The Shannon information of filtrations and the additional logarithmic utility of insiders

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A Mathematical Theory of Communication

Reprinted with corrections from The Bell System Technical Journal,V ol.27,pp.379–423,623–656,July,October,1948.A Mathematical Theory of CommunicationBy C.E.SHANNONI NTRODUCTIONHE recent development of various methods of modulation such as PCM and PPM which exchangebandwidth for signal-to-noise ratio has intensiﬁed the interest in a general theory of communication.A basis for such a theory is contained in the important papers of Nyquist1and Hartley2on this subject.In the present paper we will extend the theory to include a number of new factors,in particular the effect of noise in the channel,and the savings possible due to the statistical structure of the original message and due to the nature of theﬁnal destination of the information.The fundamental problem of communication is that of reproducing at one point either exactly or ap-proximately a message selected at another point.Frequently the messages have meaning;that is they refer to or are correlated according to some system with certain physical or conceptual entities.These semantic aspects of communication are irrelevant to the engineering problem.The signiﬁcant aspect is that the actual message is one selected from a set of possible messages.The system must be designed to operate for each possible selection,not just the one which will actually be chosen since this is unknown at the time of design.If the number of messages in the set isﬁnite then this number or any monotonic function of this number can be regarded as a measure of the information produced when one message is chosen from the set,all choices being equally likely.As was pointed out by Hartley the most natural choice is the logarithmic function.Although this deﬁnition must be generalized considerably when we consider the inﬂuence of the statistics of the message and when we have a continuous range of messages,we will in all cases use an essentially logarithmic measure.The logarithmic measure is more convenient for various reasons:1.It is practically more useful.Parameters of engineering importance such as time,bandwidth,numberof relays,etc.,tend to vary linearly with the logarithm of the number of possibilities.For example,adding one relay to a group doubles the number of possible states of the relays.It adds1to the base2logarithm of this number.Doubling the time roughly squares the number of possible messages,ordoubles the logarithm,etc.2.It is nearer to our intuitive feeling as to the proper measure.This is closely related to(1)since we in-tuitively measures entities by linear comparison with common standards.One feels,for example,thattwo punched cards should have twice the capacity of one for information storage,and two identicalchannels twice the capacity of one for transmitting information.3.It is mathematically more suitable.Many of the limiting operations are simple in terms of the loga-rithm but would require clumsy restatement in terms of the number of possibilities.The choice of a logarithmic base corresponds to the choice of a unit for measuring information.If the base2is used the resulting units may be called binary digits,or more brieﬂy bits,a word suggested by J.W.Tukey.A device with two stable positions,such as a relay or aﬂip-ﬂop circuit,can store one bit of information.N such devices can store N bits,since the total number of possible states is2N and log22N N.If the base10is used the units may be called decimal digits.Sincelog2M log10M log102332log10M1Nyquist,H.,“Certain Factors Affecting Telegraph Speed,”Bell System Technical Journal,April1924,p.324;“Certain Topics in Telegraph Transmission Theory,”A.I.E.E.Trans.,v.47,April1928,p.617.2Hartley,R.V.L.,“Transmission of Information,”Bell System Technical Journal,July1928,p.535.INFORMATIONSOURCE TRANSMITTER RECEIVER DESTINATIONNOISESOURCEFig.1—Schematic diagram of a general communication system.a decimal digit is about31physical counterparts.We may roughly classify communication systems into three main categories:discrete, continuous and mixed.By a discrete system we will mean one in which both the message and the signal are a sequence of discrete symbols.A typical case is telegraphy where the message is a sequence of lettersand the signal a sequence of dots,dashes and spaces.A continuous system is one in which the message and signal are both treated as continuous functions,e.g.,radio or television.A mixed system is one in which both discrete and continuous variables appear,e.g.,PCM transmission of speech.Weﬁrst consider the discrete case.This case has applications not only in communication theory,but also in the theory of computing machines,the design of telephone exchanges and otherﬁelds.In additionthe discrete case forms a foundation for the continuous and mixed cases which will be treated in the second half of the paper.PART I:DISCRETE NOISELESS SYSTEMS1.T HE D ISCRETE N OISELESS C HANNELTeletype and telegraphy are two simple examples of a discrete channel for transmitting information.Gen-erally,a discrete channel will mean a system whereby a sequence of choices from aﬁnite set of elementarysymbols S1S n can be transmitted from one point to another.Each of the symbols S i is assumed to have a certain duration in time t i seconds(not necessarily the same for different S i,for example the dots anddashes in telegraphy).It is not required that all possible sequences of the S i be capable of transmission on the system;certain sequences only may be allowed.These will be possible signals for the channel.Thus in telegraphy suppose the symbols are:(1)A dot,consisting of line closure for a unit of time and then line open for a unit of time;(2)A dash,consisting of three time units of closure and one unit open;(3)A letter space consisting of,say,three units of line open;(4)A word space of six units of line open.We might place the restriction on allowable sequences that no spaces follow each other(for if two letter spaces are adjacent, it is identical with a word space).The question we now consider is how one can measure the capacity of such a channel to transmit information.In the teletype case where all symbols are of the same duration,and any sequence of the32symbols is allowed the answer is easy.Each symbol representsﬁve bits of information.If the system transmits n symbols per second it is natural to say that the channel has a capacity of5n bits per second.This does not mean that the teletype channel will always be transmitting information at this rate—this is the maximum possible rate and whether or not the actual rate reaches this maximum depends on the source of information which feeds the channel,as will appear later.In the more general case with different lengths of symbols and constraints on the allowed sequences,we make the following deﬁnition:Deﬁnition:The capacity C of a discrete channel is given bylog N TC LimT∞and thereforeC log X0In case there are restrictions on allowed sequences we may