Nguyen, N., Rock, D., and Ying, V. Math 470 Spring 2011 Project 5 Problem 3.24

Transcription

1 Problem 3.24 Here are some modifications of Example 3.8 with consequences that may not be apparent at first. Consider. (a) For and, compare the Riemann approximation with the Monte Carlo approximation. Modify the method in Example 3.8 appropriately, perhaps writing a program that incorporates d = 1/2 and h = x^d, to facilitate easy changes in the parts that follow. Find. The Riemann approximation yields a value of The Monte Carlo approximation yields a value of Also, we have. Code and Output: m = a = 0; b = 1; w = (b - a)/m d = 1/2 x = seq(a + w/2, b-w/2, length=m) h = x^d rect.areas = w*h sum(rect.areas) # Riemann # set.seed(1212) u = runif(m, a, b) h = u^d; y = (b - a)*h mean(y) 2*sd(y)/sqrt(m) var(y) # Monte Carlo # MC margin of error # MC variance > sum(rect.areas) [1] > mean(y) [1] > var(y) # MC variance [1] # Riemann # Monte Carlo (b) What assumptions of Section 3.4 fails for? What is the value of? Of? Try running the two approximations. How do you explain the unexpectedly good behavior of the Monte Carlo simulation? For, the assumption of the function we want to integrate ( ) being bounded is violated. The true value of J is The value of is Running the two approximations, we get an approximation of for the Riemann method and for the Monte Carlo method. The unexpectedly good behavior of the Monte Carlo simulation has to do with the asymptotic behavior of near Page 1 of 28

2 ; only values of very close to 0 yield values of that are so large that they are troublesome? The code for this part is trivial and is not included. (c) Repeat part (b), but with. Comment. For, assumption of the function we want to integrate we want to integrate ( ) being bounded is again violated. The true value of J in this case is ] The value of is The Riemann method yields an approximation of The Monte Carlo method yields an approximation of The bad behavior of the Monte Carlo simulation has to do with the asymptotic behavior of near ; here, values of that weren t a problem in part (b) are so high that even random points are not enough to capture the area under the curve for values of x near 0. The code for this part is trivial and is not included. Page 2 of 28

3 Problem 3.25 This problem shows how the rapid oscillation of a function can affect the accuracy of a Riemann approximation. (a) Let and be a positive integer. Then use calculus to show that. Use the code below to plot k = 4 x = seq(0,1, by = 0.01); plot(x, h, type="l") ] for on. h = abs(sin(k*pi*x)) Using calculus, we have [ ] ( ) ( ) (1) as required. Plot Figure plot of h on [0,1] for k=4 Page 3 of 28

4 (b) Modify the program of Example 3.8 to integrate. Use throughout, and make separate runs with and. Compare the accuracy of the resulting Riemann and Monte Carlo approximations, and explain the behavior of the Riemann approximation. The Riemann approximation is less accurate than the Monte Carlo approximation (as can be seen in the table below) maybe because the non-random points of evaluation used in Riemann approximation are too evenly spaced to accurately evaluate the evenly spaced oscillation of the. function m Riemann approximation Riemann error Monte Carlo approximation Monte Carlo error Table comparison of accuracy of Riemann and Monte Carlo approximations Code: k = 5000 m.vector = c(2500,5000,10000,15000,20000) true.answer = 2/pi for( m in m.vector) #Riemann a = 0; b = 1; w = (b - a)/m x = seq(a + w/2, b-w/2, length=m) h = abs(sin(k*pi*x)) rect.areas = w*h riemann.answer = sum(rect.areas) # Riemann #Monte Carlo #set.seed(1212) #uncomment to reproduce results exactly u = runif(m, a, b) h = abs(sin(k*pi*u)); y = (b - a)*h monte.carlo.answer = mean(y) # Monte Carlo monte.carlo.answer #accuracy of approximations riemann.error = abs( riemann.answer - true.answer) monte.carlo.error = abs( monte.carlo.answer - true.answer) print( paste( "m=", m, ". Riemann approximation: ", riemann.answer, ". Riemann error: ", riemann.error, ". Monte Carlo approximation: ", monte.carlo.answer, ". Monte Carlo error: ", monte.carlo.error, ".", sep="")) Page 4 of 28

