Přepis řeči - AN EXTENSION OF THE ICA MODEL USING LATENT VARIABLES

0:00:13	so a good morning everyone i'm mark
0:00:15	to that's are a set
0:00:17	and uh i'm getting to
0:00:19	present you work with uh
0:00:21	i i with sample class you and which machines ski
0:00:24	and well or from that to come should but i in
0:00:28	so and the title of my tool is an extension of the ica model
0:00:33	using that and on
0:00:36	so here is the outline of my to a local a first introduce the problem
0:00:40	then a a present you the extension that type prepare of uh be independent comp and uh analysis model
0:00:47	a start with a recall in the classical know or mixture of independent sources and then
0:00:52	i will present uh what i consider the depends sources and lack and variables that that
0:00:59	a and this context i will then uh yeah
0:01:02	consider uh the problem of parameter estimation
0:01:05	in in data
0:01:06	and this will be based on a so called iterative conditional estimation method
0:01:11	so please see that there are two at an ins ice
0:01:14	a a
0:01:15	which you all know and
0:01:16	i C E which is iterative conditional estimation
0:01:20	then i present you the separation method
0:01:23	and uh and uh i show you a few simulations uh of in the case where the last and as
0:01:28	i i I D and in the case where to not
0:01:30	and finally a we conclude my to
0:01:34	so
0:01:35	you probably only blind source separation or were independent come analysis
0:01:39	as well known
0:01:41	and the use you
0:01:42	the usual assumptions are that's
0:01:44	the sources should be statistically mutually independent
0:01:47	uh they should be not ocean
0:01:49	and the observations are
0:01:52	given by a linear instantaneous mixture of the
0:01:56	however
0:01:58	uh
0:01:58	it is known that um
0:02:00	independence is not always recall
0:02:03	for example uh you may uh
0:02:06	re you may rely only on to relation in the case of a second to hold of methods and case
0:02:11	but you use non stationarity or or colour of the sources
0:02:14	and the case of uh source which belong to a finite alphabet then um
0:02:20	may also uh
0:02:21	re independence assumption that there are all other very specific cases that uh was started
0:02:26	yeah
0:02:28	the objective of this talk it is clearly to relax
0:02:32	the independence assumption in blind source separation
0:02:35	and uh for doing this that propose a new model
0:02:38	where a separation is possible
0:02:41	a thing and uh where the depends is controlled by a lot
0:02:46	so let me uh give you are limitations this is a classical i consider that i have capital T samples
0:02:53	that of the observations are denoted by X of to you know and capital and observations capital and sources given
0:02:59	by S F T
0:03:00	a a is the unknown mixing matrix a square matrix invertible and at each time instant the observations X T
0:03:07	are given by the predict eight times as of
0:03:11	i am in the blind count takes that is only a X
0:03:14	which consists of the samples uh
0:03:16	the is available
0:03:18	and the objective of a i C A is to estimate the sources
0:03:21	know the was it is equivalent to estimate D
0:03:24	which is
0:03:25	and in as the inverse of a or an inverse of a a a a to someone and be you
0:03:32	so here is a figure that probably most of you uh did uh already C
0:03:37	uh as as a way to introduce use um
0:03:39	to introduce a
0:03:40	independent component analysis
0:03:43	i consider the classical case with source whether there are two sources S one S two S one is on
0:03:48	the
0:03:48	how reasonable X as a stream and the vertical axis this is a of the left
0:03:52	uh figure
0:03:54	um and uh they are independent uniformly distribute
0:03:57	when when you makes
0:03:58	uh of the of what the matrix a which is given here
0:04:02	then you a you have a the observations on the right i project here um the different samples of
0:04:07	uh X one and X two
0:04:10	a a a a different time instants and to you C basically that
0:04:13	uh you can uh separate the sources because you the different slips of or in on of the parallelogram gives
0:04:20	you
0:04:21	uh the um
0:04:24	columns of the at mixing matrix a
0:04:26	okay this is well now and now comes
0:04:28	then there's uh
0:04:30	the central idea of my talk
0:04:33	what i'm saying is that if ica sixties in
0:04:36	separating sources which and this distribution
0:04:39	i should be able to make do something and to uh perform separation if the distribution looks something like that
