Assignment 2 Solution

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Instructions: Solutions to problems 1 and 2 are to be submitted on Quercus (PDF les only). You are strongly encouraged to do problems 3{6 but these are not to be submitted for grading.

  1. Suppose we want to generate independent Normal random variables with mean 0 and variance 1 using some sort of rejection sampling. An intuitively reasonable proposal density is a logistic density with the general form

exp(x=s)

g(x; s) = s f1 + exp(x=s)g2

where s > 0 is a parameter.

(a) Show that we can generate a random variable X from g(x; s) by X = s ln(U=(1 U)).

(b) To generate N (0; 1) random variables using g(x; s) as a proposal density, we should choose s to maximize the probability that a proposal from g(x; s) is accepted; this probability is given by 1=M(s) where

f(x)

M(s) = max

where f(x) is the N (0; 1) density. (Therefore, we want to nd s to minimize M(s).) Show

p

p

that if s 1=

2, f(x)=g(x; s) is maximized at x = 0 with M(s) minimized (for s 1= 2)

p

at s = 1= 2. (Hint: Use calculus to nd the maximum of ln(f(x)) ln(g(x; s)).)

p

  1. Given part (b), we know that M(s) must be minimized for s between 0 and 1= 2. Use some approach to nd the value of s that minimizes M(s). (Hint: You may nd it useful to use the dnorm and dlogis functions in R; for example, to look at f(x)=g(x; s) for a speci ed value of s and x between 0 and 3, we could use the following code:

  • s <- 1/sqrt(2)

  • x <- c(0:3000)/1000 # generates values 0, 0.001, 0.002 ,…, 2.999, 3

  • plot(x,dnorm(x)/dlogis(x,location=0,scale=s),type=”l”)

It is possible to use calculus to determine the optimal value of s although ad hoc approaches are ne; there is some detective work required here and you are encouraged to use R to gain some intuition for the problem.)

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2. Suppose we observe y1; ; yn where

yi = i + “i (i = 1; ; n)

where f”ig is a sequence of random variables with mean 0 and nite variance representing noise. We will assume that 1; ; n are smooth in the sense that i = g(i) for some continuous and di erentiable function g. The least squares estimates of 1; ; n are trivial

  • bi = yi for all i | but we can modify least squares in a number of ways to accommodate the \smoothness” assumption on f ig. In this problem, we will consider estimating f ig by

minimizing

  • n 1

(yi i)2

+

( i+1 2 i + i 1)2

X

Xi

i=1

=2

where > 0 is a tuning parameter that controls the smoothness of the estimates b1; ; bn.

(a) Show the estimates b1; ; bn satisfy the equations

y1 = (1 + ) 1

22+3

y

2

=

2 1

b

b

2

b

3

+

4

+(1+5 )

4

j

=

j 2b

4 j 1 + b

b

j

b

j+1

j+2

y

(1+6 )

4

+for j = 3;

; n 2

y

n 1 =

bn 3

4

bn 2

+ (1 + 5 )bn 1

b

n

b

2

y

n =

bn 2

2

bn 1

+(1+

)

nb

b

b

b

b

(Note that if we write this in matrix form y = A b, the matrix A is sparse, having at most 5 non-zero entries per row.)

  1. Show that if fyig are exactly linear, i.e. yi = a i + b for all i then bi = yi for all i. (In other words, these linear functions are eigenvectors of A with eigenvector 1.)

  1. Show (using results from class) that the Gauss-Seidel algorithm can be used to compute the estimates de ned in part (a).

  1. Write a function in R to implement the Gauss-Seidel algorithm above. (A template function is provided on Quercus but you do need to follow it.) The inputs for this function are a vector of responses y and the tuning parameter lambda. Test your function (for various tuning parameters) on data generated by the following command:

  • x <- c(1:1000)/1000

  • y <- cos(6*pi*x)*exp(-2*x) + rnorm(1000,0,0.05)

Note that the algorithm will converge more slowly as increases; the convergence can be improved by modifying the algorithm, for example, by using the Successive Over Relaxation method described in the text.

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Supplemental problems (not to hand in):

  1. To generate random variables from some distributions, we need to generate two indepen-dent two independent random variables Y and V where Y has a uniform distribution over some nite set and V has a uniform distribution on [0; 1]. It turns out that Y and V can be generated from a single Unif(0; 1) random variable U.

(a) Suppose for simplicity that the nite set is f0; 1; ; n 1g for some integer n 2. For U Unif(0; 1), de ne

Y = bnUc and V = nU Y

where bxc is the integer part of x. Show that Y has a uniform distribution on the set f0; 1; ; n 1g, V has a uniform distribution on [0; 1], and Y and V are independent.