5 (c) Use calculus to show that. How accurately is this value approximated by simulation? If, find the margin of error for the Monte Carlo approximation in part (b) based on and the Central Limit Theorem. Are your results consistent with this margin of error? To verify the equality, we first we need the following lemma: Lemma: [ ] [( ) ( )] Then, from calculus, we have ( ) ] ] [( ) ] ( ) (2) ( ) ( ) as required. This value is closely approximated by simulation, as confirmed in the table below. m Monte Carlo approximation of variance Table accuracy when approximating V(Y) For, the (95%) margin of error for the Monte Carlo approximation in part (b) based on is Note that the true answer is, so that the 95% CI based on is. Hence, Page 5 of 28

6 our Mote Carlo approximation (front part (b) with margin of error. ) of is consistent with this Code and Output: k = 5000 m.vector = c(2500,5000,10000,15000,20000) #accuracy in approximating variance using Monte Carlo method (for part c) for( m in m.vector) #Monte Carlo #set.seed(1212) #uncomment to reproduce results exactly u = runif(m, a, b) h = abs(sin(k*pi*u)); y = (b - a)*h monte.carlo.answer = mean(y) # Monte Carlo monte.carlo.answer #accuracy of approximation of variance riemann.error = abs( riemann.answer - true.answer) monte.carlo.error = abs( monte.carlo.answer - true.answer) print( paste( "m=", m, ". Monte Carlo approximatin of variance: ", var(y), ".", sep="")) #margin of error for Monte Carlo approximation in part (b) based on SD(Y) and CLT, when m=10000 m = #set.seed(1212) #uncomment to reproduce results exactly u = runif(m, a, b) h = abs(sin(k*pi*u)); y = (b - a)*h 2*sd(y)/sqrt(m) # MC margin of error > 2*sd(y)/sqrt(m)# MC margin of error [1] Page 6 of 28

7 Problem 3.26 The integral cannot be evaluated analytically, but advanced analytic methods yield. (a) Assuming this result, show that. Use R to plot both integrands on, obtaining results as in Figure We have ] ( ( )) [ ] as required. Plots: Figure plot of first integrand Page 7 of 28

8 Figure plot of second integrand Codes: par( mfrow= c(2,1)) curve( (sin(1/x))^2, from=0, to=1, n=10000, main="first Integrand") curve( (sin(x)/x)^2, from=0, to=1, n=10000, main="second Integrand") par( mfrow= c(1,1)) (b) Use both Riemann and Monte Carlo approximations to evaluate as originally defined. Then evaluate using the equation in part (a). Try both methods with, and iterations. What do you believe is the best answer? Comment on differences between methods and between equations. The result of each method and each equation are given in tables below. It is hard to determine which answer is the best. Regarding differences when using the equation for as originally defined, both methods seem to have performed similarly; when using the equation in part (a), however, the Riemann method is more consistent. m Riemann approximation Monte Carlo approximation Table Evaluating as originally defined m Riemann approximation Monte Carlo approximation Table Evaluating using equation in part (a) Page 8 of 28