0:04:47	so the idea is to consider source
0:04:51	which have
0:04:52	a distribution which is partly independent and partly depend
0:04:58	and
0:04:59	in order to uh
0:05:00	try to find another isn't the idea is to say okay
0:05:04	if i am able to your race
0:05:06	the points in green
0:05:07	or to find
0:05:08	the which are dependent and a as a sense
0:05:12	then i should be able to use
0:05:14	classical any um actually
0:05:17	be able to use any a classical ica algorithm in order to form
0:05:22	a blind source separation
0:05:25	okay so the question is how can i model such
0:05:28	type of distribution such a a of thought of sources
0:05:32	well if you are of its uh carefully we you was you see that
0:05:36	actually you have a mixture of two probability density as
0:05:40	and whether the probability P
0:05:43	between zero and one of the sources are independent and when the probability one minus P the sources are U
0:05:48	penn
0:05:50	and one possible way
0:05:52	which is
0:05:52	a classical to uh
0:05:54	to model as this is to introduce a lot and
0:05:57	a process are of T just hit
0:06:00	and to say if or if is equal to zero them
0:06:03	the sources are independent if a of T equals
0:06:06	one then the sources or depend
0:06:10	so here i'm writing the same i i uh in more uh mathematical trends
0:06:14	i consider that i have a hidden clues R of T V such that can to show you on are
0:06:20	the components of as
0:06:21	one two S
0:06:23	or or uh independent of this is given by the joint a probability distribution was is she's written
0:06:28	uh of the screen
0:06:31	and uh of T takes to use
0:06:33	either zero or one if R of T is equal to zero then
0:06:37	the usual assumptions
0:06:39	uh you made in ica appliances of the components are
0:06:44	independent
0:06:45	uh none fusion
0:06:47	and if R T then you have the dependence which
0:06:51	can choose as you want which i will process specify like
0:06:56	okay that's the model i can
0:06:58	no i'm moving to the problem of a parameter estimation
0:07:03	well first let me say that uh a you can
0:07:06	i i did not say anything about are
0:07:09	you have several possible at is
0:07:11	to possible of is to model as a or as an i I i T but only process
0:07:15	but is uh
0:07:16	or or cool zero with to P or uh equal uh one with probably one minus P and this is
0:07:22	what i i D you can also consider that are is given by a markov chain
0:07:27	well this works
0:07:28	anyway the objective is to estimate
0:07:30	a set of our me which consist of but inverse of the mixing matrix and
0:07:34	or parameters images of or
0:07:37	as i told you before if i am able to you raise to green points that this be new are
0:07:42	and it
0:07:44	well i think you i can perform ica C which it is quite
0:07:48	a easy i can estimate the power it is of a or just by counting the number of of utterances
0:07:52	of
0:07:53	uh zero the number of times
0:07:55	or is equal to zero can give me a an an estimate of P
0:07:59	i can see that the in sometimes where depends of the sources holes
0:08:03	consider only a restricted uh a sub sample of the
0:08:07	observation samples and perform my uh favourite i it
0:08:11	in other words
0:08:13	i i have a complete data estimator that is if are and X are both available i have an estimator
0:08:20	or of the clan it is but i interested
0:08:23	and fortunately or fortunately leave because a
0:08:26	i can give the stalk
0:08:27	we are a in in and complete data um
0:08:31	context that is
0:08:32	X
0:08:33	only is available on the R S and now
0:08:35	and at this point please note that a a a a a i makes to different um
0:08:40	three different I D's is the idea of complete in data we first to the fact that are is no
0:08:45	or a known and the fact that
0:08:48	a S is and known
0:08:49	is uh uh what to to to to that
0:08:52	a to say that S is in and i just say that i'm in a blind
0:08:55	come
0:08:58	okay
0:08:58	so here is the second important idea of my talk
0:09:02	for in complete data estimation there is this uh
0:09:06	quite uh if uh
0:09:07	efficient
0:09:08	i to perform
0:09:09	uh estimation
0:09:11	and
0:09:11	this algorithm is called ice E iterative conditional estimation
0:09:16	and it consists in uh the phone we first initial as the power mean value
0:09:22	oh then
0:09:23	you can you
0:09:24	the expectation of the complete data estimator
0:09:28	no i X
0:09:29	and you H rate
0:09:30	uh a it as that
0:09:32	and uh uh you but obtain converge
0:09:36	that
0:09:37	some