still often obtain a difference equation of this type andﬁnd C from the characteristic equation.In the telegraphy case mentioned aboveN t N t2N t4N t5N t7N t8N t10as we see by counting sequences of symbols according to the last or next to the last symbol occurring. Hence C is log0where0is the positive root of12457810.Solving this weﬁnd C0539.A very general type of restriction which may be placed on allowed sequences is the following:We imagine a number of possible states a1a2a m.For each state only certain symbols from the set S1S n can be transmitted(different subsets for the different states).When one of these has been transmitted the state changes to a new state depending both on the old state and the particular symbol transmitted.The telegraph case is a simple example of this.There are two states depending on whether or not a space was the last symbol transmitted.If so,then only a dot or a dash can be sent next and the state always changes. If not,any symbol can be transmitted and the state changes if a space is sent,otherwise it remains the same. The conditions can be indicated in a linear graph as shown in Fig.2.The junction points correspond totheFig.2—Graphical representation of the constraints on telegraph symbols.states and the lines indicate the symbols possible in a state and the resulting state.In Appendix1it is shown that if the conditions on allowed sequences can be described in this form C will exist and can be calculated in accordance with the following result:Theorem1:Let b s i j be the duration of the s th symbol which is allowable in state i and leads to state j. Then the channel capacity C is equal to log W where W is the largest real root of the determinant equation:∑s W bsi j i j0where i j1if i j and is zero otherwise.For example,in the telegraph case(Fig.2)the determinant is:1W2W4W3W6W2W410On expansion this leads to the equation given above for this case.2.T HE D ISCRETE S OURCE OF I NFORMATIONWe have seen that under very general conditions the logarithm of the number of possible signals in a discrete channel increases linearly with time.The capacity to transmit information can be speciﬁed by giving this rate of increase,the number of bits per second required to specify the particular signal used.We now consider the information source.How is an information source to be described mathematically, and how much information in bits per second is produced in a given source?The main point at issue is the effect of statistical knowledge about the source in reducing the required capacity of the channel,by the useof proper encoding of the information.In telegraphy,for example,the messages to be transmitted consist of sequences of letters.These sequences,however,are not completely random.In general,they form sentences and have the statistical structure of,say,English.The letter E occurs more frequently than Q,the sequence TH more frequently than XP,etc.The existence of this structure allows one to make a saving in time(or channel capacity)by properly encoding the message sequences into signal sequences.This is already done to a limited extent in telegraphy by using the shortest channel symbol,a dot,for the most common English letter E;while the infrequent letters,Q,X,Z are represented by longer sequences of dots and dashes.This idea is carried still further in certain commercial codes where common words and phrases are represented by four-orﬁve-letter code groups with a considerable saving in average time.The standardized greeting and anniversary telegrams now in use extend this to the point of encoding a sentence or two into a relatively short sequence of numbers.We can think of a discrete source as generating the message,symbol by symbol.It will choose succes-sive symbols according to certain probabilities depending,in general,on preceding choices as well as the particular symbols in question.A physical system,or a mathematical model of a system which produces such a sequence of symbols governed by a set of probabilities,is known as a stochastic process.3We may consider a discrete source,therefore,to be represented by a stochastic process.Conversely,any stochastic process which produces a discrete sequence of symbols chosen from aﬁnite set may be considered a discrete source.This will include such cases as:1.Natural written languages such as English,German,Chinese.2.Continuous information sources that have been rendered discrete by some quantizing process.Forexample,the quantized speech from a PCM transmitter,or a quantized television signal.3.Mathematical cases where we merely deﬁne abstractly a stochastic process which generates a se-quence of symbols.The following are examples of this last type of source.(A)Suppose we haveﬁve letters A,B,C,D,E which are chosen each with probability.2,successivechoices being independent.This would lead to a sequence of which the following is a typicalexample.B DC B C E C C C AD C B D D A AE C E E AA B B D A E E C A C E E B A E E C B C E A D.This was constructed with the use of a table of random numbers.4(B)Using the sameﬁve letters let the probabilities be.4,.1,.2,.2,.1,respectively,with successivechoices independent.A typical message from this source is then:A A A C D CB DC E A AD A D A CE D AE A D C A B E D A D D C E C A A A A A D.(C)A more complicated structure is obtained if successive symbols are not chosen independentlybut their probabilities depend on preceding letters.In the simplest case of this type a choicedepends only on the preceding letter and not on ones before that.The statistical structure canthen be described by a set of transition probabilities p i j,the probability that letter i is followedby letter j.The indices i and j range over all the possible symbols.A second equivalent way ofspecifying the structure is to give the“digram”probabilities p i j,i.e.,the relative frequency ofthe digram i j.The letter frequencies p i,(the probability of letter i),the transition probabilities 3See,for example,S.Chandrasekhar,“Stochastic Problems in Physics and Astronomy,”Reviews of Modern Physics,v.15,No.1, January1943,p.1.4Kendall and Smith,Tables of Random Sampling Numbers,Cambridge,1939.p i j and the digram probabilities p i j are related by the following formulas:p i ∑j p i j∑j p j i ∑j p j p j ip i j p i p i j∑j p ij ∑i p i ∑i jp i j 1As a speciﬁc example suppose there are three letters A,B,C with the probability tables:p i j A B C 045i B 21151p i jA1518270C 274135A typical message from this source is the following:A B B A B A B A B A B A B A B B B A B B B B B A B A B A B A B A B B B A C A C A BB A B B B B A B B A B AC B B B A B A.The next increase in complexity would involve trigram frequencies but no more.The choice ofa letter would depend on the preceding two letters but not on the message before that point.A set of trigram frequencies p i j k or equivalently a set of transition probabilities p i j k would be required.Continuing in this way one obtains successively more complicated stochastic pro-cesses.In the general n -gram case a set of n -gram probabilities p i 1i 2i n or of transition probabilities p i 1i 2i n 1i n is required to specify the statistical structure.