Přepis řeči - ERROR EXPONENTS FOR DECENTRALIZED DETECTION IN FEEDBACK ARCHITECTURES

0:00:13	and
0:00:15	we right don't that
0:00:16	everyone one um we pen uh and and this is a uh and so what we found since six agrees
0:00:19	in T
0:00:21	i okay so uh i first give you have a a a a brief background about uh these things rise
0:00:26	detection in the um
0:00:29	but a go problem and were looking at and then uh a power about the resulting a a message on
0:00:35	feedback architecture in which are things as
0:00:38	a a a a a lot to send two messages to the fusion center
0:00:41	and then we tell about a uh
0:00:43	the are P but i've that to the so called a message own configuration base where
0:00:47	or since as is only a lot to say uh one in the message
0:00:51	to the fusion center then will conclude sound a
0:00:54	uh remote
0:00:56	okay case
0:00:57	for red
0:00:57	a uh
0:00:59	or would be of a these things wise detection
0:01:02	basically we we have a multi sensors
0:01:05	yeah use of them making a a
0:01:07	it's all measurement
0:01:08	and then you are there's a fusion center and all the six things as a was saying
0:01:12	a quantized version of the measurements to the fusion center
0:01:15	so
0:01:18	so in um
0:01:20	in
0:01:21	in a and you know applications are
0:01:24	what
0:01:25	we talk about these are uh and that what we use it for about a wireless network so
0:01:29	there's always a a a a a a cat that's is a you can on the channel between the
0:01:33	sensor and the fusion center so that's why
0:01:35	uh a is um
0:01:37	a common to in
0:01:38	that the quantization function has to match the measurement to a finite alphabet
0:01:43	although though this is not really uh require you know problem setup
0:01:46	uh
0:01:47	but that's for for this and
0:01:50	for these cars and C we were assume these are
0:01:53	uh this this title quantization where you are mapping being to a finite of rubber
0:01:57	and not we looking at use a binary hypotheses are testing problem way by a
0:02:03	the sensors thing
0:02:05	each since a a a the sense the measurements are coming from a to a a a a different hypotheses
0:02:10	so we know that the line these solutions and the fusion center use
0:02:14	uh
0:02:15	trying to make a final decision based on the quantized measurements from the sensors
0:02:19	oh oh oh which is the correct hypotheses
0:02:22	so the the main uh a problem to solve in this case is trying to find the optimal strategy okay
0:02:28	all these quantization function
0:02:31	that were minimize on eric crack criteria
0:02:33	so uh in this will would be a uh just comes into and looking at a that area
0:02:38	a probably T it's so want to minimize the basing in error probability
0:02:42	and this a sound has basically been studied you know over the last
0:02:46	and to us by menu of those
0:02:48	by in things the be our research
0:02:51	but out of many different uh formulation
0:02:53	may different
0:02:54	things they and consider in these uh
0:02:57	these enjoyed detection
0:02:59	uh a can use that we assume that the
0:03:02	measurements okay all observations of sensors are independent
0:03:05	a a a a given a can use an on a you do hypotheses
0:03:09	so this is because many because um
0:03:11	he becomes a and B a problem you if we don't have are independent
0:03:15	assumption in the general case or i i must at that a a i'm the some special case of
0:03:20	when you have call a call we the measurements but and is on special these solutions
0:03:24	you would do
0:03:25	you
0:03:26	you why able to find you are optimal solutions
0:03:30	so the
0:03:32	main problem that the will do at ease are
0:03:34	can see the uh the pro the question of are having feed that
0:03:38	in these are
0:03:39	these things i think
0:03:40	detection architecture
0:03:42	so uh by feed that we are going to define in of or right
0:03:46	of age noise things
0:03:48	uh
0:03:49	so that if but i one to being at is basically were
0:03:52	um
0:03:54	um are all sensors have information about some all
0:03:57	or other things as messages
0:03:59	so uh in them
0:04:01	i just in the nodes uh a go kind all
0:04:04	the most people are a way that people around as in feed that you are in these kind example
0:04:08	where you have sensors us sending
0:04:11	uh in a
0:04:13	and ties measurements to a fusion center a fusion center is coming up reading the every decision
0:04:17	and then is but cost everything back to the few
0:04:20	all the things those and the sensors were saying a second message
0:04:23	making use of these on reservation as so is that's wrong information that
0:04:27	i now games right from of use and centre
0:04:30	a a to and the second message before before was saying
0:04:34	a a back to the fusion center so this
