Statistical Computing: Exam 1 Chapters 1, 3, 5

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1 Statistical Computing: Exam Chapters, 3, 5 Instructions: Read each question carefully before determining the best answer. Show all work; supporting computer code and output must be attached to the question for which it is used. Use computer output as a guide only; all questions must be answered in full on the exam paper. Report all final numerical answers to a precision of 4 units past the decimal point. There are total points on this exam. Do not discuss this exam or its components with anyone besides the course instructor. This exam is due February 27, 28, at 2:5 PM.. (5 points) Suppose U ~ U(,). Using the transformation technique from mathematical statistics, show that T = U ~ U(,). Be sure to clearly explain the reasoning underlying all of your steps. (Hint: work with indicator functions to explicitly show over what support ranges the variables are defined.) Let U ~ U(,). As per the Hint, the p.d.f. is clearly f U (u) = I (,) (u), where I A (u) is the indicator function that returns if u A and zero otherwise. Now take T = g(u) = U. To find f T (t) via the transformation technique we need the inverse function g (u), which is clearly g (t) = u = t for any t (,). Notice that the support of T is also the unit interval (,), so f T (t) will be defined using the indicator function I (,) (t). With this, apply Theorem 2..5 from Casella & Berger (22): f T (t) = f U (g (t)) g (t)/ t I (,) (t). Clearly f U (g (t)) = I (,) ( t), while g (t)/ t = ( t)/ t =. Thus we find f T (t) = I (,) ( t) I (,) (t) Since the product I (,) ( t)i (,) (t) always equals the result of I (,) (t), we find f T (t) = I (,) (t) which shows that T ~ U(,), as desired.

2 2. Consider the function ƒ X (x) = 4 x3 I (,2) (x). It can be seen that this is a valid p.d.f., so suppose a random variable X possesses this distribution. We wish to approximate moments of X via Monte Carlo integration using the simple Monte Carlo approach in Sec a. (5 points) Target the second moment = E[X ] = x2 ƒ X (x) dx = 4 x5 dx. Begin by transforming so that the range of integration is over the unit interval < u <, i.e., find a function (u) such that = g 2 (u) du. 2 Clearly, by setting u = x/2 we have x = 2u so that = 4 x5 dx = 4 (2u)5 (2du) = 4 26 u 5 du = Thus for = 64 4 u5 du = 6u5 du. (u) du, we require (u) = 6 u5. b. (5 points) Next, generate m =, pseudo-random variates from the U(,) distribution and use these to approximate the integral = (u) du. Before generating any Monte Carlo variates, set your seed to 2. Sample R code is: g2 = function(u) {6 * u^5 * (u>) * (u<)} m = set.seed( 2 ) u = runif( m, min=, max= ) round( mean(g2(u)), digits=4) an approximation for = E[X 2 ] of [] cont d

3 2. Monte Carlo integration for = E[X 2 ] (cont d). c. (5 points) From the random variates you employed in part (b), estimate the variance of your approximation for. Sample R code using a denominator of m in the sample variance is v2 = var(g2(u))/m round( v2, digits=6 ) [].592 If you choose to use a denominator of m in the variance, sample code is var(g2(u))*(m-)/m^2 essentially the same value: [].598

4 3. Return to the p.d.f. from Problem #2 and in particular to the integral = (u) du. a. ( points) Mimic the approach you took in Problem #2(b) but now approximate using antithetic variables. To do so, generate a new set of m=,/2 Monte Carlo variates after setting your seed to 3. Sample R code is: m = set.seed( 3 ) u = runif( m/2, min=, max=) w = (g2(u) + g2(-u))/2 round( mean( w ), digits=4 ) # uses only /2 = 5 variates an antithetic approximation for = [] u5 du of b. ( points) From the random variates you employed in part (a), estimate the variance of your approximation for. (Use the usual sample variance with a denominator of m.) Also calculate the percent reduction in variance over the approximation from Problem #2(c). Sample R code using a denominator of m in the sample variance is v3 = var(w)/(m/2) print( v3 ) [] or The % reduction can be found via the R code round( (v2-v3)/v2, digits=6 ) [] i.e., a % reduction in variance.