(D)Stochastic processes can also be deﬁned which produce a text consisting of a sequence of“words.”Suppose there are ﬁve letters A,B,C,D,E and 16“words”in the language with associated probabilities:.10A.16BEBE .11CABED .04DEB .04ADEB.04BED .05CEED .15DEED .05ADEE.02BEED .08DAB .01EAB .01BADD .05CA .04DAD .05EESuppose successive “words”are chosen independently and are separated by a space.A typical message might be:DAB EE A BEBE DEED DEB ADEE ADEE EE DEB BEBE BEBE BEBE ADEE BED DEED DEED CEED ADEE A DEED DEED BEBE CABED BEBE BED DAB DEED ADEB.If all the words are of ﬁnite length this process is equivalent to one of the preceding type,but the description may be simpler in terms of the word structure and probabilities.We may also generalize here and introduce transition probabilities between words,etc.These artiﬁcial languages are useful in constructing simple problems and examples to illustrate vari-ous possibilities.We can also approximate to a natural language by means of a series of simple artiﬁcial languages.The zero-order approximation is obtained by choosing all letters with the same probability and independently.The ﬁrst-order approximation is obtained by choosing successive letters independently but each letter having the same probability that it has in the natural language.5Thus,in the ﬁrst-order ap-proximation to English,E is chosen with probability .12(its frequency in normal English)and W with probability .02,but there is no inﬂuence between adjacent letters and no tendency to form the preferred5Letter,digram and trigram frequencies are given in Secret and Urgent by Fletcher Pratt,Blue Ribbon Books,1939.Word frequen-cies are tabulated in Relative Frequency of English Speech Sounds,G.Dewey,Harvard University Press,1923.digrams such as TH,ED,etc.In the second-order approximation,digram structure is introduced.After aletter is chosen,the next one is chosen in accordance with the frequencies with which the various lettersfollow theﬁrst one.This requires a table of digram frequencies p i j.In the third-order approximation, trigram structure is introduced.Each letter is chosen with probabilities which depend on the preceding twoletters.3.T HE S ERIES OF A PPROXIMATIONS TO E NGLISHTo give a visual idea of how this series of processes approaches a language,typical sequences in the approx-imations to English have been constructed and are given below.In all cases we have assumed a27-symbol “alphabet,”the26letters and a space.1.Zero-order approximation(symbols independent and equiprobable).XFOML RXKHRJFFJUJ ZLPWCFWKCYJ FFJEYVKCQSGHYD QPAAMKBZAACIBZL-HJQD.2.First-order approximation(symbols independent but with frequencies of English text).OCRO HLI RGWR NMIELWIS EU LL NBNESEBY A TH EEI ALHENHTTPA OOBTTV ANAH BRL.3.Second-order approximation(digram structure as in English).ON IE ANTSOUTINYS ARE T INCTORE ST BE S DEAMY ACHIN D ILONASIVE TU-COOWE AT TEASONARE FUSO TIZIN ANDY TOBE SEACE CTISBE.4.Third-order approximation(trigram structure as in English).IN NO IST LAT WHEY CRATICT FROURE BIRS GROCID PONDENOME OF DEMONS-TURES OF THE REPTAGIN IS REGOACTIONA OF CRE.5.First-order word approximation.Rather than continue with tetragram,,n-gram structure it is easierand better to jump at this point to word units.Here words are chosen independently but with their appropriate frequencies.REPRESENTING AND SPEEDILY IS AN GOOD APT OR COME CAN DIFFERENT NAT-URAL HERE HE THE A IN CAME THE TO OF TO EXPERT GRAY COME TO FURNISHESTHE LINE MESSAGE HAD BE THESE.6.Second-order word approximation.The word transition probabilities are correct but no further struc-ture is included.THE HEAD AND IN FRONTAL ATTACK ON AN ENGLISH WRITER THAT THE CHAR-ACTER OF THIS POINT IS THEREFORE ANOTHER METHOD FOR THE LETTERS THATTHE TIME OF WHO EVER TOLD THE PROBLEM FOR AN UNEXPECTED.The resemblance to ordinary English text increases quite noticeably at each of the above steps.Note that these samples have reasonably good structure out to about twice the range that is taken into account in their construction.Thus in(3)the statistical process insures reasonable text for two-letter sequences,but four-letter sequences from the sample can usually beﬁtted into good sentences.In(6)sequences of four or more words can easily be placed in sentences without unusual or strained constructions.The particular sequence of ten words“attack on an English writer that the character of this”is not at all unreasonable.It appears then that a sufﬁciently complex stochastic process will give a satisfactory representation of a discrete source.Theﬁrst two samples were constructed by the use of a book of random numbers in conjunction with(for example2)a table of letter frequencies.This method might have been continued for(3),(4)and(5), since digram,trigram and word frequency tables are available,but a simpler equivalent method was used.To construct(3)for example,one opens a book at random and selects a letter at random on the page.This letter is recorded.The book is then opened to another page and one reads until this letter is encountered. The succeeding letter is then recorded.Turning to another page this second letter is searched for and the succeeding letter recorded,etc.A similar process was used for(4),(5)and(6).It would be interesting if further approximations could be constructed,but the labor involved becomes enormous at the next stage.4.G RAPHICAL R EPRESENTATION OF A M ARKOFF P ROCESSStochastic processes of the type described above are known mathematically as discrete Markoff processes and have been extensively studied in the literature.6The general case can be described as follows:There exist aﬁnite number of possible“states”of a system;S1S2S n.In addition there is a set of transition probabilities;p i j the probability that if the system is in state S i it will next go to state S j.To make thisMarkoff process into an information source we need only assume that a letter is produced for each transition from one state to another.The states will correspond to the“residue of inﬂuence”from preceding letters.The situation can be represented graphically as shown in Figs.3,4and5.The“states”are thejunctionFig.3—A graph corresponding to the source in example B.points in the graph and the probabilities and letters produced for a transition are given beside the correspond-ing line.Figure3is for the example B in Section2,while Fig.4corresponds to the example C.In Fig.3Fig.4—A graph corresponding to the source in example C.there is only one state since successive letters are independent.In Fig.4there are as many states as letters. If a trigram example were constructed there would be at most n2states corresponding to the possible pairs of letters preceding the one being chosen.Figure5is a graph for the case of word structure in example D. Here S corresponds to the“space”symbol.5.E RGODIC AND M IXED S OURCESAs we have indicated above a discrete source for our purposes can be considered to be represented by a Markoff process.Among the possible discrete Markoff