9 Codes: m.vector = c(100,1000,1001,10000) #evaluating J as originally defined: for( m in m.vector) #Riemann a = 0; b = 1; w = (b - a)/m x = seq(a + w/2, b-w/2, length=m) h = (sin(1/x))^2 rect.areas = w*h riemann.answer = sum(rect.areas) # Riemann Monte Carlo #Monte Carlo set.seed(1212) #uncomment to reproduce results exactly u = runif(m, a, b) h = (sin(1/u))^2; y = (b - a)*h monte.carlo.answer = mean(y) # monte.carlo.answer #accuracy of approximations print( paste( "m=", m, ". Riemann approximation: ", riemann.answer, ". Monte Carlo approximation: ", monte.carlo.answer, ".", sep="")) #evaluating J as defined in part (a): for( m in m.vector) #Riemann a = 0; b = 1; w = (b - a)/m x = seq(a + w/2, b-w/2, length=m) h = pi/2 - (x^-2)*(sin(x))^2 rect.areas = w*h riemann.answer = sum(rect.areas) # Riemann Monte Carlo #Monte Carlo set.seed(1212) #uncomment to reproduce results exactly u = runif(m, a, b) h = pi/2 - (u^-2)*(sin(u))^2; y = (b - a)*h monte.carlo.answer = mean(y) # monte.carlo.answer #accuracy of approximations print( paste( "m=", m, ". Riemann approximation: ", riemann.answer, ". Monte Carlo approximation: ", monte.carlo.answer, ".", sep="")) Page 9 of 28

10 Problem 3.27 Modify the program of Example 3.9 to approximate the volume beneath the bivariate standard normal density surface and above two additional regions of integration as specified below. Use both the Riemann and Monte Carlo methods in parts (a) and (b), with. (a) Evaluate. Because and are independent standard normal random variables, we know that this probability is. For each method, say whether it would have been better to use points to find and then square the answer. For the Riemann approximation method, it would have been better to use points to find and then square the answer. However, for the Monte Carlo method it would have been better to find the answer directly. Surprisingly, the Riemann method performed better than the Monte Carlo method. algorithm Riemann approx. Riemann error MC approx. MC error Table Comparison of methods and algorithms Code and Outputs: true.answer = (pnorm(1) - 0.5)^2 #== ## METHOD OF DIRECTLY FINDING P0 < Z1 <= 1, 0 < Z2 <= 1: # Riemann approximation m = g = round(sqrt(m)) # no. of grid pts on each axis x1 = rep((1:g - 1/2)/g, times=g) # these two lines give x2 = rep((1:g - 1/2)/g, each=g) # coordinates of grid points hx = dnorm(x1)*dnorm(x2) riemann.approx = sum(hx[x1 <= 1 & x2 <= 1])/g^2 # Riemann approximation riemann.approx riemann.error = abs( true.answer - riemann.approx) riemann.error > riemann.approx [1] > riemann.error [1] e-07 # Monte Carlo approximation set.seed(pi) # uncomment to reproduce results exactly u1 = runif(m) # these two lines give a random u2 = runif(m) # point in the unit square hu = dnorm(u1)*dnorm(u2) hu.acc = hu[u1 <= 1 & u2 <= 1] # heights above accepted points mc.approx = mean(hu.acc) # Monte Carlo result mc.approx mc.error = abs( true.answer - mc.approx) Page 10 of 28

11 mc.error > mc.approx [1] > mc.error [1] e-05 ## # METHOD OF FINDING P0 < Z <= 1 AND THEN SQUARING: # Riemann approximation (works) m = a = 0; b = 1; w = (b - a)/m x = seq(a + w/2, b - w/2, length=m) h = dnorm(x) rect.areas = w*h riemann.approx = (sum(rect.areas))^2 riemann.error = abs( true.answer - riemann.approx) riemann.error > riemann.approx [1] > riemann.error [1] e-11 # Alternative Riemann approximation (equivalent to above) m=10000 x1 = (1:m - 1/2)/m hx = dnorm(x1) riemann.approx = (sum(hx[x1 < 1])/m)^2 riemann.approx riemann.error = abs( true.answer - riemann.approx) riemann.error > riemann.approx [1] > riemann.error [1] e-11 # MC approximation: #set.seed(pi) # uncomment to reproduce results exactly m=10000 u1 = runif(m,0,1) hu = dnorm(u1) hu.acc = hu[u1 < 1] mc.approx = mean(hu.acc)^2 # mc.approx = (sum( hu[u1 < 1] ) / sum(u1<1))^2 #same as above mc.approx mc.error = abs( true.answer - mc.approx) mc.error > mc.approx [1] Page 11 of 28