0:09:38	uh of the components of a a T to are not computable if you can compute the uh
0:09:44	the a posteriori
0:09:45	the posterior or uh expectation
0:09:47	then you have another possibility
0:09:49	you can just draw a random hard labels according to the little of are no an X
0:09:54	and then replace
0:09:56	the egg
0:09:56	dictation
0:09:57	well uh uh uh just a some a seven divided by a K which is number of real as it
0:10:03	relies H
0:10:05	so
0:10:05	if you have a look at i see it has quite a we uh requirements
0:10:11	you need a complete data estimator here just an ica algorithm
0:10:15	will we use the job
0:10:17	and you need to be able to you to calculate the posterior expectation or
0:10:21	much weaker
0:10:22	to draw are according to
0:10:24	the new of or no
0:10:27	and here in the model i
0:10:28	but
0:10:29	i propose the the problem is to calculate the probably of are and X
0:10:34	well if our "'cause" i I D it's just a even by the base room she's uh written here
0:10:39	if or is a mark of chain you can also do it you just uh a need to uh two
0:10:46	resort to uh
0:10:47	a one in forward backward algorithm which are is you to copy
0:10:51	the probably of are no
0:10:54	okay
0:10:56	so let me one you have uh a a a a a a problem that you may encounter a here
0:11:00	is it's
0:11:01	the distribution of the sources
0:11:03	however
0:11:04	you all do that i has some be you to this so
0:11:09	when you perform ica
0:11:11	uh at a given my yeah i
0:11:13	E
0:11:14	a iteration you don't new
0:11:16	which
0:11:17	uh
0:11:17	permutation mutation you obtain
0:11:19	so the as the uh inverse metrics you found correspond to
0:11:23	left or through the right distribution
0:11:25	so to you have all this problem
0:11:27	we decided it
0:11:28	to
0:11:29	consider for the ice
0:11:31	E algorithm that
0:11:33	the uh
0:11:34	distribution
0:11:35	off uh the sources look
0:11:38	like this this is
0:11:39	uh invariant
0:11:40	by us a permutation and it avoids any problem of course this is not true distribution that true distribution is
0:11:48	uh at the top and
0:11:50	for a calculating
0:11:52	the little of are no X with used to consider
0:11:55	that this is this to this to
0:11:59	so um
0:12:00	just you well i just fruit it's more uh
0:12:03	more and more mathematical terms
0:12:05	you have a uh the joint density of are are S ask just given by a yeah
0:12:10	top equation
0:12:12	and the simulated data is uh as a written in the
0:12:16	uh column on the
0:12:17	the middle and the assume distribution for i C E is a written on the right to you see you
0:12:23	have a
0:12:24	uh
0:12:25	well and is a normal do L is it had not that's a double exponential low
0:12:30	and uh we just so much was it uh
0:12:32	in
0:12:33	the case for
0:12:34	the assume distribution
0:12:38	okay so no columns
0:12:40	my combine i C a a i C E estimation algorithm
0:12:44	so you first initialize the power
0:12:49	uh i zero and T zero
0:12:51	and then you achieve the the
0:12:53	the following steps
0:12:54	you first calculate the put the posterior of are new X
0:12:59	having this do
0:13:00	you can draw far tables
0:13:02	are
0:13:03	uh
0:13:04	according to the post your a little of are X
0:13:08	there there uh whether the current value of the parameter
0:13:10	then
0:13:11	you select the instant
0:13:13	times where R is equal to zero
0:13:16	and you perform an ica algorithm
0:13:19	when
0:13:20	considering only if these samples where R is equal to zero and you just ignore the part
0:13:25	and you're also
0:13:26	a
0:13:27	the power to P which is the probability that
0:13:30	uh
0:13:31	what that uh the sources are are or or not
0:13:34	a condition
0:13:36	and it work depend
0:13:38	um in the case of a markov chain is it's uh almost of the same thing you just can have
0:13:43	a more complicated step in order to do
0:13:45	a power just but it's
0:13:47	it's basically the same
0:13:49	and you can use
0:13:50	any uh favourite algorithm in our simulations we could use J but the other
0:13:55	uh the all those you work well
0:13:58	so here here yeah a a a few simulations
0:14:02	uh you have here
0:14:04	the average mean square error on the sources depending on P so the probability of that uh the sources are
0:14:11	independent depends P E one is the case where
0:14:14	uh the sources are effectively independent
0:14:17	the assumptions of i C holes
0:14:19	so we have green
0:14:20	the case of complete got a that is are is known
0:14:24	i perform just a a a a a a a or ica on uh on the
