Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (61.14 KB, 18 trang )
<span class='text_page_counter'>(1)</span><div class='page_container' data-page=1>
1. Review: The periodogram
The periodogram is defined as
I(ν) = <sub>|</sub>X(ν)<sub>|</sub>2
= 1
n
n
X
t=1
e−2πitνx
t
2
= X<sub>c</sub>2(ν) + X<sub>s</sub>2(ν).
n
n
X
t=1
cos(2πtν)x<sub>t</sub>,
X<sub>s</sub>(ν) = <sub>√</sub>1
n
n
X
We want to understand the asymptotic behavior of the periodogram I(ν) at
a particular frequency ν, as n increases. We’ll see that its expectation
converges to f(ν).
We’ll start with a simple example: Suppose that X<sub>1</sub>, . . . , X<sub>n</sub> are
i.i.d. N(0, σ2) (Gaussian white noise). From the definitions,
Xc(νj) =
1
n
n
X
t=1
cos(2πtνj)xt, Xs(νj) =
1
√
n
n
X
t=1
sin(2πtνj)xt,
Also,
Var(X<sub>c</sub>(ν<sub>j</sub>)) = σ
2
n
X
t=1
cos2(2πtν<sub>j</sub>)
= σ
2
2n
n
X
t=1
(cos(4πtνj) + 1) =
σ2
2 .
Also,
Cov(X<sub>c</sub>(ν<sub>j</sub>), X<sub>s</sub>(ν<sub>j</sub>)) = σ
2
n
n
X
t=1
cos(2πtν<sub>j</sub>) sin(2πtν<sub>j</sub>)
= σ
2
2n
n
X
t=1
sin(4πtν<sub>j</sub>) = 0,
Cov(X<sub>c</sub>(ν<sub>j</sub>), X<sub>c</sub>(ν<sub>k</sub>)) = 0
Cov(X<sub>s</sub>(ν<sub>j</sub>), X<sub>s</sub>(ν<sub>k</sub>)) = 0
That is, if X<sub>1</sub>, . . . , X<sub>n</sub> are i.i.d. N(0, σ2)
(Gaussian white noise; f(ν) = σ2), then the X<sub>c</sub>(ν<sub>j</sub>) and X<sub>s</sub>(ν<sub>j</sub>) are all
i.i.d. N(0, σ2/2). Thus,
2
σ2 I(νj) =
2
σ2 X
2
c(νj) + Xs2(νj)
∼ χ2<sub>2</sub>.
So for the case of Gaussian white noise, the periodogram has a chi-squared
distribution that depends on the variance σ2 (which, in this case, is the
Under more general conditions (e.g., normal <sub>{</sub>Xt}, or linear process {Xt}
with rapidly decaying ACF), the Xc(νj), Xs(νj) are all asymptotically
independent and N(0, f(νj)/2).
Consider a frequency ν. For a given value of n, let νˆ(n) be the closest
Fourier frequency (that is, νˆ(n) = j/n for a value of j that minimizes
|ν <sub>−</sub> j/n<sub>|</sub>). As n increases, νˆ(n) <sub>→</sub> ν, and (under the same conditions that
ensure the asymptotic normality and independence of the sine/cosine
transforms), f(ˆν(n)) <sub>→</sub> f(ν). (picture)
In that case, we have
2
f(ν)I(ˆν
(n)<sub>) =</sub> 2
f(ν)
Thus,
EI(ˆν(n)) = f(ν)
2 E
2
f(ν)
X<sub>c</sub>2(ˆν(n)) + X<sub>s</sub>2(ˆν(n))
→ f(<sub>2</sub>ν)E(Z<sub>1</sub>2 + Z<sub>2</sub>2) = f(ν),
Since we know its asymptotic distribution (chi-squared), we can compute
approximate confidence intervals:
Pr
2
f(ν)I(ˆν
(n)<sub>)</sub> <sub>> χ</sub>2
2(α)
→ α,
where the cdf of a χ2<sub>2</sub> at χ2<sub>2</sub>(α) is 1 <sub>−</sub> α. Thus,
Pr
2I(ˆν(n))
χ2<sub>2</sub>(α/2) ≤ f(ν) ≤
2I(ˆν(n))
χ2<sub>2</sub>(1 <sub>−</sub> α/2)
Unfortunately, Var(I(ˆν(n))) → f(ν)2Var(Z<sub>1</sub>2 + Z<sub>2</sub>2)/4, where Z<sub>1</sub>, Z<sub>2</sub> are
i.i.d. N(0,1), that is, the variance approaches a constant.