12 > mc.error [1] (b) Evaluate. Here the region of integration does not have area 1, so remember to multiply by an appropriate constant. Because CHISQ(2), the exact answer can be found with pchisq(1,2). Note that the true answer is pchisq(1,2)= The Riemann method yields an approximation of ; the error of this approximation is The Monte Carlo method yields an approximation of ; the error of this approximation is Codes: #true answer: true.answer = pchisq(1,2) > true.answer [1] # riemann method (correct?): m = g = round(sqrt(m)) x = rep((1:g)/g,times=g) y = rep((1:g)/g,each=g) hx = dnorm(x)*dnorm(y) riemann.approx = sum( hx[ (x^2) + (y^2) < 1 ]) / m * 4 riemann.approx #== riemann.error = abs( riemann.approx - true.answer) riemann.error #== > riemann.approx [1] > riemann.error [1] # MC approximation 1 (works): set.seed(1237) #uncomment to reproduce results exactly m = u1 = runif(m,-1,1) u2 = runif(m,-1,1) hu = dnorm(u1)*dnorm(u2) hu.acc = hu[ (u1^2) + (u2^2) < 1 ] mc.approx = mean(hu.acc) * pi mc.approx #== mc.error = abs( mc.approx - true.answer) mc.error #== > mc.approx #== [1] > mc.error #== [1] Page 12 of 28

13 # MC approximation 2 (this works even better than above): set.seed(1237) #uncomment to reproduce results exactly m = u1 = runif(m,0,1) u2 = runif(m,0,1) hu = dnorm(u1)*dnorm(u2) hu.acc = hu[ (u1^2) + (u2^2) < 1 ] mc.approx = mean(hu.acc) * pi / 4 * 4 mc.approx #== mc.error = abs( mc.approx - true.answer) mc.error #== > mc.approx #== [1] > mc.error #== [1] (c) The joint density function of has circular contour lines centered at the origin, so that probabilities of regions do not change if they are rotated about the origin. Use this fact to argue, which was approximated in Example 3.9, can be found that the exact value of with (pnorm(1/sqrt(2)) 0.5)^2. Consider the series of rotations about the origin illustrated in Figures through This response will continue after the break Figure Page 13 of

14 Figure = 1/ = 1/ = 1/ = 1/ Figure Since probabilities of regions do not change if they are rotated about the origin, then the probability of the triangular region in Figure , must be equal to the probability of the square region in Figure This probability is just ( ), where standard normal random variable. However, this is equivalent to ( is the density function of the ) (, where is the cdf of a standard normal random variable), which in turn is equivalent to ( )., which was approximated in Example 3.9, can be found with Thus the exact value of (pnorm(1/sqrt(2)) 0.5)^2, as required. Page 14 of 28

15 Problem 3.28 Here we extend the idea of Example 3.9 to three dimensions. Suppose three items are drawn at random from a population of items with weights (in grams) distributed as NORM(100,10). (a) Using the R function pnorm (cumulative distribution function of a standard normal), find the probability that the sum of the three weights is less than 310 g. Also find the probability that the minimum weight of these three items exceeds 100 g. For the probability that the sum is less than 310 g, note that since then and so we have (3.28.1) where is a standard normal random variable. pnorm(1/sqrt(3)) = , as required. Thus the answer for the sum is ( ) For the probability that the minimum exceeds 100 g, we have where the last equality comes from the R command pnorm(100, mean=100, sd=10, lower.tail=f)^3. Similarly, in terms of standard normal random variables, we could have used ( ) as required. Page 15 of 28 (3.28.2)

