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For each of the exercises in this section that requires a simulation, see if you can think of a way to use control variates or antithetic variates to reduce the variance of the simulation estimator.

Short Answer

Expert verified

Antithetic variates to reduce the variance of the simulation estimator are considered in exercises 4, 6, and 10.

Step by step solution

01

Reduce the variance of the simulation estimator

Consider exercises 4, 6, and 10. Perhaps in another exercise, one of the methods could also be used. See solutions for exercises 14 and 16.

In exercise 4, part (b), the method could be used in this case set function h(x) to be\({e^{ - x}}\).

In exercise 6, part(a) and part(b), the quotient of the two functions is monotone, implying that the antithetic variates method could be used. See the solution to exercise 6.

02

Exercises 6 in this section

The method could be used in exercise 6, part (a). In this case, set function h(x) to be\(x{e^{ - {x^2}/2}}\). See the solution to exercise 6.

Exercise 10 is based on exercise 6, so one could use both methods described earlier.

Here, the final result is considered for exercises 4, 6, and 10.

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Most popular questions from this chapter

Show how to simulate Cauchy random variables using the probability integral transformation.

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.

Let \({\bf{f}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{ = cg}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) be a joint p.d.f for \(\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{,}}\)each \({x_{2\,}}\)let\({{\bf{h}}_{{\bf{2}}\,}}\left( {{{\bf{x}}_{{\bf{1}}\,}}} \right){\bf{ = g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) that is what we get by considering \({\bf{g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) as a function of \({{\bf{x}}_{{\bf{1}}\,}}\)for fixed \({{\bf{x}}_{2\,}}\)show that there is a multiplicative factor \({{\bf{c}}_{{\bf{2}}\,}}\)that does not depend on such that is the conditional p.d.f of \({{\bf{x}}_{{\bf{1}}\,}}\) given \(\left( {{{\bf{x}}_{{\bf{2}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\)

The method of antithetic variates is a technique for reducing the variance of simulation estimators. Antithetic variates are negatively correlated random variables with an expected mean and variance. The variance of the average of two antithetic variates is smaller than the variance of the average of two i.i.d. variables. In this exercise, we shall see how to use antithetic variates for importance sampling, but the method is very general. Suppose that we wish to compute \(\smallint \,g\left( x \right)\,\,dx\), and we wish to use the importance function f. Suppose that we generate pseudo-random variables with the p.d.f. f using the integral probability transformation. For \(\,{\bf{i = 1,2,}}...{\bf{,\nu ,}}\,\)let \({{\bf{X}}^{\left( {\bf{i}} \right)}}{\bf{ = }}{{\bf{F}}^{{\bf{ - 1}}}}\left( {{\bf{1 - }}{{\bf{U}}^{\left( {\bf{i}} \right)}}} \right)\), where \({{\bf{U}}^{\left( {\bf{i}} \right)}}\)has the uniform distribution on the interval (0, 1) and F is the c.d.f. Corresponding to the p.d.f. f . For each \(\,{\bf{i = 1,2,}}...{\bf{,\nu ,}}\,\) define

\(\begin{aligned}{l}{{\bf{T}}^{\left( {\bf{i}} \right)}}{\bf{ = }}{{\bf{F}}^{ - {\bf{1}}}}\left( {{\bf{1}} - {{\bf{U}}^{\left( {\bf{i}} \right)}}} \right)\,\,{\bf{.}}\\{{\bf{W}}^{\left( {\bf{i}} \right)}}{\bf{ = }}\frac{{{\bf{g}}\left( {{{\bf{X}}^{\left( {\bf{i}} \right)}}} \right)}}{{{\bf{f}}\left( {{{\bf{X}}^{\left( {\bf{i}} \right)}}} \right)}}\\{{\bf{V}}^{\left( {\bf{i}} \right)}}{\bf{ = }}\frac{{{\bf{g}}\left( {{{\bf{T}}^{\left( {\bf{i}} \right)}}} \right)}}{{{\bf{f}}\left( {{{\bf{T}}^{\left( {\bf{i}} \right)}}} \right)}}\\{{\bf{Y}}^{\left( {\bf{i}} \right)}}{\bf{ = 0}}{\bf{.5}}\left( {{{\bf{W}}^{\left( {\bf{i}} \right)}}{\bf{ + k}}{{\bf{V}}^{\left( {\bf{i}} \right)}}} \right){\bf{.}}\end{aligned}\)

Our estimator of\(\smallint \,{\bf{g}}\left( {\bf{x}} \right)\,\,{\bf{dx}}\)is then\({\bf{Z = }}\frac{{\bf{I}}}{{\bf{\nu }}}\sum\nolimits_{{\bf{i = 1}}}^{\bf{\nu }} {{{\bf{Y}}^{\left( {\bf{i}} \right)}}{\bf{.}}} \).

a. Prove that\({T^{\left( i \right)}}\)has the same distribution as\({X^{\left( i \right)}}\).

b. Prove that\({\bf{E}}\left( {\bf{Z}} \right){\bf{ = }}\smallint \,\,{\bf{g}}\left( {\bf{x}} \right)\,\,{\bf{dx}}\).

c. If\({\bf{g}}\left( {\bf{x}} \right)\,{\bf{/f}}\left( {\bf{x}} \right)\)it is a monotone function, explain why we expect it \({{\bf{V}}^{\left( {\bf{i}} \right)}}\)to be negatively correlated.

d. If \({{\bf{W}}^{\left( {\bf{i}} \right)}}\) and \({{\bf{V}}^{\left( {\bf{i}} \right)}}\)are negatively correlated, show that Var(Z) is less than the variance one would get with 2v simulations without antithetic variates.

Use the data consisting of 30 lactic acid concentrations in cheese,10 from example 8.5.4 and 20 from Exercise 16 in sec.8,6, Fit the same model used in Example 8.6.2 with the same prior distribution, but this time use the Gibbs sampling algorithm in Example 12.5.1. simulate 10,000 pairs of \(\left( {{\bf{\mu ,\tau }}} \right)\) parameters. Estimate the posterior mean of \({\left( {\sqrt {{\bf{\tau \mu }}} } \right)^{ - {\bf{1}}}}\), and compute the standard simulation error of the estimator.

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