processes there is a group with special properties of signiﬁcance in communication theory.This special class consists of the“ergodic”processes and we shall call the corresponding sources ergodic sources.Although a rigorous deﬁnition of an ergodic process is somewhat involved,the general idea is simple.In an ergodic process every sequence produced by the process 6For a detailed treatment see M.Fr´e chet,M´e thode des fonctions arbitraires.Th´e orie des´e v´e nements en chaˆıne dans le cas d’un nombreﬁni d’´e tats possibles.Paris,Gauthier-Villars,1938.is the same in statistical properties.Thus the letter frequencies,digram frequencies,etc.,obtained from particular sequences,will,as the lengths of the sequences increase,approach deﬁnite limits independent of the particular sequence.Actually this is not true of every sequence but the set for which it is false has probability zero.Roughly the ergodic property means statistical homogeneity.All the examples of artiﬁcial languages given above are ergodic.This property is related to the structure of the corresponding graph.If the graph has the following two properties7the corresponding process will be ergodic:1.The graph does not consist of two isolated parts A and B such that it is impossible to go from junctionpoints in part A to junction points in part B along lines of the graph in the direction of arrows and also impossible to go from junctions in part B to junctions in part A.2.A closed series of lines in the graph with all arrows on the lines pointing in the same orientation willbe called a“circuit.”The“length”of a circuit is the number of lines in it.Thus in Fig.5series BEBES is a circuit of length5.The second property required is that the greatest common divisor of the lengths of all circuits in the graph beone.If theﬁrst condition is satisﬁed but the second one violated by having the greatest common divisor equal to d1,the sequences have a certain type of periodic structure.The various sequences fall into d different classes which are statistically the same apart from a shift of the origin(i.e.,which letter in the sequence is called letter1).By a shift of from0up to d1any sequence can be made statistically equivalent to any other.A simple example with d2is the following:There are three possible letters a b c.Letter a isfollowed with either b or c with probabilities13respectively.Either b or c is always followed by lettera.Thus a typical sequence isa b a c a c a c a b a c a b a b a c a cThis type of situation is not of much importance for our work.If theﬁrst condition is violated the graph may be separated into a set of subgraphs each of which satisﬁes theﬁrst condition.We will assume that the second condition is also satisﬁed for each subgraph.We have in this case what may be called a“mixed”source made up of a number of pure components.The components correspond to the various subgraphs.If L1,L2,L3are the component sources we may writeL p1L1p2L2p3L37These are restatements in terms of the graph of conditions given in Fr´e chet.where p i is the probability of the component source L i.Physically the situation represented is this:There are several different sources L1,L2,L3which are each of homogeneous statistical structure(i.e.,they are ergodic).We do not know a priori which is to be used,but once the sequence starts in a given pure component L i,it continues indeﬁnitely according to the statistical structure of that component.As an example one may take two of the processes deﬁned above and assume p12and p28.A sequence from the mixed sourceL2L18L2would be obtained by choosingﬁrst L1or L2with probabilities.2and.8and after this choice generating a sequence from whichever was chosen.Except when the contrary is stated we shall assume a source to be ergodic.This assumption enables one to identify averages along a sequence with averages over the ensemble of possible sequences(the probability of a discrepancy being zero).For example the relative frequency of the letter A in a particular inﬁnite sequence will be,with probability one,equal to its relative frequency in the ensemble of sequences.If P i is the probability of state i and p i j the transition probability to state j,then for the process to be stationary it is clear that the P i must satisfy equilibrium conditions:P j∑iP i p i jIn the ergodic case it can be shown that with any starting conditions the probabilities P j N of being in state j after N symbols,approach the equilibrium values as N∞.6.C HOICE,U NCERTAINTY AND E NTROPYWe have represented a discrete information source as a Markoff process.Can we deﬁne a quantity which will measure,in some sense,how much information is“produced”by such a process,or better,at what rate information is produced?Suppose we have a set of possible events whose probabilities of occurrence are p1p2p n.These probabilities are known but that is all we know concerning which event will occur.Can weﬁnd a measure of how much“choice”is involved in the selection of the event or of how uncertain we are of the outcome?If there is such a measure,say H p1p2p n,it is reasonable to require of it the following properties:1.H should be continuous in the p i.2.If all the p i are equal,p i12,p216.On the right weﬁrst choose between two possibilities each withprobability13,1216H121312is because this second choice only occurs half the time.In Appendix2,the following result is established:Theorem2:The only H satisfying the three above assumptions is of the form:H Kn∑i1p i log p iwhere K is a positive constant.This theorem,and the assumptions required for its proof,are in no way necessary for the present theory.It is given chieﬂy to lend a certain plausibility to some of our later deﬁnitions.The real justiﬁcation of these deﬁnitions,however,will reside in their implications.Quantities of the form H∑p i log p i(the constant K merely amounts to a choice of a unit of measure) play a central role in information theory as measures of information,choice and uncertainty.The form of H will be recognized as that of entropy as deﬁned in certain formulations of statistical mechanics8where p i isthe probability of a system being in cell i of its phase space.H is then,for example,the H in Boltzmann’s famous H theorem.We shall call H∑p i log p i the entropy of the set of probabilities p1p n.If x is a chance variable we will write H x for its entropy;thus x is not an argument of a function but a label for anumber,to differentiate it from H y say,the entropy of the chance variable y.The entropy in the case of two possibilities with probabilities p and q1p,namelyH p log p q log qis plotted in Fig.7as a function of p.HBITSpFig.7—Entropy in the case of two possibilities with probabilities p and1p.The quantity H has a number of interesting properties which further substantiate it as a reasonable measure of choice or information.1.H0if and only if all the p i but one are zero,this one having the value unity.Thus only when weare certain of the outcome does H vanish.Otherwise H is positive.2.For a given n,H is a maximum and equal to log n when all the p i are equal(i.e.,1。