(b) What happens to the precision of V de ned in part (a) as n increases? (For example, if

  • has 16 decimal digits and n = 106, how many decimal digits will V have?) Is the method in part (a) particularly feasible if n is very large?

  1. One issue with rejection sampling is a lack of e ciency due to the rejection of random variables generated from the proposal density. An alternative is the acceptance-complement (A-C) method described below.

Suppose we want to generate a continuous random variable from a density f(x) and that

f(x) = f1(x) + f2(x) (where both f1 and f2 are non-negative) where f1(x) g(x) for some density function g. Then the A-C method works as follows:

  1. Generate two independent random variables U Unif(0; 1) and X with density g.

  1. If U f1(X)=g(X) then return X.

  1. Otherwise (that is, if U > f1(X)=g(X)) generate X from the density

f2 (x) =

f2(x)

:

Z 1 f2(t) dt

Note that we must be able to easily sample from g and f2 in order for the A-C method to be e cient; in some cases, they can both be taken to be uniform distributions.

  1. Show that the A-C method generates a random variable with a density f. What is the probability that the X generated in step 1 (from g) is \rejected” in step 2?

  1. Suppose we want to sample from the truncated Cauchy density

f(x) =

2

( 1 x 1)

(1 + x2)

3

2

1

x

2

using the A-C method with f (x) = k, a constant, for

1 (so that f (x) = 1=2 is a

uniform density on [ 1; 1]) with

f1(x) = f(x) f2(x) = f(x) k ( 1 x 1):

If g(x) is also a uniform density on [ 1; 1] for what range of values of k can the A-C method be applied?

(c) De ning f1, f2, and g as in part (b), what value of k minimizes the probability that X generated in step 1 of the A-C algorithm is rejected?

  1. Suppose we want to generate a random variable X from the tail of a standard normal distribution, that is, a normal distribution conditioned to be greater than some constant b. The density in question is

exp( x2=2)

f(x) = p2 (1 (b))

for x b

with f(x) = 0 for x < b where (x) is the standard normal distribution function. Consider rejection sampling using the shifted exponential proposal density

g(x) = b exp( b(x b)) for x b.

  1. De ne Y be an exponential random variable with mean 1 and U be a uniform random variable on [0; 1] independent of Y . Show that the rejection sampling scheme de nes X =

b + Y =b if

Y 2

2 ln(U) b2 :

(Hint: Note that b + Y =b has density g.)

(b) Show the probability of acceptance is given by

p

2 b(1 (b)):

exp( b2=2)

What happens to this probability for large values of b? (Hint: You need to evaluate M = max f(x)=g(x).)

(c) Suppose we replace the proposal density g de ned above by

g (x) = exp( (x b)) for x b.

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(Note that g is also a shifted exponential density.) What value of maximizes the proba-bility of acceptance? (Hint: Note that you are trying to solve the problem

min max f(x)

>0 x b g (x)

for . Because the density g (x) has heavier tails, the minimax problem above will have the same solution as the maximin problem

max min f(x)

x b >0 g (x)

which may be easier to solve.)

6. Another interesting variation of rejection sampling is the ratio of uniforms method. We Z 1

start by taking a bounded function h with h(x) 0 for all x and h(x) dx < 1. We then de ne the region

q

Ch = (u; v) : 0 u h(v=u)

and generate (U; V ) uniformly distributed on Ch. We then de ne the random variable X = V=U.

(a) The joint density of (U; V ) is

f(u; v) =

1

for (u; v) 2 Ch

jChj

where jChj is the area of Ch. Show that the joint density of (U; X) is

u

for 0 u q

g(u; x) =

h(x)

h

jC

j

and that the density of X is h(x) for some > 0.

  1. The implementation of this method is somewhat complicated by the fact that it is typically di cult to sample (U; V ) from a uniform distribution on Ch. However, it is usually

possible to nd a rectangle of the form Dh = f(u; v) : 0 u u+; v v v+g such that Ch is contained within Dh. Thus to draw (U; V ) from a uniform distribution on Ch, we can use rejection sampling where we draw proposals (U ; V ) from a uniform distribution on the rectangle Dh; note that the proposals U and V are independent random variables with

Unif(0; u+) and Unif(v

; v+) distributions, respectively. Show that we can de ne u+, v and

v+ as follows:

+ =

x

q

x

q

+

x

q

u

h(x) v

h(x):

max

h(x) v = min x

= max x

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(Hint: It su ces to show that if (u; v) 2 Ch then (u; v) 2 Dh where Dh is de ned using u+,

  • , and v+ above.)

(c) Implement (in R) the method above for the standard normal distribution taking h(x) =

exp( x2=2). In this case, u+ = 1, v =

q

= 0:8577639, and v+ = q

2=e

2=e = 0:8577639.

What is the probability that proposals are accepted?

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