5 4. Return to the p.d.f. from Problem #2 and in particular to the integral = (u) du. a. (5 points) Now approximate using control variates. The target integrand here is (u) from Problem #2(a). For the control variate use h(u) = 6u 6. To find the correlation between (U) and h(u), generate a preliminary Monte Carlo sample of m = 5, variates, using a seed of 4. Report (i) the approximated Correlation between (U) and h(u) = 6U 6 and (ii) the consequent value of c* you will use for the control-variate approximation of. We will need the population mean of the control variate: μ = E[h(U)] = h(u) du = 6 u6 dx = 6 7 u7 = 6 7. This is employed in the control variate expression (U) + c*[h(u) μ] where (U) = 6U 5. Sample R code is: m = 5 h = function(u) {6 * u^6 * (u>) * (u<)}; mu = 6/7 set.seed( 4 ); u = runif( m ) cstar = -cov(g2(u),h(u))/var(h(u)) #c* for future use c( cor(g2(u),h(u)), cstar) #print out requested values [] i.e., (6U 5,6U 6 ) =.9964 (near ±., as desired) and c* =.57. (Note that we can determine the Covariance and hence the Correlation analytically for this problem, due to the elementary forms of and h. The actual value is [6U 5,6U 6 ] = ) b. ( points) Next, to employ the control variate algorithm generate a new set of m=, Monte Carlo variates after re-setting your seed to 4. Use these to compute a control-variate approximation for. Sample R code [including c* =.57 and μ = 6/7 from part (a)] is: m = set.seed( 4 ); u = runif( m ) theta.g = g2(u) theta.control = theta.g + cstar*(h(u) - mu) round( c(mean(theta.g), mean(theta.control)), digits=4 ) [] That is, another approximation for using simple Monte Carlo integration as in Problem #2(b) is the first value of , while a control-variate approximation for is cont d

6 4. Control variate integration for = (u) du (cont d). c. ( points) From the random variates you employed in parts (a)-(b), calculate the variance of your estimate of. (Use the usual sample variance with a denominator of m.) Also calculate the percent reduction in variance over the simple Monte Carlo approximation you found in Problem #2(c). Sample R code using a denominator of m in the sample variance is v4 = var(theta.control)/m print( v4 ) [].39866e-5 i.e., The % reduction can be found via the R code round( (v2-v4)/v2, digits=6) [] i.e., a sizeable % reduction in variance.

7 5. Return to the p.d.f. from Problem #2 and in particular to the integral = (x) dx. Now approximate using importance sampling. Take (u) from Problem #2(a) and for the importance function ƒ(x) consider each of the following possible choices. In each case, (i) approximate using a Monte Carlo sample of m =, simulated variates and (ii) estimate the variance of that estimate from the Monte Carlo sample. a. ( points) ƒ(x) = 2( x) I (,) (x). Set your seed to 5. The importance function here is the Beta(,2) density. So, sample from X ~ Beta(,2) and calculate the objective ratio (x) ƒ(x) = 6x 5 2( x) = 8 x 5 ( x). Sample R code (fairly simple, since X ~ Beta(,2) has the same target support): m = set.seed(5) x = rbeta(m,,2) gfratio = 8 * (x^5) / (-x) c( mean(gfratio), var(gfratio)/m ) produces [] i.e. (i) estimate as with (ii) variance =.455. If instead using the inverse probability transform to generate X ~ Beta(,2), find the c.d.f. as F(x) = x, so that F (u) = ( u) /2 produces the desired X from U ~ U(,). Sample R code: m = set.seed(5) x = - sqrt( - runif(m)) gfratio = 8 * (x^5) / (-x) c( mean(gfratio), var(gfratio)/m ) produces [] i.e. (i) estimate as with (ii) variance =.24. cont d

8 5. Importance sampling integration for = (u) du (cont d). b. ( points) ƒ(x) = e x I (, ) (x). Set your seed to 52. The importance function here is the Exp() density. So, sample from X ~ Exp() and calculate the objective ratio (x) ƒ(x) = 6x5 I (,) (x) e x I (, ) (x) = 6 x 5 e x I (,) (x). Notice the use of an indicator function to discard any values of x >. Sample R code is: m = set.seed(52) x = rexp(m,) gfratio = 6 * (x^5) * exp(x) * (x>) * (x<) c( mean(gfratio), var(gfratio)/m ) [] i.e. (i) estimate as with (ii) variance = cont d

9 5. Importance sampling integration for = (u) du (cont d). c. (5 points) ƒ(x) = 6x 5 I (,) (x). Use any seed you wish. Explain the unfamiliar result that results. The importance function here is the Beta(6,) density, so we sample from X ~ Beta(6,) and calculate the objective ratio (x) ƒ(x) = 6x5 6x 5 = 6 6 But this is exactly constant and not a function of x! The corresponding R code is rather trivial (for any choice of seed, the answer is always the same): m = set.seed( ) x = rbeta( m,6, ) #this is never used! gfratio = 6/6 c( mean(gfratio), var(gfratio)/m ) [] NA What has happened? If we can find an importance function ƒ(x) that produces an exactly constant importance ratio, (x)/ƒ(x), over a finite range of integration, importance sampling integrates that constant over the finite range. This should produce the exact answer for the integral. Here, it is 6/6 = Notice the NA for the variance: there is only value for gfratio, as it doesn t depend upon x, and R cannot compute a variance for value. In effect, all the simulated values are the same so there is no observed variation in the simulated sample.