人工智能英文参考文献(最新120个)

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下面是搜索整理的人工智能英文参考文献的分享，供大家借鉴参考。

人工智能英文参考文献一：[1]Lars Egevad,Peter Str?m,Kimmo Kartasalo,Henrik Olsson,Hemamali Samaratunga,Brett Delahunt,Martin Eklund. The utility of artificial intelligence in the assessment of prostate pathology[J]. Histopathology,2020,76(6).[2]Rudy van Belkom. The Impact of Artificial Intelligence on the Activities ofa Futurist[J]. World Futures Review,2020,12(2).[3]Reza Hafezi. How Artificial Intelligence Can Improve Understanding in Challenging Chaotic Environments[J]. World Futures Review,2020,12(2).[4]Alejandro Díaz-Domínguez. How Futures Studies and Foresight Could Address Ethical Dilemmas of Machine Learning and Artificial Intelligence[J]. World Futures Review,2020,12(2).[5]Russell T. Warne,Jared Z. Burton. Beliefs About Human Intelligence in a Sample of Teachers and Nonteachers[J]. Journal for the Education of the Gifted,2020,43(2).[6]Russell Belk,Mariam Humayun,Ahir Gopaldas. Artificial Life[J]. Journal of Macromarketing,2020,40(2).[7]Walter Kehl,Mike Jackson,Alessandro Fergnani. Natural Language Processing and Futures Studies[J]. World Futures Review,2020,12(2).[8]Anne Boysen. Mine the Gap: Augmenting Foresight Methodologies with Data Analytics[J]. World Futures Review,2020,12(2).[9]Marco Bevolo,Filiberto Amati. The Potential Role of AI in Anticipating Futures from a Design Process Perspective: From the Reflexive Description of “Design” to a Discussion of Influences by the Inclusion of AI in the Futures Research Process[J]. World Futures Review,2020,12(2).[10]Lan Xu,Paul Tu,Qian Tang,Dan Seli?teanu. Contract Design for Cloud Logistics (CL) Based on Blockchain Technology (BT)[J]. Complexity,2020,2020.[11]L. Grant,X. Xue,Z. Vajihi,A. Azuelos,S. Rosenthal,D. Hopkins,R. Aroutiunian,B. Unger,A. Guttman,M. Afilalo. LO32: Artificial intelligence to predict disposition to improve flow in the emergency department[J]. CJEM,2020,22(S1).[12]A. Kirubarajan,A. Taher,S. Khan,S. Masood. P071: Artificial intelligence in emergency medicine: A scoping review[J]. 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A New Approach to Linear Filtering and Prediction Problems

(2) A nonlinear difference (or differential) equation is derived for the covariance matrix of the optimal estimation error. From the solution of this equation the coefficients of the difference (or differential) equation of the optimal linear filter are obtained without further calculations.
In all these works, the objective is to obtain the specification of a linear dynamic system (Wiener filter) which accomplishes the prediction, separation, or detection of a random signal.4 ———
In his pioneering work, Wiener [1]3 showed that problems (i) and (ii) lead to the so-called Wiener-Hopf integral equation; he also gave a method (spectral factorization) for the solution of this integral equation in the practically important special case of stationary statistics and rational spectra.

