Question

Use the method of antithetic variates that was described in Exercise 15. Let g(x) be the function that we tried to integrate into Example 12.4.1. Let f (x) be the function\({f_3}\)in Example 12.4.1. Estimate Var\(\left( {{V^{\left( i \right)}}} \right)\), and compare it to\(\mathop \sigma \limits\ _32\)Example 12.4.1.

Short answer

Estimator is \(Z = \frac{1}{\nu }\sum\limits_{i = 1}^\nu {{Y^{\left( i \right)}}.} \)

The estimate of the integral in the first simulation is 0.52466, and the simulation standard error is \(2.398\,\, \cdot \,\,{10^{ - 6}}\).

Step-by-step solution

Find the estimator

From the exercise and the example,

\(\begin{aligned}{l}g\left( x \right) = \frac{{{e^{ - x}}}}{{1 + {x^2}}},\,\,\,0 < x < 1,\\{f_3}\left( x \right) = \frac{{{e^{ - x}}}}{{1 - {e^{ - 1}}}},\,\,\,0 < x < 1.\end{aligned}\)

First, simulate\({X^{\left( i \right)}},\,\,i = 1,2,....,\nu \). To do that, use integral probability transformation. The cumulative density function is

\({F_3}\left( x \right) = \,\,\,\smallint _0^X\,\,\frac{{{e^{ - t}}}}{{1 - {e^{ - 1}}}}dt = \frac{{1 - {e^{ - x}}}}{{1 - {e^{ - 1}}}},\,\,\,\,\,\,\,0 < x < 1.\)

Let U be from the uniform distribution on the interval (0, 1). It follows that X values can be simulated using U as

\(X = - \log \left( {1 - U\left( {1 - {e^{ - 1}}} \right)} \right)\)

which is obtained from

\(U = \frac{{1 - {e^{ - x}}}}{{1 - {e^{ - 1}}}}.\)

The random variable T is

\({T^{\left( i \right)}} = {F^{ - 1}}\left( {1 - {U^{\left( i \right)}}} \right) = - \log \,\left( {1 - \left( {1 - {U^{\left( i \right)}}} \right)\,\,\left( {1 - {e^{ - 1}}} \right)} \right)\,\,,\,\,\,\,\,i = 1,2,...,\nu ,\)

Simulations of W and V are obtained as

\(\begin{aligned}{l}{W^{\left( i \right)}} &= \frac{{g\left( {{X^{\left( i \right)}}} \right)}}{{{f_3}\left( {{X^{\left( i \right)}}} \right)}} &= \frac{{1 - {e^{ - 1}}}}{{1 + {X^{\left( i \right)\,2}}}},\,\,\,\,\,i &= 1,2,...,\nu ,\\{V^{\left( i \right)}} &= \frac{{g\left( {{T^{\left( i \right)}}} \right)}}{{{f_3}\left( {{T^{\left( i \right)}}} \right)}} &= \frac{{1 - {e^{ - 1}}}}{{1 + {T^{\left( i \right)\,2}}}},\,\,\,\,\,i = 1,2,...,\nu ,\end{aligned}\)

Using those, simulations of Y are

\({Y^{\left( i \right)}} = 0.5\,\,.\,\,\left( {{W^{\left( i \right)}} + {V^{\left( i \right)}}} \right)\)

And the estimator

\(Z = \frac{1}{\nu }\sum\limits_{i = 1}^\nu {{Y^{\left( i \right)}}.} \)

Use the following code in R to estimate the integral. The estimate of the integral in the first simulation is 0.52466, and the simulation standard error is \(2.398\,\, \cdot \,\,{10^{ - 6}}\). The simulation standard deviation is smaller than in the example.

Size of simulation

#Size of simulation

N = 10000

#Sample from uniform

sample.u = runif(N,0,1)

#Simulation of X and T

sample.x = -log(1-sample.u*(1-exp(-1)))

sample.t = -log(1-(1-sample.u)*(1-exp(-1)))

#Simulation of W and V

sample.w = (1-exp(-1))/(1+sample.x^2)

sample.v = (1-exp(-1))/(1+sample.t^2)

#Simulation of Y

sample.y = 0.5 * (sample.w + sample.v)

#Estimate

estimate.int = mean(sample.y)

#simulation standard deviation

estimate.sd = sd(sample.y)/N

Here, the estimate of the integral in the first simulation is 0.52466, and the simulation standard error is \(2.398\,\, \cdot \,\,{10^{ - 6}}\).