曝光臺(tái) 注意防騙 網(wǎng)曝天貓店富美金盛家居專營(yíng)店坑蒙拐騙欺詐消費(fèi)者

the requirement for the algorithm to update the estimator ˆk in the sense of shifting
the estimator in the right direction, towards k.
We would like to start with stating the following lemma:
Denote g(u, k) = ES (Yk, u).
Lemma 3.5.1.1 Assume that the conditions (A1) - (A4) are satisfied. For any u 2
R, n 2 N, 0  k  n, the representation g(u, k) = −G(u, k)(u−k) holds uniformly
over  2  for some function G(u, k), such that G1  G(u, k)  G2 with some
positive constants G1 and G2.
Proof. We try to prove the lemma in cases.
Case 1: |u|  H + L + h
Lets first compute the function g(u, k). Remember, it is the expectation of the
function S (Yk, u):
g(u, k) = ES (Yk, u) = E − d(Yk − u) = −1{Yk − u  0} + ( − 1)1{Yk − u = 0}
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= E(1{Yk − u  0}) + E(( − 1)1{Yk − u < 0})
Using the basic properties of expectation, one gets:
g(u, k) = P(Yk − u  0) + ( − 1)P(Yk − u < 0)
Now, say, q = P(Yk − u  0), which implies 1 − q = P(Yk − u < 0), and rewriting
the above equation:
g(u, k) = ES (Yk, u) = q + ( − 1)(1 − q) = q − q + 1 − q
=  − (1 − q) =  − P(Yk − u < 0)
Using the equation (1), Yk = k + k, one gets:
ES (Yk, u) =  − P(Yk − u < 0) =  − P(k < u − k) =  − F(u − k)
Now, it is left to prove that there exist a function G(u, k) satisfying the conditions
in the lemma with:
G(u, k) = −g(u, k)
(u − k) =
F(u − k) − 
(u − k)
We know F, being a distribution function, is monotone increasing and by assumption
(A1) F = ; therefore,
if (u − k) < 0, F(u − k)   and F(u − k)/(u − k)  0, and
if (u − k) > 0, F(u − k)   and F(u − k)/(u − k)  0.
Hence, G(u, k) is always non-negative. Now, denote (u − k) by v. If |v|  ,
with constant  from assumption (A2), we can use a Taylor series expansion to
approximate F around 0 upto the first order term.
F(v) = F(0) + F0 (v)(v − 0)
with F0(v) = f(v), where |v|  |v|  . Using (A1), F(0) = , we can express G
as:
G(u, k) =
 + f(v)v − 
v = f(v)
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Now we can conclude by using the assumption (A2) and (A3) that:
0 < p  G(u, k)  sup
|v|
f(v) < 1
If |u − k| = |v| > , then
G(u, k) =
F(u − k) − 
u − k 
1

Moreover, since  2 ,
|(u)| = |(u)|−|(0)|+|(0)|  |(u)−(0)|+|(0)|  L|u−0|+H = Lu+H  H+L
for any u 2 [0, 1]. Using this result:
|u − k|  |u| + |k|  2(H + L) + h
This leads us to:
G(u, k) =
F(u − k) − 
u − k 
min{F() − ,  − F(−)}
2(H + L) + h
Lets again use Taylor expansions around 0 upto the first order tirm for F() and
F(−) to examine min{F() − ,  − F(−)}.
F() = F(0) + F0() with ||  ||
) F() −  =  + f () −   p by (A1) and (A2)
and
F(−) = F(0) + F0 ()(−) with |  |  ||
)  − F(−) =  −  + f ()  p by (A1) and (A2)
Therefore;
G(u, k) =
F(u − k) − 
u − k 
min{F() − ,  − F(−)}
2(H + L) + h 
p
2(H + L) + h
> 0
Case 2: |u| > H + L + h
In this case S (Yk, u) = −u so that g(u, k) = ES (Yk, u) = −u which givesG(u, k) =
u/(u − k). Using the basic inequality |u − k|  |u| − |k|  h, conclude that:
G(u, k) =
u
u − k
=
u − k + k
u − k
= 1 +
k
u − k  |1| + |k|
|u − k|  1 +
H + L
h
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On the other hand, in order to find a lower bound for G(u, k), lets look at the signs
of u and k:
• if u and k are of the same sign then clearly G(u, k)  1.
• if u and k are of opposite signs, observe that,
|u − k|  H + L + h + |k| ( 2(H + L) + h)
then:
G(u, k) =
u
u − k
=
u − k + k
u − k
= 1+
k
u − k  1− |k|
|u − k|  1− |k|
H + L + h + |k|
=
H + L + h
H + L + h + |k| 
H + L + h
2(H + L) + h 
1
2
To sum up;
• if |u|  H + L + h ) 1/2  G(u, k)  1 + [(H + L)/h]
• if |u| > H + L + h
　
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