Thus, I(ˆν(n)) is not a consistent estimator of f(ν). In particular, if
f(ν) > 0, then for ǫ > 0, as n increases,
Pr n
I(ˆν
(n)<sub>)</sub>
− f(ν)
> ǫ
o
This means that the approximate confidence intervals we obtain are
typically wide.
The source of the difficulty is that, as n increases, we have additional data
(the n values of x<sub>t</sub>), but we use it to estimate additional independent
random variables, (the n independent values of X<sub>c</sub>(ν<sub>j</sub>), X<sub>s</sub>(ν<sub>j</sub>)).
How can we reduce the variance? The typical approach is to average
independent observations. In this case, we can take an average of “nearby”
values of the periodogram, and hope that the spectral density at the
1. Review: The periodogram
Define a band of frequencies
νk −
L
2n, νk +
L
2n
of bandwidth L/n. Suppose that f(ν) is approximately constant in this
frequency band.
<i>Consider the following smoothed spectral estimator.</i> (assume <sub>L</sub> is odd)
ˆ
f(ν<sub>k</sub>) = 1
L
(L<sub>−</sub>1)/2
X
l=<sub>−</sub>(L<sub>−</sub>1)/2
I(ν<sub>k</sub> <sub>−</sub> l/n)
= 1
L
(L<sub>−</sub>1)/2
X
l= (L 1)/2
For a suitable time series (e.g., Gaussian, or a linear process with
sufficiently rapidly decreasing autocovariance), we know that, for large n,
all of the X<sub>c</sub>(ν<sub>k</sub> <sub>−</sub> l/n) and X<sub>s</sub>(ν<sub>k</sub> <sub>−</sub> l/n) are approximately independent
and normal, with mean zero and variance f(ν<sub>k</sub> <sub>−</sub> l/n)/2. From the
assumption that f(ν) is approximately constant across all of these
frequencies, we have that, asymptotically,
ˆ
f(ν<sub>k</sub>) <sub>∼</sub> f(ν<sub>k</sub>)χ
2
2L
Thus,
Efˆ(ˆν(n)) ≈ f(ν)
2L E
2L
X
i=1
Z<sub>i</sub>2
!
= f(ν),
Varfˆ(ˆν(n)) <sub>≈</sub> f
2<sub>(</sub><sub>ν</sub><sub>)</sub>
4L2 Var
2L
X
i=1
Z<sub>i</sub>2
!
= f
2<sub>(</sub><sub>ν</sub><sub>)</sub>
2L Var(Z
2
1),
From the asymptotic distribution, we can define approximate confidence
intervals as before:
Pr
(
2Lfˆ(ˆν(n))
χ2<sub>2L</sub>(α/2) ≤ f(ν) ≤
2Lfˆ(ˆν(n))
χ2<sub>2L</sub>(1 <sub>−</sub> α/2)
)
≈ 1 <sub>−</sub> α.
<i>Notice the bias-variance trade off:</i>
For bandwidth B = L/n, we have Varfˆ(ν<sub>k</sub>) <sub>≈</sub> c/(Bn) for some constant c.
So we want a bigger bandwidth B <i>to ensure low variance (bandwidth</i>
<i>stability).</i>
But the larger the bandwidth, the more questionable the assumption that
Since the asymptotic mean and variance of fˆ(ˆν(n)) are proportional to f(ν)
and f2(ν)<i>, it is natural to consider the logarithm of the estimator. Then we</i>
can define approximate confidence intervals as before:
Pr
(
2Lfˆ(ˆν(n))
χ2<sub>2L</sub>(α/2) ≤ f(ν) ≤
2Lfˆ(ˆν(n))
χ2<sub>2L</sub>(1 <sub>−</sub> α/2)
)
≈ 1 <sub>−</sub> α,
Pr
logfˆ(ˆν(n)) + log
2L
χ2<sub>2L</sub>(α/2)
≤ log(f(ν)) <sub>≤</sub> log fˆ(ˆν(n)) + log
2L
χ2<sub>2L</sub>(1 <sub>−</sub> α/2)
≈ 1 <sub>−</sub> α.