信息论英文课后部分习题答案

本答案是英文原版的配套答案，与翻译的中文版课本题序不太一样但内容一样。

翻译的中文版增加了题量。

2.2、Entropy of functions. Let X be a random variable taking on a finite number of values. What is the (general) inequality relationship of ()H X and ()H Y if(a) 2X Y =?(b) cos Y X =?Solution: Let ()y g x =. Then():()()x y g x p y p x ==∑.Consider any set of x ’s that map onto a single y . For this set()():():()()log ()log ()log ()x y g x x y g x p x p x p x p y p y p y ==≤=∑∑,Since log is a monotone increasing function and ():()()()x y g x p x p x p y =≤=∑.Extending this argument to the entire range of X (and Y ), we obtain():()()log ()()log ()x y x g x H X p x p x p x p x =-=-∑∑∑()log ()()yp y p y H Y ≥-=∑,with equality iff g if one-to-one with probability one.(a) 2X Y = is one-to-one and hence the entropy, which is just a function of the probabilities does not change, i.e., ()()H X H Y =.(b) cos Y X =is not necessarily one-to-one. Hence all that we can say is that ()()H X H Y ≥, which equality if cosine is one-to-one on the range of X .2.16. Example of joint entropy. Let (,)p x y be given byFind(a) ()H X ,()H Y .(b) (|)H X Y ,(|)H Y X . (c) (,)H X Y(d) ()(|)H Y H Y X -.(e) (;)I X Y(f) Draw a Venn diagram for the quantities in (a) through (e).Solution:Fig. 1 Venn diagram(a) 231()log log30.918 bits=()323H X H Y =+=.(b) 12(|)(|0)(|1)0.667 bits (/)33H X Y H X Y H X Y H Y X ==+===((,)(|)()p x y p x y p y =)((|)(,)()H X Y H X Y H Y =-)(c) 1(,)3log3 1.585 bits 3H X Y =⨯=(d) ()(|)0.251 bits H Y H Y X -=(e)(;)()(|)0.251 bits=-=I X Y H Y H Y X(f)See Figure 1.2.29 Inequalities. Let X,Y and Z be joint random variables. Prove the following inequalities and find conditions for equality.(a) )ZHYXH≥X(Z()|,|(b) )ZIYXI≥X((Z);,;(c) )XYXHZ≤Z-H-XYH),(,)(((X,,)H(d) )XYIZIZII+-XZY≥Y(););(|;;(Z|)(XSolution:(a)Using the chain rule for conditional entropy,HZYXHZXH+XH≥XYZ=),(|(Z)(||,()|)With equality iff 0YH,that is, when Y is a function of X andXZ,|(=)Z.(b)Using the chain rule for mutual information,ZIXXIZYX+=,I≥IYZ(|;)X);)(,;;(Z)(With equality iff 0ZYI, that is, when Y and Z areX)|;(=conditionally independent given X.(c)Using first the chain rule for entropy and then definition of conditionalmutual information,XZHYHIXZYX==-H-XHYYZ)()(;Z)|,|),|(X(,,)(XHXZH-Z≤,=,)()()(X|HWith equality iff 0ZYI, that is, when Y and Z areX(=|;)conditionally independent given X .(d) Using the chain rule for mutual information,);()|;();,();()|;(Z X I X Y Z I Z Y X I Y Z I Y Z X I +==+And therefore this inequality is actually an equality in all cases.4.5 Entropy rates of Markov chains.(a) Find the entropy rate of the two-state Markov chain with transition matrix⎥⎦⎤⎢⎣⎡--=1010010111p p p p P (b) What values of 01p ,10p maximize the rate of part (a)?(c) Find the entropy rate of the two-state Markov chain with transition matrix⎥⎦⎤⎢⎣⎡-=0 1 1p p P(d) Find the maximum value of the entropy rate of the Markov chain of part (c). We expect that the maximizing value of p should be less than 2/1, since the 0 state permits more information to be generated than the 1 state.Solution:(a) T he stationary distribution is easily calculated.10010*********,p p p p p p +=+=ππ Therefore the entropy rate is10011001011010101012)()()()()|(p p p H p p H p p H p H X X H ++=+=ππ(b) T he entropy rate is at most 1 bit because the process has only two states. This rate can be achieved if( and only if) 2/11001==p p , in which case the process is actually i.i.d. with2/1)1Pr()0Pr(====i i X X .(c) A s a special case of the general two-state Markov chain, the entropy rate is1)()1()()|(1012+=+=p p H H p H X X H ππ.(d) B y straightforward calculus, we find that the maximum value of)(χH of part (c) occurs for 382.02/)53(=-=p . The maximum value isbits 694.0)215()1()(=-=-=H p H p H (wrong!)5.4 Huffman coding. Consider the random variable⎪⎪⎭⎫ ⎝⎛=0.02 0.03 0.04 0.04 0.12 0.26 49.0 7654321x x x x x x x X (a) Find a binary Huffman code for X .(b) Find the expected codelength for this encoding.(c) Find a ternary Huffman code for X .Solution:(a) The Huffman tree for this distribution is(b)The expected length of the codewords for the binary Huffman code is 2.02 bits.( ∑⨯=)()(i p l X E )(c) The ternary Huffman tree is5.9 Optimal code lengths that require one bit above entropy. The source coding theorem shows that the optimal code for a random variable X has an expected length less than 1)(+X H . Given an example of a random variable for which the expected length of the optimal code is close to 1)(+X H , i.e., for any 0>ε, construct a distribution for which the optimal code has ε-+>1)(X H L .Solution: there is a trivial example that requires almost 1 bit above its entropy. Let X be a binary random variable with probability of 1=X close to 1. Then entropy of X is close to 0, but the length of its optimal code is 1 bit, which is almost 1 bit above its entropy.5.25 Shannon code. Consider the following method for generating a code for a random variable X which takes on m values {}m ,,2,1 with probabilities m p p p ,,21. Assume that the probabilities are ordered so thatm p p p ≥≥≥ 21. Define ∑-==11i k i i p F , the sum of the probabilities of allsymbols less than i . Then the codeword for i is the number ]1,0[∈i Frounded off to i l bits, where ⎥⎥⎤⎢⎢⎡=i i p l 1log . (a) Show that the code constructed by this process is prefix-free and the average length satisfies 1)()(+<≤X H L X H .(b) Construct the code for the probability distribution (0.5, 0.25, 0.125, 0.125).Solution:(a) Since ⎥⎥⎤⎢⎢⎡=i i p l 1log , we have 11log 1log +<≤i i i p l pWhich implies that 1)()(+<=≤∑X H l p L X H i i .By the choice of i l , we have )1(22---<≤ii l i l p . Thus j F , i j > differs from j F by at least il -2, and will therefore differ from i F is at least one place in the first i l bits of the binary expansion of i F . Thus thecodeword for j F , i j >, which has length i j l l ≥, differs from thecodeword for i F at least once in the first i l places. Thus no codewordis a prefix of any other codeword.(b) We build the following table3.5 AEP. Let ,,21X X be independent identically distributed random variables drawn according to theprobability mass function {}m x x p ,2,1),(∈. Thus ∏==n i i n x p x x x p 121)(),,,( . We know that)(),,,(log 121X H X X X p n n →- in probability. Let ∏==n i i n x q x x x q 121)(),,,( , where q is another probability mass function on {}m ,2,1.