0:04:36	this would be a um a a a a a a not to do the kind of feedback there
0:04:40	a a on this thing
0:04:42	uh about will be looking at a
0:04:45	so now and in
0:04:46	to this is that we call a two message some configuration
0:04:50	oh we also do at uh some one mess X about these eating configuration it don't
0:04:55	but the that thing to note you use that the message messages are not independent anymore because um
0:05:00	now the second message is going to depend on
0:05:03	ah
0:05:04	information
0:05:05	that a actually you is that car really some of the things of measurements of with you
0:05:10	a and we want to and so close of how to design all these are quantization function singer know optimal
0:05:15	away
0:05:16	K where the optimal a i in the there's use would do that
0:05:21	okay
0:05:22	so yeah so this key the problem
0:05:25	that
0:05:25	people that we want solve is to to of find all these are team uh
0:05:29	on use and functions to minimize or error a probably the
0:05:32	a we know that using a a likelihood ratio quantizers is
0:05:36	i use the a team in to do so not surprise hearing C's
0:05:40	as you some by uh
0:05:41	a king and a fast knee ninety six
0:05:44	so they provide a uh
0:05:46	so call
0:05:46	P uh as and by an optimal solutions
0:05:49	whereby if you fix
0:05:51	uh the
0:05:53	the quantization functions all file all the other things thus
0:05:56	okay except for the one are interested in you fix all the rising day
0:06:00	you
0:06:01	you can uh find the optimal quantizer for that particular things that uh you want to optimize
0:06:07	okay
0:06:07	and you can can do this iterative the so is
0:06:10	by a side of process
0:06:12	and to fine two
0:06:13	uh to will find the optimal solution to compose of a the solution
0:06:18	a at the end
0:06:20	they about unfortunately there's not close once a reason in uh
0:06:23	we don't really know why use the
0:06:25	uh performance uh the ads
0:06:28	actual performance
0:06:30	uh uh and then to good performance of the uh the or um put
0:06:34	particular architecture
0:06:36	okay so all these word depend on numerical michael computation
0:06:40	uh so that is not a price
0:06:42	okay
0:06:42	so even but we out feed that we we know that
0:06:46	find a a a uh these optimal quantizers is that they've got
0:06:50	problem
0:06:51	K uh you do we even if we assume that the sensors are making i i'd observations
0:06:57	okay it turns out that a a the a D much as was okay so
0:07:01	but we we know that all the cup uh since as to be using that you
0:07:05	racial quantizers so base you want to find a too much wrestle
0:07:09	okay oh oh ah a
0:07:11	oh of these are likelihood ratio quantizers and that's wrestles kind
0:07:15	can be different
0:07:16	okay
0:07:16	you
0:07:17	well you use of a basically you to find stress soon use of a a system a couple you questions
0:07:22	and can sound this so
0:07:24	can be a different even five we've i idea some sense so
0:07:28	um
0:07:29	the if you want to find a a the optimal solution
0:07:33	i
0:07:34	so i missed the problem right in uh i
0:07:37	intractable tractable and uh use difficult got to compare the performance of a
0:07:41	with a if you have it down here
0:07:43	you have no feed that was the difference in in the performance or even one compatible
0:07:47	performance across different kinds of a on that were architectures
0:07:51	okay so that's why that's the it this motivation for looking at these also quite a respondent
0:07:56	okay so what we you
0:07:58	one to do is on you still trying to find a the a team was we use and we want
0:08:01	to
0:08:02	a fine i
0:08:03	strategies disease that uh
0:08:05	that mean all these these errors buildings an egg you want to use or want to find
0:08:10	uh
0:08:11	a team uh strategy is to minimize these error exponent of maximise the F so we value all these error
0:08:16	exponent
0:08:17	can so that a a power configuration we all feed that
0:08:20	we we see that the um
0:08:23	a team error exponent is given by a uh
0:08:27	finding the quantizer the that a a a minus the such an a an over here
0:08:32	okay and
0:08:34	and this so we so we see that is actually a team of for all the things is to use
0:08:38	the same quantizer a K two
0:08:41	uh to quantized a measurement before is saying send you back to the fusion center
0:08:46	okay so now look as as a if we use
0:08:48	we use the same kind of or how are we going to uh a that these uh
0:08:52	a problem of that
0:08:53	we a
0:08:54	how the the i don't have a spend and then D for from the
0:08:58	case of these parallel configuration ralph
0:09:02	so the first a a that's so are looking at use the
0:09:05	uh to two its architecture which are already uh
0:09:09	this a brief these us now also
0:09:11	so um i didn't
