, and so is g(x). Besides, under (C3), we can see that both

$$\begin{aligned} \log _2 \left( \frac{2^a}{\epsilon }\right) \ge 1 \end{aligned}$$

(C10)

and

$$\begin{aligned} \log _2 \left( \frac{2^a}{\epsilon }\right) \ge \frac{a}{\log 2} \end{aligned}$$

(C11)

hold by simple algebra, and therefore,

$$\begin{aligned} \tilde{x}_{a,\epsilon }=\log _2\left( \frac{2^a}{\epsilon }\right) +a\log _2\left( \log _2\left( \frac{2^a}{\epsilon }\right) \right) \ge \frac{a}{\log 2}. \end{aligned}$$

(C12)

Hence, we obtain

$$\begin{aligned} \frac{x^a}{2^x} \le \frac{(\tilde{x}_{a,\epsilon })^a}{2^{\tilde{x}_{a,\epsilon }}} \le \epsilon \end{aligned}$$

(C13)

for $x\ge \tilde{x}_{a,\epsilon }$. $\square$

Now, let us prove Lemma 3.

Proof of Lemma 3

Under (28) and (30), Lemma 5 implies that

$$\begin{aligned} \frac{(2n+1)^{m\sigma ^+}}{2^{2n+1}} \le \epsilon ^\prime . \end{aligned}$$

(C14)

Then, we see that

$$\begin{aligned}&\left| R_f(x,n,m,h)h^{2n-m+1}\right| \nonumber \\&\quad \le Ac^{2n+1}\left( (2n+1)!\right) ^{\sigma }m \left( \frac{em}{2}\right) ^{2n}h^{2n-m+1} \nonumber \\&\quad \le Ac^{2n+1}\left( (2n+1)^{\sigma ^+}\right) ^{2n+1}m \left( \frac{em}{2}\right) ^{2n}h^{2n-m+1} \nonumber \\&\quad = \frac{2A}{e}\left( \frac{ecm(2n+1)^{\sigma ^+} h}{2}\right) ^{2n+1}h^{-m} \nonumber \\&\quad \le \frac{2(ecm)^mA}{e}\frac{(2n+1)^{m\sigma ^+}}{2^{2n+1}} \nonumber \\&\quad \le \frac{2(ecm)^mA}{e}\epsilon ^\prime \nonumber \\&\quad = \epsilon . \end{aligned}$$

(C15)

Here, at the first inequality, we use (25) with $M=Ac^{2n+1}\left( (2n+1)!\right) ^{\sigma }$, since $f\in \mathcal {G}_{A,C,\sigma }$. We also use $((2n+1)!)^\sigma \le ((2n+1)^{\sigma ^+})^{2n+1}$ at the second inequality, (29) at the third inequality, (C14) at the fourth inequality, and (31) at the last equality. Because of Theorem 1, (C15) indicates that (32) holds. $\square$