Importance Sampling Examples

from qmcpy import *
from numpy import *
import pandas as pd
pd.options.display.float_format = '{:.2e}'.format
from matplotlib import pyplot as plt
import matplotlib
%matplotlib inline

Game Example

Consider a game where X_1, X_2 \overset{\textrm{IID}}{\sim} \mathcal{U}(0,1) are drawn with a payoff of

Y = \text{payoff}(X_1,X_2) = \begin{cases} \$10, & 1.7 \le X_1 + X_2 \le 2, \\ 0, & 0 \le X_1 + X_2 < 1.7, \end{cases}

What is the expected payoff of this game?

payoff = lambda x: 10*(x.sum(1)>1.7)
abs_tol = 1e-3

Vanilla Monte Carlo

With ordinary Monte Carlo we do the following:

\mu = \mathbb{E}(Y) = \int_{[0,1]^2} \text{payoff}(x_1,x_2) \,
\mathrm{d} x_1 \mathrm{d}x_2

d = 2
integral = CustomFun(
    true_measure = Uniform(Lattice(d)),
    g = payoff)
solution1,data1 = CubQMCLatticeG(integral, abs_tol).integrate()
data1
Solution: 0.4498
CustomFun (Integrand Object)
Lattice (DiscreteDistribution Object)
    d               2^(1)
    randomize       1
    order           natural
    seed            519533
    mimics          StdUniform
Uniform (TrueMeasure Object)
    lower_bound     0
    upper_bound     1
CubQMCLatticeG (StoppingCriterion Object)
    abs_tol         0.001
    rel_tol         0
    n_init          2^(10)
    n_max           2^(35)
LDTransformData (AccumulateData Object)
    n_total         2^(16)
    solution        0.450
    error_bound     6.78e-04
    time_integrate  0.057

Monte Carlo with Importance Sampling

We may add the importance sampling to increase the number of samples with positive payoffs. Let

\boldsymbol{Z} = (X_1^{1/(p+1)}, X_2^{1/(p+1)}), \qquad \boldsymbol{X} \sim \mathcal{U}(0,1)^2

This means that Z_1 and Z_2 are IID with common CDF F(z) =z^{p+1} and common PDF \varrho(z) = (p+1)z^{p}. Thus,

\mu = \mathbb{E}(Y) =  \int_{[0,1]^2} \frac{\text{payoff}(z_1,z_2)}{(p+1)^2(z_1z_2)^{p}} \, \varrho(z_1)
\varrho(z_2) \, \mathrm{d} z_1 \mathrm{d}z_2 \\
= \int_{[0,1]^2}
\frac{\text{payoff}(x_1^{1/(p+1)},x_2^{1/(p+1)})}{(p+1)^2(x_1x_2)^{p/(p+1)}}
\, \mathrm{d} x_1 \mathrm{d}x_2

p = 1
d = 2
integral = CustomFun(
    true_measure = Uniform(Lattice(d)),
    g = lambda x: payoff(x**(1/(p+1))) / ((p+1)**2 * (x.prod(1))**(p/(p+1))))
solution2,data2 = CubQMCLatticeG(integral, abs_tol).integrate()
data2
Solution: 0.4497
CustomFun (Integrand Object)
Lattice (DiscreteDistribution Object)
    d               2^(1)
    randomize       1
    order           natural
    seed            659232
    mimics          StdUniform
Uniform (TrueMeasure Object)
    lower_bound     0
    upper_bound     1
CubQMCLatticeG (StoppingCriterion Object)
    abs_tol         0.001
    rel_tol         0
    n_init          2^(10)
    n_max           2^(35)
LDTransformData (AccumulateData Object)
    n_total         2^(14)
    solution        0.450
    error_bound     6.56e-04
    time_integrate  0.020
print('Imporance Sampling takes %.3f the time and %.3f the samples'%\
     (data2.time_integrate/data1.time_integrate,data2.n_total/data1.n_total))
Imporance Sampling takes 0.359 the time and 0.250 the samples

Asian Call Option Example

The stock price must raise significantly for the payoff to be positive. So we will give a upward drift to the Brownian motion that defines the stock price path. We can think of the option price as the multidimensional integral