0:09:13	well is happening here is the first message saying by use things the okay so we is the
0:09:18	oh okay is quantized by some quantization function
0:09:22	and the that do we sent to the fusion center
0:09:25	so now i it was an a is gonna car it a feed that message
0:09:28	so this can be a of and no if that message a which can see some of V are sending
0:09:33	a feedback that two things are K
0:09:35	so what you need to do we you
0:09:38	you only need to take all the measurements uh or the messages from the all the other things as
0:09:42	use some cell function to from the message
0:09:45	and feed that to since okay maybe has
0:09:47	sense K already we he's all measurements we we
0:09:50	or we know like a so we doesn't need
0:09:52	you don't need to consider that inside is on feedback comes
0:09:56	and uh the second that is is going to be for use things that can using a old measurement okay
0:10:01	so these nice alone
0:10:02	doesn't change that may only one single measurement each sensor
0:10:05	and
0:10:06	also also based on the as strong information that the fusion center has provided
0:10:11	to form a second message
0:10:12	before passing me back to a some something
0:10:16	and finally a fusion center of make a final decision based on all the message it was all the first
0:10:21	message as a sell like the second message
0:10:23	a to make the final decision
0:10:25	so that are two types of it that that we can do get
0:10:28	the first one is the so for three that way by you
0:10:31	there's a you don't want eyes any of the
0:10:33	information the fusion center doesn't compress any other information you has
0:10:38	so we sense of for every information back to the
0:10:40	or the sensors
0:10:42	uh use no light uh a a a a a a a it because up because um
0:10:46	in in in actual a because he's using you won't be able to
0:10:49	i be doing to do that because we we are doing use a where
0:10:53	increasing the number of things as like
0:10:56	so
0:10:57	this is that sometimes them as a character
0:10:59	okay that go set up
0:11:01	and then map in that case is when you have a re should the feed that where you are use
0:11:05	of your are um feedback information is as you compress
0:11:08	okay to some find for but
0:11:11	in don't in here is that we are assumed that a fusion center is gonna retain all these values or
0:11:16	the values of the first
0:11:18	messages okay
0:11:19	used used retaining that okay
0:11:21	a a compressed person
0:11:24	so that is the the first result that we have all the um
0:11:28	to to message architecture and turns out that
0:11:31	as you
0:11:32	the average spend that under the for feed that
0:11:34	case
0:11:35	is is send S are at every S more under the restricted that case
0:11:39	and better than not
0:11:40	the same as the error and
0:11:42	and the the parallel configuration we
0:11:44	that
0:11:45	so the
0:11:47	how that that is an our as known are talking about yeah is uh
0:11:50	when we do at
0:11:51	the is saying so actually a lot to send
0:11:54	a a a a a message is as you or you can do not be says
0:11:58	having a a a a that message about of it
0:12:00	the power the number of be
0:12:02	so we the original you can string of quantization functions to
0:12:06	we have any uh
0:12:07	i have a for the size of a diesel
0:12:09	down have the power configuration were
0:12:11	be a a
0:12:12	it to send a message of of of a but size of two D
0:12:16	okay
0:12:17	so it
0:12:17	lot these results is that a that is
0:12:20	oh you you so not useful
0:12:22	we we are talking about a exponent
0:12:25	um
0:12:27	that we is a big and in to give a uh maybe a i can try they spend these
0:12:32	no
0:12:34	when we first did is the result i also the hidden facts bet
0:12:38	that feedback back as you doesn't uh improve the error exponent
0:12:41	but you know as we've moves the results way a set oh
0:12:45	i i have some intuitive explanation for that
0:12:47	so uh
0:12:49	in this case of what is happening these are all the feed that you know a finite when you have
0:12:54	find that number things as your feed that
0:12:56	can improve your error exponent
0:12:58	and no sorry can improve your detection performance
0:13:01	uh
0:13:02	uh is that in to a of of the the power configuration soon use
0:13:06	not increasing in sporting city fast
0:13:09	so we were talking about
0:13:10	about uh a rose bone were talking about uh
0:13:13	yeah how to here
0:13:15	a a as should with the the every race on the king sporting you fast
0:13:19	well we see is the feed that improvement
0:13:23	he's
0:13:23	no in uh use now at the saying this point is a scale where the
0:13:27	as the main every self so that's i
0:13:30	so even though when you doing the for see that you are not going and