16 (b) Using both Riemann and Monte Carlo methods, approximate the probability that (simultaneously) the minimum of the weights exceeds 100 g and their sum is less than 310 g. The suggested procedure is to (i) express this problem in terms of three standard normal random variables, and (ii) modify appropriate parts of the program in Example 3.9 to approximate the required integral over a triangular cone (of area 1/6) in the unit cube using. Here is R code to make the three grid vectors needed for the Riemann approximation: [for the R code, see page 84 of the text] The probability that the minimum of the weights exceeds 100 g can be expressed as: (3.28.3) where the last equality follows since. The probability that the sum is less than 310 g can be expressed as: (3.28.4) where the s are standard normal random variables. Since the three grid vectors provided in the problem statement contain points with components all greater than 0, we don t need to worry about defining a region of integration that satisfies equation (3.28.3). We only need to worry about equation (3.28.4). Hence, in our R code below, we only need to worry about accepting points that satisfy the condition x1 + x2 + x3 < 1. Here are our results: True answer from book Riemann Approximation MC Approximation Approximately Table probability that (simultaneously) the minimum of the weights exceeds 100 g and their sum is less than 310 g. Code: ### Riemann approximation ## g = 25 m = g^3 x1 = rep((1:g-1/2)/g,each=g^2) x2 = rep(rep((1:g-1/2)/g,each=g),times=g) x3 = rep((1:g-1/2)/g,times=g^2) hx = dnorm(x1)*dnorm(x2)*dnorm(x3) riemann.approx = sum(hx[x1 + x2 + x3 < 1])/g^3 riemann.approx #== ### Monte Carlo Simulation ### set.seed(pi) #uncomment to reporduce results exactly m = u1 = runif(m) u2 = runif(m) u3 = runif(m) hu = dnorm(u1)*dnorm(u2)*dnorm(u3) hu.acc = hu[u1 + u2 + u3 < 1] mc.approx = (1/6) * mean(hu.acc) mc.approx #== Page 16 of 28

17 Problem 3.29 For and, use Monte Carlo approximation with m= to find the probability that a variate (independent) standard normal distribution places in the part of the -dimensional unit ball with all coordinates positive. To do this, imitate the program in Example 3.9, and use the fact that the entire 4-dimensional unit ball has hypervolume. Compare the results with appropriate values computed using the chi-squared distribution with degrees of freedom; use the R function pchisq. Results are presented in Table Code is after that exact result using pchisq Monte Carlo result Monte Carlo error e Table Code: fark = function(d) # Input: d - the dimention of the unit ball whose hypervolume we wish to calculuate # Returns: the hypervolume of the d-dimensional unit ball hypervolume = (pi^(d/2)) / gamma( (d+2)/2) return( hypervolume) m = ## # for d = 2 d=2 # true answer true.answer = pchisq(1,d) #== true.answer # MC answer (works): set.seed(pi) #uncomment to reproduce results exactly u1 = runif(m,-1,1) u2 = runif(m,-1,1) hu = dnorm(u1)*dnorm(u2) hu.acc = hu[ (u1^2) + (u2^2) < 1 ] mc.approx = mean(hu.acc) * fark(d) mc.approx > mc.approx [1] > mc.error [1] e-05 ## # for d = 3 Page 17 of 28

18 d=3 # true answer true.answer = pchisq(1,d) #== true.answer # MC answer (works): set.seed(pi) u1 = runif(m,-1,1) u2 = runif(m,-1,1) u3 = runif(m,-1,1) hu = dnorm(u1)*dnorm(u2)*dnorm(u3) hu.acc = hu[ (u1^2) + (u2^2) + (u3^2) < 1 ] mc.approx = mean(hu.acc) * fark(d) mc.approx > mc.approx [1] > mc.error [1] ## # for d = 4 d=4 # true answer true.answer = pchisq(1,d) #== true.answer # MC answer (works): set.seed(pi) u1 = runif(m,-1,1) u2 = runif(m,-1,1) u3 = runif(m,-1,1) u4 = runif(m,-1,1) hu = dnorm(u1)*dnorm(u2)*dnorm(u3)*dnorm(u4) hu.acc = hu[ (u1^2) + (u2^2) + (u3^2) + (u4^2) < 1 ] mc.answer = mean(hu.acc) * fark(d) mc.answer > mc.answer [1] > mc.error [1] Page 18 of 28