\mu = \mathbb{E}[f(\boldsymbol{X})] = \int_{\mathbb{R}^d}
  f(\boldsymbol{x})
  \frac{\exp\bigl(-\frac{1}{2} \boldsymbol{x}^T\mathsf{\Sigma}^{-1}
  \boldsymbol{x}\bigr)}
  {\sqrt{(2 \pi)^{d} \det(\mathsf{\Sigma})}} \, \mathrm{d} \boldsymbol{x}

where

\boldsymbol{X}  \sim \mathcal{N}(\boldsymbol{0}, \mathsf{\Sigma}), \qquad
\mathsf{\Sigma} = \bigl(\min(j,k)T/d \bigr)_{j,k=1}^d,  \qquad
d  =  13, \\
f(\boldsymbol{x})  = \max\biggl(K - \frac 1d \sum_{j=1}^d
S(jT/d,\boldsymbol{x}), 0 \biggr) \mathrm{e}^{-rT}, \\
S(jT/d,\boldsymbol{x}) = S(0) \exp\bigl((r - \sigma^2/2) jT/d +
\sigma x_j\bigr).

We will replace \boldsymbol{X} by

\boldsymbol{Z} \sim \mathcal{N}(\boldsymbol{a}, \mathsf{\Sigma}),
\qquad \boldsymbol{a} = (aT/d)(1, \ldots, d)

where a positive a will create more positive payoffs. This corresponds to giving our Brownian motion a drift. To do this we re-write the integral as

\mu = \mathbb{E}[f_{\mathrm{new}}(\boldsymbol{Z})]
= \int_{\mathbb{R}^d}
f_{\mathrm{new}}(\boldsymbol{z})
\frac{\exp\bigl(-\frac{1}{2} (\boldsymbol{z}-\boldsymbol{a})^T
\mathsf{\Sigma}^{-1}
(\boldsymbol{z} - \boldsymbol{a}) \bigr)}
{\sqrt{(2 \pi)^{d} \det(\mathsf{\Sigma})}} \, \mathrm{d} \boldsymbol{z} ,
\\
f_{\mathrm{new}}(\boldsymbol{z}) =
f(\boldsymbol{z})
\frac{\exp\bigl(-\frac{1}{2} \boldsymbol{z}^T
\mathsf{\Sigma}^{-1} \boldsymbol{z} \bigr)}
{\exp\bigl(-\frac{1}{2} (\boldsymbol{z}-\boldsymbol{a})^T
\mathsf{\Sigma}^{-1}
(\boldsymbol{z} - \boldsymbol{a}) \bigr)}
= f(\boldsymbol{z}) \exp\bigl((\boldsymbol{a}/2 - \boldsymbol{z})^T
\mathsf{\Sigma}^{-1}\boldsymbol{a} \bigr)

Finally note that

\mathsf{\Sigma}^{-1}\boldsymbol{a} = \begin{pmatrix} 0 \\ 0 \\ \vdots
\\ 0 \\ a \end{pmatrix}, \qquad f_{\mathrm{new}}(\boldsymbol{z}) =
f(\boldsymbol{z}) \exp\bigl((aT/2 - z_d)a \bigr)

This drift in the Brownian motion may be implemented by changing the drift input to the BrownianMotion object.