0:14:28	restricting a a set consisting of on a uh a a of the
0:14:32	of the samples where independence whole so it's a a lower bound of the mse that uh you can tame
0:14:38	you have a in but to the case were just ignore
0:14:42	and depends to make as if the sources were
0:14:45	independent although they are not just apply J
0:14:48	and you can see that
0:14:49	a a one P decreases is that is one that's also become no longer depend
0:14:54	of course J
0:14:55	phase to uh to separate the source
0:14:58	and in read red
0:15:00	you have a uh the results given by my i agree with them
0:15:04	which means uh it's is it in complete data estimation
0:15:08	and you can see that we are somehow uh we have um
0:15:12	quite interesting performance
0:15:14	in the case where P is greater than say
0:15:17	do you point for two point five
0:15:19	depends what you
0:15:20	expel
0:15:22	here you have basically the same a results
0:15:25	the only thing you can use as as that of the results uh remain true one um
0:15:30	whatever whatever the sample size maybe
0:15:33	and finally a lets me uh out that a you can uh use the same at that
0:15:38	in the case where R is a markov chain
0:15:41	and in this case
0:15:43	you can see that if you uh take into account the fact that are as of of close you are
0:15:48	able to improve
0:15:49	the performance
0:15:50	which is you here we uh we we generated a lack the is going to a of change
0:15:56	and a separate the sources either ignoring the smog uh property or
0:16:01	uh using it and we obtain better performance with uh
0:16:06	using the mark of prob
0:16:08	so a conclusion confusion proposed in your model which consist of a linear mixture of source
0:16:13	the sources are dependent
0:16:16	and
0:16:17	uh there exists a lack and were hidden process which controls the depends what the dependence of the source
0:16:24	and the separation methods method
0:16:25	but was
0:16:26	uh
0:16:28	relies and i two different uh
0:16:31	to different methods of first i C A which
0:16:34	uh is applied to the independent part sources distribution
0:16:38	and i C E
0:16:39	which estimates
0:16:40	uh the power meter or uh the meters
0:16:43	and are incomplete data
0:16:46	with into that
0:16:47	i think you're for your attention
0:17:05	you have a a a case of for for more
0:17:07	what a very good question i have no application of that's time but i'm interested in anyone having me if
0:17:13	there is an interesting application
0:17:16	and if uh if i understand well a
0:17:19	do you want to estimate a
0:17:21	just an independent sense
0:17:23	oh okay
0:17:24	when you want to estimate the mixture
0:17:27	i i by some
0:17:29	at the and you only used it depend
0:17:33	you do you find a
0:17:34	yes i i i i i uh i fine
0:17:37	the in sometimes where in depends holes
0:17:40	and i ignore the instance
0:17:42	the samples where in depends does not hold
0:17:45	a a a a you still mix
0:17:47	a the other
0:17:49	i i i i set that to a when do you have no it mean penn this
0:17:53	is to have to so six that a mixture
0:17:56	i i
0:17:59	uh i don't know how i don't understand you your questions so okay you have to to under lighting process
0:18:04	yes
0:18:05	and i think uh a in and one you have independence of sources
0:18:08	and the know that you don't have to depend S
0:18:10	but you still have the mixture
0:18:12	in impulse case
0:18:14	is i i i didn't understand
0:18:17	i i i i have just one product one vector process
0:18:20	okay K as one of the of T
0:18:22	instant i a different and sent either they are in and or not and i mix them what the the
0:18:27	same matter
0:18:30	really the in the case where with and then
0:18:32	yes i makes them
0:18:34	i have a S one of T S of T
0:18:36	and i mix them
0:18:37	whether they are dependent or not
0:18:39	okay yeah i disk one than uh
0:18:42	a lot if you could be possible to to all something for my something that case
0:18:46	where the the the they are the because
0:18:49	you have a also the mixture in that case so maybe some
0:18:52	this information
0:18:54	uh a could be a not ultimately yes of course is you are some so yes
0:18:58	uh i i i i agree if you know more about the distribution then you can
0:19:04	probably can perform better i'd just ignore a part of the information yes that's

AN EXTENSION OF THE ICA MODEL USING LATENT VARIABLES

Source Separation and Applications

Přednášející: Marc Castella, Autoři: Selwa Rafi, Marc Castella, Wojciech Pieczynski, Institut Telecom / Telecom SudParis, France