Description
1. An interesting variation of rejection sampling is the ratio of uniforms method. We start
by taking a bounded function h with h(x) ≥ 0 for all x and Z ∞
−∞
h(x) dx < ∞. We then
define the region
Ch =
(u, v) : 0 ≤ u ≤
q
h(v/u)
and generate (U, V ) uniformly distributed on Ch. We then define the random variable X =
V/U.
(a) The joint density of (U, V ) is
f(u, v) = 1
|Ch|
for (u, v) ∈ Ch
where |Ch| is the area of Ch. Show that the joint density of (U, X) is
g(u, x) = u
|Ch|
for 0 ≤ u ≤
q
h(x)
and that the density of X is γ h(x) for some γ > 0.
(b) The implementation of this method is somewhat complicated by the fact that it is
typically difficult to sample (U, V ) from a uniform distribution on Ch. However, it is usually
possible to find a rectangle of the form Dh = {(u, v) : 0 ≤ u ≤ u+, v− ≤ v ≤ v+} 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 define u+, v− and
v+ as follows:
u+ = max
x
q
h(x) v− = min
x
x
q
h(x) v+ = max
x
x
q
h(x).
(Hint: It suffices to show that if (u, v) ∈ Ch then (u, v) ∈ Dh where Dh is defined using u+,
v−, and v+ above.)
(c) Implement (in R) the method above for the standard normal distribution taking h(x) =
exp(−x
2/2). In this case, u+ = 1, v− = −
q
2/e = −0.8577639, and v+ =
q
2/e = 0.8577639.
What is the probability that proposals are accepted?
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2. Suppose we observe y1, · · · , yn where
yi = θi + εi (i = 1, · · · , n)
where {εi} is a sequence of random variables with mean 0 and finite variance representing
noise. We will assume that θ1, · · · , θn are smooth in the sense that θi = g(i) for some
continuous and differentiable function g. The least squares estimates of θ1, · · · , θn are trivial
— bθi = yi
for all i — but we can modify least squares in a number of ways to accommodate
the “smoothness” assumption on {θi}. In this problem, we will consider estimating {θi} by
minimizing
Xn
i=1
(yi − θi)
2 + λ
nX−1
i=2
(θi+1 − 2θi + θi−1)
2
where λ > 0 is a tuning parameter that controls the smoothness of the estimates bθ1, · · · ,
bθn.
(This method is known as Whittaker graduation in actuarial science and the HodrickPrescott filter in economics.)
(a) Show that if {yi} are exactly linear, i.e. yi = a × i + b for all i then bθi = yi
for all i.
(b) In principal, we could compute bθ1, · · · ,
bθn using ordinary least squares estimation. Show
that bθ = (bθ1, · · · ,
bθn)
T minimizes
ky
∗ − Xθk
2
where y
∗
is a vector of length 2n−2 and X is an (2n−2)×n matrix. What are y
∗ and X?
(c) When n is large, computing bθ1, · · · ,
bθn directly, for example, using the OLS formulation
in part (b) is computationally expensive when n is large. Alternatively, we could use the
Gauss-Seidel algorithm but it converges slowly, particularly for larger values of λ. One
possible alternatively is a randomized modification of the Gauss-Seidel algorithm, which at
each stage solves a p( n) variable least squares problem.
The basic algorithm is as follows:
0. Initialize bθ.
1. Randomly sample a subset w of size p from the integers 1, · · · , n. Define Xw to be
the submatrix of X with column indices w and Xw¯ to be the submatrix of X with
column indices in the complement of w; define θw and θw¯ analogously so that Xθ =
Xwθw + Xw¯θw¯.
2. Define bθw to minimize
y
∗ − Xw¯
bθw¯ − Xwθw
with respect to θw.
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3. Repeat steps 1 and 2 until convergence.
Show that the objective function is non-increasing from one iteration to the next.