abs_tol = 1e-2
dimension = 32

Vanilla Monte Carlo

integrand = AsianOption(Sobol(dimension))
solution1,data1 = CubQMCSobolG(integrand, abs_tol).integrate()
data1
Solution: 1.7899
AsianOption (Integrand Object)
    volatility      2^(-1)
    call_put        call
    start_price     30
    strike_price    35
    interest_rate   0
    mean_type       arithmetic
    dimensions      2^(5)
    dim_fracs       0
Sobol (DiscreteDistribution Object)
    d               2^(5)
    randomize       1
    graycode        0
    seed            370922
    mimics          StdUniform
    dim0            0
BrownianMotion (TrueMeasure Object)
    time_vec        [0.031 0.062 0.094 ... 0.938 0.969 1.   ]
    drift           0
    mean            [0. 0. 0. ... 0. 0. 0.]
    covariance      [[0.031 0.031 0.031 ... 0.031 0.031 0.031]
                    [0.031 0.062 0.062 ... 0.062 0.062 0.062]
                    [0.031 0.062 0.094 ... 0.094 0.094 0.094]
                    ...
                    [0.031 0.062 0.094 ... 0.938 0.938 0.938]
                    [0.031 0.062 0.094 ... 0.938 0.969 0.969]
                    [0.031 0.062 0.094 ... 0.938 0.969 1.   ]]
    decomp_type     pca
CubQMCSobolG (StoppingCriterion Object)
    abs_tol         0.010
    rel_tol         0
    n_init          2^(10)
    n_max           2^(35)
LDTransformData (AccumulateData Object)
    n_total         2^(12)
    solution        1.790
    error_bound     0.008
    time_integrate  0.029

Monte Carlo with Importance Sampling

drift = 1
integrand = AsianOption(BrownianMotion(Sobol(dimension),drift=drift))
solution2,data2 = CubQMCSobolG(integrand, abs_tol).integrate()
data2
Solution: 1.7802
AsianOption (Integrand Object)
    volatility      2^(-1)
    call_put        call
    start_price     30
    strike_price    35
    interest_rate   0
    mean_type       arithmetic
    dimensions      2^(5)
    dim_fracs       0
Sobol (DiscreteDistribution Object)
    d               2^(5)
    randomize       1
    graycode        0
    seed            140003
    mimics          StdUniform
    dim0            0
BrownianMotion (TrueMeasure Object)
    time_vec        [0.031 0.062 0.094 ... 0.938 0.969 1.   ]
    drift           0
    mean            [0. 0. 0. ... 0. 0. 0.]
    covariance      [[0.031 0.031 0.031 ... 0.031 0.031 0.031]
                    [0.031 0.062 0.062 ... 0.062 0.062 0.062]
                    [0.031 0.062 0.094 ... 0.094 0.094 0.094]
                    ...
                    [0.031 0.062 0.094 ... 0.938 0.938 0.938]
                    [0.031 0.062 0.094 ... 0.938 0.969 0.969]
                    [0.031 0.062 0.094 ... 0.938 0.969 1.   ]]
    decomp_type     pca
    transform       BrownianMotion (TrueMeasure Object)
                       time_vec        [0.031 0.062 0.094 ... 0.938 0.969 1.   ]
                       drift           1
                       mean            [0.031 0.062 0.094 ... 0.938 0.969 1.   ]
                       covariance      [[0.031 0.031 0.031 ... 0.031 0.031 0.031]
                                       [0.031 0.062 0.062 ... 0.062 0.062 0.062]
                                       [0.031 0.062 0.094 ... 0.094 0.094 0.094]
                                       ...
                                       [0.031 0.062 0.094 ... 0.938 0.938 0.938]
                                       [0.031 0.062 0.094 ... 0.938 0.969 0.969]
                                       [0.031 0.062 0.094 ... 0.938 0.969 1.   ]]
                       decomp_type     pca
CubQMCSobolG (StoppingCriterion Object)
    abs_tol         0.010
    rel_tol         0
    n_init          2^(10)
    n_max           2^(35)
LDTransformData (AccumulateData Object)
    n_total         2^(11)
    solution        1.780
    error_bound     0.007
    time_integrate  0.027
print('Imporance Sampling takes %.3f the time and %.3f the samples'%\
     (data2.time_integrate/data1.time_integrate,data2.n_total/data1.n_total))
Imporance Sampling takes 0.922 the time and 0.500 the samples