(a) Evaluate ),,,(log 1lim 21n X X X q n-, where ,,21X X are i.i.d. ~ )(x p . Solution: Since the n X X X ,,,21 are i.i.d., so are )(1X q ,)(2X q ,…，)(n X q ，and hence we can apply the strong law of large numbers to obtain∑-=-)(log 1lim ),,,(log 1lim 21i n X q n X X X q n 1..))((log p w X q E -=∑-=)(log )(x q x p∑∑-=)(log )()()(log )(x p x p x q x p x p )()||(p H q p D +=8.1 Preprocessing the output. One is given a communication channel withtransition probabilities )|(x y p and channel capacity );(max )(Y X I C x p =.A helpful statistician preprocesses the output by forming )(_Y g Y =. He claims that this will strictly improve the capacity.(a) Show that he is wrong.(b) Under what condition does he not strictly decrease the capacity? Solution:(a) The statistician calculates )(_Y g Y =. Since _Y Y X →→ forms a Markov chain, we can apply the data processing inequality. Hence for every distribution on x ,);();(_Y X I Y X I ≥. Let )(_x p be the distribution on x that maximizes );(_Y X I . Then__)()()(_)()()();(max );();();(max __C Y X I Y X I Y X I Y X I C x p x p x p x p x p x p ==≥≥===.Thus, the statistician is wrong and processing the output does not increase capacity.(b) We have equality in the above sequence of inequalities only if we have equality in data processing inequality, i.e., for the distribution that maximizes );(_Y X I , we have Y Y X →→_forming a Markov chain.8.3 An addition noise channel. Find the channel capacity of the following discrete memoryless channel:Where {}{}21Pr 0Pr ====a Z Z . The alphabet for x is {}1,0=X . Assume that Z is independent of X . Observe that the channel capacity depends on the value of a . Solution: A sum channel.Z X Y += {}1,0∈X , {}a Z ,0∈We have to distinguish various cases depending on the values of a .0=a In this case, X Y =,and 1);(max =Y X I . Hence the capacity is 1 bitper transmission.1,0≠≠a In this case, Y has four possible values a a +1,,1,0. KnowingY ,we know the X which was sent, and hence 0)|(=Y X H . Hence thecapacity is also 1 bit per transmission.1=a In this case Y has three possible output values, 0,1,2, the channel isidentical to the binary erasure channel, with 21=f . The capacity of this channel is 211=-f bit per transmission.1-=a This is similar to the case when 1=a and the capacity is also 1/2 bit per transmission.8.5 Channel capacity. Consider the discrete memoryless channel)11 (mod Z X Y +=, where ⎪⎪⎭⎫ ⎝⎛=1/3 1/3, 1/3,3 2,,1Z and {}10,,1,0 ∈X . Assume thatZ is independent of X .(a) Find the capacity.(b) What is the maximizing )(*x p ?Solution: The capacity of the channel is );(max )(Y X I C x p =)()()|()()|()();(Z H Y H X Z H Y H X Y H Y H Y X I -=-=-=bits 311log)(log );(=-≤Z H y Y X I , which is obtained when Y has an uniform distribution, which occurs when X has an uniform distribution.(a)The capacity of the channel is bits 311log /transmission.(b) The capacity is achieved by an uniform distribution on the inputs.10,,1,0for 111)( ===i i X p 8.12 Time-varying channels. Consider a time-varying discrete memoryless channel. Let n Y Y Y ,,21 be conditionally independent givenn X X X ,,21 , with conditional distribution given by ∏==ni i i i x y p x y p 1)|()|(.Let ),,(21n X X X X =, ),,(21n Y Y Y Y =. Find );(max )(Y X I x p . Solution:∑∑∑∑∑=====--≤-≤-=-=-=-=ni i n i i i n i i ni i i ni i i n p h X Y H Y H X Y H Y H X Y Y Y H Y H X Y Y Y H Y H X Y H Y H Y X I 111111121))(1()|()()|()(),,|()()|,,()()|()();(With equlity ifnX X X ,,21 is chosen i.i.d. Hence∑=-=ni i x p p h Y X I 1)())(1();(max .10.2 A channel with two independent looks at Y . Let 1Y and 2Y be conditionally independent and conditionally identically distributed givenX .(a) Show );();(2),;(21121Y Y I Y X I Y Y X I -=. (b) Conclude that the capacity of the channelX(Y1,Y2)is less than twice the capacity of the channelXY1Solution:(a) )|,(),(),;(212121X Y Y H Y Y H Y Y X I -=)|()|();()()(212121X Y H X Y H Y Y I Y H Y H ---+=);();(2);();();(2112121Y Y I Y X I Y Y I Y X I Y X I -=-+=(b) The capacity of the single look channel 1Y X → is );(max 1)(1Y X I C x p =.Thecapacityof the channel ),(21Y Y X →is11)(211)(21)(22);(2max );();(2max ),;(max C Y X I Y Y I Y X I Y Y X I C x p x p x p =≤-==10.3 The two-look Gaussian channel. Consider the ordinary Shannon Gaussian channel with two correlated looks at X , i.e., ),(21Y Y Y =, where2211Z X Y Z X Y +=+= with a power constraint P on X , and ),0(~),(221K N Z Z ,where⎥⎦⎤⎢⎣⎡=N N N N K ρρ. Find the capacity C for (a) 1=ρ (b) 0=ρ (c) 1-=ρSolution:It is clear that the two input distribution that maximizes the capacity is),0(~P N X . Evaluating the mutual information for this distribution,),(),()|,(),()|,(),(),;(max 212121212121212Z Z h Y Y h X Z Z h Y Y h X Y Y h Y Y h Y Y X I C -=-=-==Nowsince⎪⎪⎭⎫⎝⎛⎥⎦⎤⎢⎣⎡N N N N N Z Z ,0~),(21ρρ,wehave)1()2log(21)2log(21),(222221ρππ-==N e Kz e Z Z h.Since11Z X Y +=, and22Z X Y +=, wehave ⎪⎪⎭⎫⎝⎛⎥⎦⎤⎢⎣⎡++++N N P N N P N Y Y P P ,0~),(21ρρ, And ))1(2)1(()2log(21)2log(21),(222221ρρππ-+-==PN N e K e Y Y h Y .Hence⎪⎪⎭⎫⎝⎛++=-=)1(21log 21),(),(21212ρN P Z Z h Y Y h C(a) 1=ρ.In this case, ⎪⎭⎫⎝⎛+=N P C 1log 21, which is the capacity of a single look channel.(b) 0=ρ. In this case, ⎪⎭⎫⎝⎛+=N P C 21log 21, which corresponds to using twice the power in a single look. The capacity is the same as the capacity of the channel )(21Y Y X +→.(c) 1-=ρ. In this case, ∞=C , which is not surprising since if we add1Y and 2Y , we can recover X exactly.10.4 Parallel channels and waterfilling. Consider a pair of parallel Gaussianchannels,i.e.,⎪⎪⎭⎫⎝⎛+⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛212121Z Z X X Y Y , where⎪⎪⎭⎫ ⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛222121 00 ,0~σσN Z Z , And there is a power constraint P X X E 2)(2221≤+. Assume that 2221σσ>. At what power does the channel stop behaving like a single channel with noise variance 22σ, and begin behaving like a pair of channels? Solution: We will put all the signal power into the channel with less noise until the total power of noise+signal in that channel equals the noise power in the other channel. After that, we will split anyadditional power evenly between the two channels. Thus the combined channel begins to behave like a pair of parallel channels when the signal power is equal to the difference of the two noise powers, i.e., when 22212σσ-=P .。