0:13:34	to get any a uh it's true a
0:13:36	gain in terms of error experiment
0:13:39	so
0:13:40	in terms of the proof so uh and just going go through a a few a a few steps of
0:13:45	the proof
0:13:46	so the first thing to know use that we know that the full feedback cases
0:13:50	of is going to be better than of the received that case right
0:13:53	in the from that case of a feeding back
0:13:55	the fruit phonation
0:13:57	okay so in this case you can the for feedback case you are able to see me the the received
0:14:02	a that case
0:14:03	a ease of the things by T
0:14:05	uh a in the
0:14:07	the for information in compressing then
0:14:09	and then a that the receipt a feedback is a is a is a better than the
0:14:14	the parallel configuration because of
0:14:16	in this case we can just know
0:14:18	the feedback that can achieve the same performance
0:14:21	so that we will do need to so be in this case is that the from feed that
0:14:25	a a a a can not go for and as well as a parallel configuration
0:14:29	and the
0:14:31	the way to go about doing these Z uh using a lot so uh that a real the last deviation
0:14:35	so we had
0:14:36	from the trend that by we have a a little about that depends on
0:14:40	um this low moment generating function i so i is the second and they're be for the
0:14:45	lot more more joint thing to my function
0:14:48	so
0:14:49	to
0:14:49	well we need to do is uh
0:14:51	as actually the the means that used to trying to a these on the value of these directive
0:14:56	the
0:14:57	where and we have i the observations disease
0:14:59	as you quite easy you do not when
0:15:02	when our observations are no longer a i D
0:15:05	because a
0:15:06	and the second message is as recall rate that
0:15:09	with each other
0:15:11	okay so
0:15:12	that that would be the
0:15:13	this is the main problem down here that we need to about
0:15:17	so
0:15:17	if we do that is known in generating function seems about the do talking about a a likelihood ratio yours
0:15:23	alright these are lot and renting in and has this phone is a convex curve that goes
0:15:27	uh take the value is you i'm during one
0:15:30	uh
0:15:31	well we have these uh this
0:15:32	uh this result that says that
0:15:35	if i E do at the
0:15:37	a is just planning on this curve or a of is gonna change when the
0:15:41	the number of nodes increases
0:15:43	okay you would get sampling shopper
0:15:45	uh what happens he's
0:15:47	if i do that
0:15:48	uh fixed once low on the curve
0:15:50	okay and i do at the second if
0:15:54	at the point that
0:15:55	that has no that's not T
0:15:57	then these a second derivative can not increase uh a starting point not really
0:16:03	okay so the
0:16:05	a that's that's he's actually
0:16:07	uh you
0:16:08	in not in there then um
0:16:10	applying james and in quality at um
0:16:14	a a a a in is and uh
0:16:16	cases
0:16:17	in a by trying to all uh
0:16:19	the
0:16:20	second directly
0:16:22	the first and second or these with their are quantizer portions
0:16:25	i of the uh all of a wide a function that you know
0:16:28	trying to copy
0:16:30	okay
0:16:32	so once so we have that the and once you can buy the secondary T if and then you see
0:16:36	that uh these
0:16:38	this funds this square a function will were go to zero if you take the a we have a angle
0:16:43	a on both sides
0:16:45	okay and you are left with these thing
0:16:47	and and in are we need to find a a a a lower by of for these are uh moments
0:16:51	renting function which can be easy D down okay is not to do got here
0:16:55	and in the there is uh
0:16:57	the is uh a lot of and in function can be shown to be
0:17:01	ah
0:17:03	to be the little bottle by these are
0:17:05	i respond and from the parallel configuration
0:17:08	okay so one
0:17:10	a a a a week
0:17:11	one thought of that some and that we were making use that the fusion center a
0:17:14	readings the
0:17:16	for the information the message so as the first set of messages
0:17:19	so i can see if you has only limited memory and can only
0:17:23	a a a a reading of compressible version
0:17:25	oh of these are a them at macy's message was and ten out that this is as you a got
0:17:29	problem
0:17:30	K uh
0:17:31	so that the top how in night to use a all tax approach that uh because it is a matter
0:17:37	of types of you need to assume
0:17:39	find a of about observations i didn't a good no quantization function so one
0:17:43	okay to to to so like in this case is uh the feed that also doesn't help
0:17:48	we a to open problem that
0:17:50	in the more general uh scenario
0:17:53	with the these uh
0:17:55	how does a few that contribute to the error spend then
0:17:58	a so you know that to kind of try to gain some insight into that problem than we do a
0:18:02	a really the problem