Importance Sampling MC vs QMC

Test Parameters

  • dimension = 16

  • abs_tol = .025

  • trials = 3

df = pd.read_csv('../workouts/mc_vs_qmc/out/importance_sampling.csv')
df['Problem'] = df['Stopping Criterion'] + ' ' + df['Distribution'] + ' (' + df['MC/QMC'] + ')'
df = df.drop(['Stopping Criterion','Distribution','MC/QMC'],axis=1)
problems = ['CubMCCLT IIDStdUniform (MC)',
            'CubMCG IIDStdUniform (MC)',
            'CubQMCCLT Sobol (QMC)',
            'CubQMCLatticeG Lattice (QMC)',
            'CubQMCSobolG Sobol (QMC)']
df = df[df['Problem'].isin(problems)]
mean_shifts = df.mean_shift.unique()
df_samples = df.groupby(['Problem'])['n_samples'].apply(list).reset_index(name='n')
df_times = df.groupby(['Problem'])['time'].apply(list).reset_index(name='time')
df.loc[(df.mean_shift==0) | (df.mean_shift==1)].set_index('Problem')
# Note: mean_shift==0 --> NOT using importance sampling
mean_shift solution n_samples time
Problem
CubMCCLT IIDStdUniform (MC) 0.00e+00 1.77e+00 3.15e+05 1.25e+00
CubMCCLT IIDStdUniform (MC) 1.00e+00 1.79e+00 8.79e+04 3.34e-01
CubMCG IIDStdUniform (MC) 0.00e+00 1.77e+00 4.12e+05 1.71e+00
CubMCG IIDStdUniform (MC) 1.00e+00 1.80e+00 1.28e+05 5.03e-01
CubQMCCLT Sobol (QMC) 0.00e+00 1.78e+00 8.19e+03 5.45e-01
CubQMCCLT Sobol (QMC) 1.00e+00 1.79e+00 4.10e+03 2.89e-01
CubQMCLatticeG Lattice (QMC) 0.00e+00 1.78e+00 1.02e+03 7.08e-03
CubQMCLatticeG Lattice (QMC) 1.00e+00 1.78e+00 1.02e+03 6.91e-03
CubQMCSobolG Sobol (QMC) 0.00e+00 1.78e+00 2.05e+03 8.36e-03
CubQMCSobolG Sobol (QMC) 1.00e+00 1.78e+00 1.02e+03 4.56e-03
fig,ax = plt.subplots(nrows=1, ncols=2, figsize=(20, 6))
idx = arange(len(problems))
width = .35
ax[0].barh(idx+width,log(df.loc[df.mean_shift==0]['n_samples'].values),width)
ax[0].barh(idx,log(df.loc[df.mean_shift==1]['n_samples'].values),width)
ax[1].barh(idx+width,df.loc[df.mean_shift==0]['time'].values,width)
ax[1].barh(idx,df.loc[df.mean_shift==1]['time'].values,width)
fig.suptitle('Importance Sampling Comparison by Stopping Criterion on Asian Call Option')
xlabs = ['log(Samples)','Time']
for i in range(len(ax)):
    ax[i].set_xlabel(xlabs[i])
    ax[i].spines['top'].set_visible(False)
    ax[i].spines['bottom'].set_visible(False)
    ax[i].spines['right'].set_visible(False)
    ax[i].spines['left'].set_visible(False)
    ax[1].legend(['Vanilla Monte Carlo','Importance Sampling\nMean Shift=1'],loc='upper right',frameon=False)
ax[1].get_yaxis().set_ticks([])
ax[0].set_yticks(idx)
ax[0].set_yticklabels(problems)
plt.tight_layout();
../_images/importance_sampling_18_0.png
fig,ax = plt.subplots(nrows=1, ncols=2, figsize=(22, 8))
df_samples.apply(lambda row: ax[0].plot(mean_shifts,row.n,label=row['Problem']),axis=1)
df_times.apply(lambda row: ax[1].plot(mean_shifts,row.time,label=row['Problem']),axis=1)
ax[1].legend(frameon=False, loc=(-.85,.7),ncol=1)
ax[0].set_ylabel('Total Samples')
ax[0].set_yscale('log')
ax[1].set_yscale('log')
ax[1].set_ylabel('Runtime')
for i in range(len(ax)):
    ax[i].set_xlabel('mean shift')
    ax[i].spines['top'].set_visible(False)
    ax[i].spines['right'].set_visible(False)
fig.suptitle('Comparing Mean Shift Across Problems');
../_images/importance_sampling_19_0.png