Integration Examples using QMCPy package

In this demo, we show how to use qmcpy for performing numerical multiple integration of two built-in integrands, namely, the Keister function and the Asian call option payoff. To start, we import the qmcpy module and the function arrange() from numpy for generating evenly spaced discrete vectors in the examples.

from qmcpy import *
from numpy import arange

Keister Example

We recall briefly the mathematical definitions of the Keister function, the Gaussian measure, and the Sobol distribution:

  • Keister integrand: y_j = \pi^{d/2} \cos(||x_j||_2)

  • Gaussian true measure: \mathcal{N}(0,\frac{1}{2})

  • Sobol discrete distribution: x_j \overset{lLD}{\sim} \mathcal{U}(0,1)

integrand = Keister(Sobol(dimension=3,seed=7))
solution,data = CubQMCSobolG(integrand,abs_tol=.05).integrate()
print(data)
Solution: 2.1661
Keister (Integrand Object)
Sobol (DiscreteDistribution Object)
    d               3
    randomize       1
    graycode        0
    seed            7
    mimics          StdUniform
    dim0            0
Gaussian (TrueMeasure Object)
    mean            0
    covariance      2^(-1)
    decomp_type     pca
CubQMCSobolG (StoppingCriterion Object)
    abs_tol         0.050
    rel_tol         0
    n_init          2^(10)
    n_max           2^(35)
LDTransformData (AccumulateData Object)
    n_total         2^(10)
    solution        2.166
    error_bound     0.011
    time_integrate  0.002

Arithmetic-Mean Asian Put Option: Single Level

In this example, we want to estimate the payoff of an European Asian put option that matures at time T. The key mathematical entities are defined as follows:

  • Stock price at time t_j := jT/d for j=1,\dots,d is a function of its initial price S(0), interest rate r, and volatility \sigma: S(t_j) = S(0)e^{\left(r-\frac{\sigma^2}{2}\right)t_j + \sigma\mathcal{B}(t_j)}

  • Discounted put option payoff is defined as the difference of a fixed strike price K and the arithmetic average of the underlying stock prices at d discrete time intervals in [0,T]: \max \left(K-\frac{1}{d}\sum_{j=1}^{d} S(t_j), 0 \right) e^{-rT}

  • Brownian motion true measure: \mathcal{B}(t_j) = \mathcal{B}(t_{j-1}) + Z_j\sqrt{t_j-t_{j-1}} \; for \;Z_j \sim \mathcal{N}(0,1)

  • Lattice discrete distribution: \:\: x_j \overset{LD}{\sim} \mathcal{U}(0,1)

integrand = AsianOption(
    sampler = IIDStdUniform(dimension=16, seed=7),
    volatility = 0.5,
    start_price = 30,
    strike_price = 25,
    interest_rate = 0.01,
    mean_type = 'arithmetic')
solution,data = CubMCCLT(integrand, abs_tol=.025).integrate()
print(data)
Solution: 6.2662
AsianOption (Integrand Object)
    volatility      2^(-1)
    call_put        call
    start_price     30
    strike_price    25
    interest_rate   0.010
    mean_type       arithmetic
    dimensions      2^(4)
    dim_fracs       0
IIDStdUniform (DiscreteDistribution Object)
    d               2^(4)
    seed            7
    mimics          StdUniform
BrownianMotion (TrueMeasure Object)
    time_vec        [0.062 0.125 0.188 ... 0.875 0.938 1.   ]
    drift           0
    mean            [0. 0. 0. ... 0. 0. 0.]
    covariance      [[0.062 0.062 0.062 ... 0.062 0.062 0.062]
                    [0.062 0.125 0.125 ... 0.125 0.125 0.125]
                    [0.062 0.125 0.188 ... 0.188 0.188 0.188]
                    ...
                    [0.062 0.125 0.188 ... 0.875 0.875 0.875]
                    [0.062 0.125 0.188 ... 0.875 0.938 0.938]
                    [0.062 0.125 0.188 ... 0.875 0.938 1.   ]]
    decomp_type     pca
CubMCCLT (StoppingCriterion Object)
    inflate         1.200
    alpha           0.010
    abs_tol         0.025
    rel_tol         0
    n_init          2^(10)
    n_max           10000000000
MeanVarData (AccumulateData Object)
    levels          1
    solution        6.266
    n               874562
    n_total         875586
    error_bound     0.026
    confid_int      [6.241 6.292]
    time_integrate  2.133

Arithmetic-Mean Asian Put Option: Multi-Level

This example is similar to the last one except that we use a multi-level method for estimation of the option price. The main idea can be summarized as follows:

Y_0 = 0

Y_1 = \text{ Asian option monitored at } t = [\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1]

Y_2 = \text{ Asian option monitored at } t= [\frac{1}{16}, \frac{1}{8}, ... , 1]

Y_3 = \text{ Asian option monitored at } t= [\frac{1}{64}, \frac{1}{32}, ... , 1]

Z_1 = \mathbb{E}[Y_1-Y_0] + \mathbb{E}[Y_2-Y_1] + \mathbb{E}[Y_3-Y_2] = \mathbb{E}[Y_3]

integrand = AsianOption(
        sampler = IIDStdUniform(seed=7),
        volatility = 0.5,
        start_price = 30,
        strike_price = 25,
        interest_rate = 0.01,
        mean_type = 'arithmetic',
        multi_level_dimensions = [4,8,16])
solution,data = CubMCCLT(integrand, abs_tol=.025).integrate()
print(data)
Solution: 6.2891
AsianOption (Integrand Object)
    volatility      2^(-1)
    call_put        call
    start_price     30
    strike_price    25
    interest_rate   0.010
    mean_type       arithmetic
    dimensions      [ 4  8 16]
    dim_fracs       [0. 2. 2.]
IIDStdUniform (DiscreteDistribution Object)
    d               2^(4)
    seed            7
    mimics          StdUniform
BrownianMotion (TrueMeasure Object)
    time_vec        [0.062 0.125 0.188 ... 0.875 0.938 1.   ]
    drift           0
    mean            [0. 0. 0. ... 0. 0. 0.]
    covariance      [[0.062 0.062 0.062 ... 0.062 0.062 0.062]
                    [0.062 0.125 0.125 ... 0.125 0.125 0.125]
                    [0.062 0.125 0.188 ... 0.188 0.188 0.188]
                    ...
                    [0.062 0.125 0.188 ... 0.875 0.875 0.875]
                    [0.062 0.125 0.188 ... 0.875 0.938 0.938]
                    [0.062 0.125 0.188 ... 0.875 0.938 1.   ]]
    decomp_type     pca
CubMCCLT (StoppingCriterion Object)
    inflate         1.200
    alpha           0.010
    abs_tol         0.025
    rel_tol         0
    n_init          2^(10)
    n_max           10000000000
MeanVarData (AccumulateData Object)
    levels          3
    solution        6.289
    n               [1693278.   34067.   30106.]
    n_total         1760523
    error_bound     0.025
    confid_int      [6.264 6.315]
    time_integrate  1.261

Keister Example using Bayesian Cubature

This examples repeats the Keister using cubBayesLatticeG and cubBayesNetG stopping criterion:

  • Keister integrand: y_j = \pi^{d/2} \cos(||x_j||_2)

  • Gaussian true measure: \mathcal{N}(0,\frac{1}{2})

  • Lattice discrete distribution: x_j \overset{lLD}{\sim} \mathcal{U}(0,1)

dimension=3
abs_tol=.001
integrand = Keister(Lattice(dimension=dimension, order='linear'))
solution,data = CubBayesLatticeG(integrand,abs_tol=abs_tol).integrate()
print(data)
Solution: 2.1683
Keister (Integrand Object)
Lattice (DiscreteDistribution Object)
    d               3
    randomize       1
    order           linear
    seed            894913
    mimics          StdUniform
Gaussian (TrueMeasure Object)
    mean            0
    covariance      2^(-1)
    decomp_type     pca
CubBayesLatticeG (StoppingCriterion Object)
    abs_tol         0.001
    rel_tol         0
    n_init          2^(8)
    n_max           2^(22)
    order           2^(1)
LDTransformBayesData (AccumulateData Object)
    n_total         4096
    solution        2.168
    error_bound     7.12e-04
    time_integrate  0.092
dimension=3
abs_tol=.001
integrand = Keister(Sobol(dimension=dimension, graycode=False))
solution,data = CubBayesNetG(integrand,abs_tol=abs_tol).integrate()
print(data)
Solution: 2.1683
Keister (Integrand Object)
Sobol (DiscreteDistribution Object)
    d               3
    randomize       1
    graycode        0
    seed            612251
    mimics          StdUniform
    dim0            0
Gaussian (TrueMeasure Object)
    mean            0
    covariance      2^(-1)
    decomp_type     pca
CubBayesNetG (StoppingCriterion Object)
    abs_tol         0.001
    rel_tol         0
    n_init          2^(8)
    n_max           2^(22)
LDTransformBayesData (AccumulateData Object)
    n_total         16384
    solution        2.168
    error_bound     5.06e-04
    time_integrate  0.208