0:18:03	the so i one i six are architecture way by
0:18:06	now
0:18:07	we have
0:18:08	we do but the sensors into two to good of things as so the first goal was so as is
0:18:12	uh things the first message
0:18:14	uh to the fusion center and fusion center fixed back
0:18:18	these are compressed
0:18:19	i don't the for information or compressed version to the second most things sensors
0:18:24	okay
0:18:26	ooh
0:18:27	in this case if you
0:18:29	you will uh the architecture use actually equivalent to a
0:18:34	to this kind of that so
0:18:36	a i that so that's why is a call you are uh daisy chain architecture
0:18:40	okay where you
0:18:42	you have a first the the
0:18:44	i so as we at which can be compressed or or you and then before use been fact that
0:18:48	to the a a second group most things are
0:18:52	they so that didn't we can see that two types of it that the for that
0:18:55	okay where you there's no loss in the information that is been fact that you know the risk a feedback
0:19:00	case
0:19:02	and again we have to a uh you know go T which is uh in media
0:19:06	so that in the full of feedback case
0:19:08	the for three that
0:19:09	every and then
0:19:10	has to be at you have at least ask us as the we that one
0:19:14	okay but in this case we need to compare we've the tree configuration
0:19:18	so in this case if we don't have any you back to the second group a basically are getting a
0:19:22	three
0:19:23	okay so
0:19:24	uh a reason that thing to do is to compare these we've the
0:19:27	i or ten minutes three
0:19:30	so okay a again if we have a feed that the actually the error S one
0:19:35	a same as the parallel
0:19:37	but
0:19:38	E if we
0:19:40	if we have we should that feed that in this case
0:19:43	one just doing didn't happens use that you receipt of feed that
0:19:46	is she the you worse than your parallel configuration
0:19:50	okay okay use a by the by your tree error exponents so
0:19:53	in some cases this the receipt a feed that have one and can be the same as a three N
0:19:58	in some cases
0:19:59	which we can uh find uh put to good example
0:20:03	do so that the received a feed that use um
0:20:07	i actually were saying the street K
0:20:09	so that have a smaller than you this case and that for the received the feedback
0:20:13	is has a given by these uh
0:20:15	this expression suppression what here
0:20:16	if you do that these one B C's actually um
0:20:20	is happening here is the first school are so as is going to use
0:20:24	the same quantization function come mad
0:20:26	and then
0:20:27	at the
0:20:29	at the first stage the fusion center is con
0:20:31	compared the ah
0:20:33	the receive are likelihood ratio
0:20:36	we've
0:20:37	that's that's T
0:20:38	okay
0:20:39	before you saying um
0:20:41	before you compress the all the
0:20:44	the a a first group of information
0:20:47	i i first group model of a
0:20:50	sense this so green into a really decisions you are one
0:20:54	so in this case if a the second row things the received a zero fun of use think that is
0:20:59	going to use a different
0:21:00	quantization as compact use uh receiving a one
0:21:04	okay but in in either case the second be is going also going to use the same quantization function
0:21:10	okay
0:21:11	the which depends on the feedback message
0:21:15	oh there so i
0:21:16	in this case under the uh is sounds
0:21:18	we also provide some special conditions
0:21:21	we by a receipt of feed that i read or use the same as all tree at right one then
0:21:27	okay
0:21:27	so in conclusion
0:21:29	that that a S messages
0:21:31	this is
0:21:32	uh
0:21:34	uh
0:21:35	we had that binary hypothesis testing
0:21:38	no by i mean the decentralized binary hypothesis testing problem
0:21:42	and we so that
0:21:43	she that in the two configurations that we talk about
0:21:46	uh
0:21:48	does not improve the error exponent but
0:21:50	that's because uh the fusion center has access to a full information that is available
0:21:55	a a lot of can get from these it's seems memory she right uh the performance gain due to feed
0:22:01	that may not
0:22:02	so i you're increasing the communication cost
0:22:05	when you're are you're do have feedback
0:22:08	that we see that feed that can as improve on her from in the received that feed that uh
0:22:13	well i six a busy chain architecture
0:22:16	okay and we provide a characterization for the exponent
0:22:19	so that L i if the top of that mean be we've one open problem
0:22:24	okay
0:22:25	i
0:22:26	okay

ERROR EXPONENTS FOR DECENTRALIZED DETECTION IN FEEDBACK ARCHITECTURES

Sensor Networks

Přednášející: Wee Peng Tay, Autoři: Wee Peng Tay, Nanyang Technological University, Singapore; John Tsitsiklis, Massachusetts Institute of Technology, United States