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elements of information theory中文版

Elements of Information Theory（信息论要素）1. 简介《Elements of Information Theory》是由Thomas M. Cover和Joy A. Thomas所著的信息论经典教材。

信息论是一门研究信息传输、压缩和加密等问题的学科，它在通信、数据压缩、密码学等领域具有重要的应用价值。

本书从基本概念出发，深入浅出地介绍了信息论的核心内容，包括熵、条件熵、互信息、信道容量等。

通过学习本书，读者可以全面了解和掌握信息论的基本原理和方法。

2. 内容概述2.1 信息与不确定性•信息的概念：介绍了Shannon如何定义和衡量信息，并引入了熵作为衡量不确定性的度量。

•熵：详细讨论了离散随机变量和连续随机变量的熵，并给出了计算熵的方法。

•条件熵：介绍了条件熵的概念及其计算方法，用于描述在给定某些条件下随机变量的不确定性。

•相对熵：引入相对熵作为度量两个概率分布之间差异程度的指标。

2.2 信息编码与压缩•信息编码：介绍了无失真编码和有损编码的基本原理和方法，包括霍夫曼编码、香农-费诺编码等。

•压缩理论：讨论了数据压缩的基本概念和方法，包括无失真压缩和有损压缩，以及压缩算法的效率。

2.3 信道与信道容量•信道模型：介绍了离散信道和连续信道的基本概念，并给出了常见的几种信道模型。

•信息传输定理：阐述了香农信息传输定理，即在给定一定带宽和信噪比条件下，可靠地传输信息的极限。

•信道容量：定义了信道容量，并介绍了如何计算离散信道和连续信道的容量。

2.4 错误检测与纠正•奇偶校验：介绍了奇偶校验作为一种简单的错误检测方法。

•海明码：详细讨论了海明码作为一种常用的错误检测和纠正方法。

3. 学习收获通过学习《Elements of Information Theory》，读者可以获得以下方面的知识和技能：•了解信息论的基本概念和原理，掌握信息论的核心内容。

•掌握熵、条件熵、互信息等概念及其计算方法，能够衡量信息的不确定性。

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