19 Problem 6.1 In each part below, consider the three Markov chains specified as follows: (i), ; (ii), ; and (iii),. (a) Find and, for. For we have and Hence, we have (i) and. (ii) and. (iii) and. (b) Use the given values of and and means of geometric distributions to find the average cycle length for each chain. The average cycle length of a chain is (i) The average cycle length is (ii) The average cycle length is (iii) The average cycle length is Hence for (c) For each chain, modify the program of Example 6.1 to find the long-run fraction of steps in state 1. The long-run fraction of steps in state 1 are: (i) , (ii) 0.273, and (iii) Codes (for parts (c) and (d)) Pause <- function () cat("hit <enter> to continue...") readline() invisible() # Preliminaries set.seed(1237) #uncomment to reproduce results exactly m = 2000; n = 1:m; x = numeric(m); x[1] = 0 #alpha = 1; beta = 0.44 alpha = c(0.3,0.15,0.03); beta = c(0.7,0.35,0.07) label = c( "(i) alpha = 0.3, beta = 0.7", "(ii) alpha = 0.15, beta = 0.35", "(iii) alpha = 0.03, beta = 0.07", sep="") # Simulation for (j in 1:3) for (i in 2:m) Page 19 of 28

20 if (x[i-1]==0) x[i] = rbinom(1, 1, alpha[j]) else x[i] = rbinom(1, 1, 1 - beta[j]) y = cumsum(x)/n # Running fractions of Long eruptions # Results lrf = y[m]; # long run fraction of steps in state 1 among m. Same as: mean(x) print( paste( "long run fraction of steps in state 1 for ", label[j], ": ", lrf, sep="")) Pause() a = sum(x[1:(m-1)]==0 & x[2:m]==1); a # No. of cycles acl = m/a; acl # Average cycle length plot(x[1:round(6*acl)], type="b", xlab="step", ylab="state", main=label[j]) # Similar to Fig. 6.3 Pause() # ---- (New plot, similar to Fig. 6.4) plot(y, type="l", ylim=c(0,1), xlab="step", ylab="proportion Long", main=label[j]) Pause() # ---- (New plot, similar to Fig. 6.5; additional output) acf(x, main=label[j]) # Autocorrelation plot acf(x, plot=f) # Printed output Pause() (d) For each chain, make and interpret plots similar to Figures 6.3 (where the number of steps is chosen to illustrate the behavior clearly), 6.4, and 6.5. For (i) we have: Page 20 of 28

21 For (ii), we have: For (iii), we have Codes See part (c). (e) In summary, what do these three chains have in common, and in what respects are they most remarkably different. Is any one of these chains an independent process, and how do you know? In summary, all three chains have in common the fact that they converge to a stationary distribution. They are most remarkably different in terms of their cycle lengths. The only chain that is an independent process is the one from part (i); we know this from the essentially zero autocorrelation. Page 21 of 28

22 Problem 6.2 There is more information in the joint distribution of two random variables than can be discerned by looking only at their marginal distributions. Consider two random variables and, each distributed as BINOM(1, ), where. (a) In general, show that. In particular, evaluate in three cases: where and are independent, where, and where. Case 1: When and are independent, we have (where the last equality follows since the. Then, since, we have. Hence we have Case 2: When, we have. Hence we have where the left-hand (in)equality follows since. Case 3: When, we have Hence we have where the right-hand (in)equality follows since. Thus, in each case, we have as required. (b) For each case in part (a), evaluate. Case 1: When and are independent, we have where the last equality follows since the. Case 2: When, we have. Case 3: When, we have. Page 22 of 28