(d) On Quercus, there is a function HP in a file HP.txt, which implements the randomized
block Gauss-Seidel algorithm outlined in part (c). This function can be used as follows
> r <- HP(x,lambda=2000,p=20,niter=100)
where lambda is the value of λ, p is the value of p, and niter specifies the number of
iterations of the algorithm. The output of the function (contained here in r) consists of two
components: the estimates of {θ} in r$xhat and the values of the objective function for each
iteration in r$ss.
Use this function (with λ = 2000) on data on monthly yields on short-term British securities
over a 21 year period (252 months). Try out various values of p between 5 and 50 (using
1000 iterations). Describe how the objective function decreases with each iteration as p
varies between 5 and 50.
(e) (Optional) Methods such as the randomized block Gauss-Seidel algorithm are essential
in problems where the number of unknown parameters is extremely large. The goal is not to
minimize the objective function but merely to find a solution that’s close to the minimum.
In the context of the randomized block Gauss-Seidel algorithm, what factors should you
consider in choosing p, the numbers of parameters being optimized at each step? Keep in
mind that for least squares, the number of floating point operations needed increases with p
like p
2
.
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Supplemental problems (not to hand in):
3. To generate random variables from some distributions, we need to generate two independent two independent random variables Y and V where Y has a uniform distribution over
some finite 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 finite set is {0, 1, · · · , n − 1} for some integer n ≥ 2. For
U ∼ Unif(0, 1), define
Y = bnUc and V = nU − Y
where bxc is the integer part of x. Show that Y has a uniform distribution on the set
{0, 1, · · · , n − 1}, V has a uniform distribution on [0, 1], and Y and V are independent.
(b) What happens to the precision of V defined in part (a) as n increases? (For example, if
U 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?
4. One issue with rejection sampling is a lack of efficiency 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.
2. If U ≤ f1(X)/g(X) then return X.
3. Otherwise (that is, if U > f1(X)/g(X)) generate X from the density
f
∗
2
(x) = f2(x)
Z ∞
−∞
f2(t) dt
.
Note that we must be able to easily sample from g and f
∗
2
in order for the A-C method to
be efficient; in some cases, they can both be taken to be uniform distributions.
(a) 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?
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(b) Suppose we want to sample from the truncated Cauchy density
f(x) = 2
π(1 + x
2
)
(−1 ≤ x ≤ 1)
using the A-C method with f2(x) = k, a constant, for −1 ≤ x ≤ 1 (so that f
∗
2
(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) Defining 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?
5. 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
f(x) = exp(−x
2/2)
√
2π(1 − Φ(b))
for x ≥ b
with f(x) = 0 for x < b where Φ(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.
(This proposal density is used by the Monty Python algorithm to sample from the tail of
the normal distribution.)
(a) Define 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 defines X =
b + Y/b if
−2 ln(U) ≥
Y
2
b
2
.
(Hint: Note that b + Y/b has density g.)
(b) Show the probability of acceptance is given by
√
2πb(1 − Φ(b))
exp(−b
2/2) .
What happens to this probability for large values of b? (Hint: You need to evaluate M =
max f(x)/g(x).)
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(c) Suppose we replace the proposal density g defined above by
gλ(x) = λ exp(−λ(x − b)) for x ≥ b.
(Note that gλ is also a shifted exponential density.) What value of λ maximizes the probability of acceptance? (Hint: Note that you are trying to solve the problem
min
λ>0
max
x≥b
f(x)
gλ(x)
for λ. Because the density gλ(x) has heavier tails, the minimax problem above will have the
same solution as the maximin problem
max
x≥b
min
λ>0
f(x)
gλ(x)
which may be easier to solve.)
6. (a) Suppose that E1, E2, · · · are independent Exponential random variables with density
f(x) = λ exp(−λx) for x ≥ 0. Then the distribution of Tk = E1 + · · · + Ek is a Gamma
distribution whose density is
gk(x) = λ
kx
k−1
exp(−λx)
(k − 1)! for x ≥ 0.
Show that
P(Tk ≥ 1) = Z ∞
1
gk(x) dx =
k
X−1
j=0
λ
j
exp(−λ)
j!
(Hint: Use integration by parts.)
(b) How might you use the result of part (a) to generate random variables from a Poisson
distribution with mean λ?
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