代写bandwidth留学生、代做R设计、R程序语言调试

日期：2019-03-19 09:20

Problem sheet 5

The next question is optional, yet it will be a bonus to your final score.

2. If the density function is compactly supported, standard kernel density estimators are invalid/inconsistent at or

near boundary points. Below we introduce a new estimator that automatically adapts to the (possibly unknown)

boundaries of the support of the density without requiring specific data modification or additional tuning parameters.

Let be a random sample, where is a continuous random variable with a smooth CDF over its possibly

unknown support for . If and , then is supported on the whole real line. The

density function is , where the derivative is interpreted as a one-sided derivative at a boundary

point of . Let be the empirical distribution function. For every , solve

where is a bandwidth, and denotes a nonnegative, symmetric and continuous kernel function supported on

. The proposed boundary adaptive density estimator is then defined as

.

X1,…, Xn Xi

[a, b] a < b a = ∞ b = ∞ Xi

f (x) = d

d x(Xi ≤ x)

[a, b] F

(x) = (1/n)∑ni=1

I(Xi ≤ x) x ∈

= argminβ=(β0,β1,β2)

(Xi) β0β1(Xi x) β2(Xi x)2}2K(Xi xh ),

h > 0 K[1,1]f ba(x) = β?

1(x)

1

(a) Write an R function to implement the above estimator, with as an input parameter. Use the triangular

kernel , .

(b) Conduct simulation study based on exponential distribution with density function if and

if . This density function is then supported on . We want to estimate the density at

, where is the boundary, is a near-boundary point and is an interior point. Take the sample

size . For various choices of the bandwidth , conduct 5000 Monte Carlo repetitions, and report the

average estimation error and the corresponding standard deviation at .

(c) Use 5-fold cross-validation to select the bandwidth for the above estimator. Again, for exponential distribution,

take and conduct 5000 Monte Carlo repetitions to compare the performance of this estimator with

standard kernel density estimator (bandwidth chosen by default) at .

h > 0

K(u) = max{0,1 |u |} u ∈f (x) = e?x x ≥ 0

f (x) = 0 x < 0 [0,∞)

x ∈ {0,0.5,1} 0 0.5 1

n = 1000 h| f ba(x) f (x)| x ∈ {0,0.5,1}

n = 1000

x ∈ {0,0.5,1}