.., x(i)
k represents the measurement at Si1, Si2, ..., Sik respectively. A
comparison between xi and x(i)
1 , x(i)
2 , ..., x(i)
k can be made by calculating the
di?®erence between xi and the median of x(i)
1 , x(i)
2 , ..., x(i)
k :
di = xi ??’ medi. (18)
The reason why the comparison is made with the median instead of the
mean is that the value of a mean could easily be biased due to the extreme
readings of faulty sensors.
To quantify the extremeness of di, let??™s consider another set of sensors
N?¤(Si) which contains Si and its n ??’ 1 neighbors. Based on the notion of
equation (18), sensors in N?¤(Si) yield a set of D = {d1, d2, ..., dn}.
?†?µ =
1
n
n
Xi=1
di, (19)
?†?? = vuut
1
n ??’ 1
n
Xi=1
(di ??’ ?†?µ)2. (20)
157
Ren-Shiou Liu, Lifeng Sang, and Prasun Sinha
Fig. 5. An example of N?¤ neighborhood N?¤(Si) of sensor Si and N neighborhood
N(Si) of sensor Si inside N?¤(Si). While the N neighborhood is used to compute di
and the N?¤ neighborhood is used to quantify the extremeness of di, one can simply
select N?¤(Si) = N(Si).
With the mean (19) and standard deviation (20) of the set D, one can
standardize D to obtain Y = {y1, y2, ..., yn}, where:
y1 =
d1 ??’ ?†?µ
?†??
, ..., yi =
di ??’ ?†?µ
?†??
, ..., yn =
dn ??’ ?†?µ
?†??
. (21)
Since the standard deviation for a set of statistic values denotes the typical
number of steps away from their mean, the standardized set {y1, y2, .
Pages:
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274