23 (c) If and, then express,, and in terms of and. First, we have (3) as required. Second, we have (4) Last, we have ( ) (5) (d) In part (c), find the correlation, recalling that. First, we have where the last equality follows from (5). Second (since ), we have, so that Page 23 of 28 ( ) where the last equality comes from equation (3). Third (since ), we have, so that (6) (7)

24 Thus the correlation is ( ) ( ) (8) ( ) ( ) ( ) ( ) ( ) ( ) as required. (9) Page 24 of 28

25 Problem 6.3 Geometric distributions. Consider a coin with. A geometric random variable can be used to model the number of independent tosses required until the first Head is seen. (a) Show that the probability function For any, we have, for as required. (b) Show that the geometric series The sum of the geometric series sums to 1, so that one is sure to see a Head eventually. is (12) Multiplying equation (12) by, yields (13) Finally, subtracting (13) from (12) yields ] Thus the geometric series sums to 1, and one is sure to see a Head eventually. (c) Show that the moment generating function of hence that. ], and is (You may assume that the limits involved in differentiation and summing an infinite series can be interchanged as required.) The mgf of is (14) Multiplying (14) by yields (15) Finally, subtracting (15) from (14) and solving for and hence, we get, as required. Page 25 of 28

26 Problem 6.4 Suppose the weather for a day is either Dry (0) or Rainy (1) according to a homogeneous 2-state Markov chain with and. Today is Monday and the weather is Dry. (a) What is the probability that both tomorrow and Wednesday will be Dry? We have (b) What is the probability that it will be Dry on Wednesday? We have (c) Use equation (6.5) [(sic)] to find the probability that it will be Dry two weeks from Wednesday. The probability that it will be Dry two weeks from Wednesday ( ), given that it is dry today ( ) can be found using the 16-step transition matrix. Specifically, this probably,, is the upper-left entry of the following 16-step transition matrix [ ] * + Thus the probability that it will be Dry two weeks from Wednesday is (d) Modify the R code of Example 6.2 to find the probability that it will be Dry two weeks from Wednesday. The answer is, again, Code and Outputs: P = matrix(c( 0.9, 0.1, 0.5, 0.5), nrow=2, ncol=2, byrow=t) P Page 26 of 28

27 P2 = P %*% P; P2 P4 = P2 %*% P2; P4 P8 = P4 %*% P4; P8 P16 = P8 %*% P8; P16 > P = matrix(c( 0.9, 0.1, + 0.5, 0.5), nrow=2, ncol=2, byrow=t) > P [,1] [,2] [1,] [2,] > P2 = P %*% P; P2 [,1] [,2] [1,] [2,] > P4 = P2 %*% P2; P4 [,1] [,2] [1,] [2,] > P8 = P4 %*% P4; P8 [,1] [,2] [1,] [2,] > P16 = P8 %*% P8; P16 [,1] [,2] [1,] [2,] (e) Over the long run, what will be the proportion of Rainy days? Modify the R code of Example 6.1 to simulate the chain and find an approximate answer. The predicted proportion of Rainy days we get from simulation is The modification to the R code of Example 6.1 is rudimentary and so it is not included here. (f) What is the average length of runs of Rainy days? Note that we have. Thus the average length of a run of Rainy (1) days is. (g) How do the answers above change if and? If and, then (a') the probability that both tomorrow and Wednesday will be Dry becomes. (b') the probability that it will be Dry on Wednesday becomes. (c') the probability that it will be Dry two weeks from Wednesday becomes (d') modifying the R code of Example 6.2 to find the probability that it will be Dry two weeks from Wednesday yields an answer of The changes to the code in part (d) are rudimentary and so the code is not included. Page 27 of 28

28 (e') modifying the code of Example 6.1 to simulate the chain suggests that, over the long run, the proportion of Rainy days will be approximately The changes to the code in part (e) is rudimentary and so the code is not included. (f') we have., so that the average length of a run of Rainy (1) days is Page 28 of 28