1、2.2、Entropy of functions. Let be a random variable taking on a finite Xnumber of values. What is the (general) inequality relationship of HXand if HY(a) ?2X(b) ?cosSolution: Let . Then ygx.:()xygppConsider any set of s that map onto a single . For this set xy,:() :()loglo()log()xygxygp pySince is a
2、monotone increasing function and . lo :()()xygExtending this argument to the entire range of (and ), we obtain XY:()()log()lo()x yxgHXppx,()l()yyHYwith equality iff if one-to-one with probability one.g(a) is one-to-one and hence the entropy, which is just a function of 2XYthe probabilities does not
3、change, i.e., .()HXY(b) is not necessarily one-to-one. Hence all that we can say is costhat , which equality if cosine is one-to-one on the range ()HXYof .2.16. Example of joint entropy. Let be given by (,)pxyXY0 10 1/3 1/31 0 1/3Find(a) , .()HXY(b) , .|(|)(c) ()(d) .|HYX(e) (;)I(f) Draw a Venn diag
4、ram for the quantities in (a) through (e).Solution: H ( X | Y )H ( X ) H ( Y )H ( Y | X )I ( X ; Y )H ( X ； Y )Fig. 1 Venn diagram(a) .231()logl0.98bits=()HX(b) (2|(|)|067 bits/)3YXYHYX)( ),)(|)(pxy|(,)(H(c) 1,3log.58 bitsHXY(d) ()|)02(e) ;(|).1 itsIYX(f) See Figure 1.2.29 Inequalities. Let , and be
5、 joint random variables. Prove the XYZfollowing inequalities and find conditions for equality.(a) )|()|,(ZHYX(b) ;II(c) )(,(),(),( XH(d) ;|;|; ZIYIXZIYXI Solution:(a) Using the chain rule for conditional entropy, )|(),|()|()|,( ZXHYZXHYWith equality iff ,that is, when is a function of and 0,|.Z(b)Us
6、ing the chain rule for mutual information,);()|;();();,( ZXIYIZXIYI With equality iff , that is, when and are conditionally 0|independent given .(c) Using first the chain rule for entropy and then definition of conditional mutual information, )|;()|(),|(),(),( XZYIHYXZHZYX,|With equality iff , that
7、is, when and are conditionally 0)|;(Iindependent given .X(d)Using the chain rule for mutual information, );()|;();,();()|;( ZXIYZIXIYZIXI 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
8、 transition matrix 1010 pP(b) What values of , maximize the rate of part (a)?01p(c) Find the entropy rate of the two-state Markov chain with transition matrix 0 1pP(d)Find the maximum value of the entropy rate of the Markov chain of part (c). We expect that the maximizing value of should be less tha
9、np, since the 0 state permits more information to be generated than 2/1the 1 state.Solution:(a)The stationary distribution is easily calculated.101010,ppTherefore the entropy rate is 10101012 )()()()()|( pHpHXH(b)The entropy rate is at most 1 bit because the process has only two states. This rate ca
10、n be achieved if( and only if) , in 2/101pwhich case the process is actually i.i.d. with .2/1)Pr()0r(iiX(c)As a special case of the general two-state Markov chain, the entropy rate is .1)()()|(1012 pHpXH(d)By straightforward calculus, we find that the maximum value of )(Hof part (c) occurs for . The
11、 maximum value is3820/)53(p(wrong!)bits 694.0)215()1()Hp5.4 Huffman coding. Consider the random variable 0.2 3.4 0.12 649.076531 xxxX(a) Find a binary Huffman code for .X(b)Find the expected codelength for this encoding.(c) Find a ternary Huffman code for .Solution:(a) The Huffman tree for this dist
12、ribution is(b)The expected length of the codewords for the binary Huffman code is 2.02 bits.( )()(iplXE(c) The ternary Huffman tree is 5.9 Optimal code lengths that require one bit above entropy. The source coding theorem shows that the optimal code for a random variable has Xan expected length less
13、 than . Given an example of a random 1)(XHvariable for which the expected length of the optimal code is close to , i.e., for any , construct a distribution for which the optimal 1)(XH0code has .1)(LSolution: there is a trivial example that requires almost 1 bit above its entropy. Let be a binary ran
14、dom variable with probability of X 1Xclose to 1. Then entropy of is close to 0, but the length of its optimal Xcode 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 which takes on values with Xmm,21probabil
15、ities . Assume that the probabilities are ordered so that mp,21. Define , the sum of the probabilities of all p21 1ikFsymbols less than . Then the codeword for is the number i i 1,0iFrounded off to bits, where .il iipl1og(a) Show that the code constructed by this process is prefix-free and the avera
16、ge length satisfies .1)()(XHL(b) Construct the code for the probability distribution (0.5, 0.25, 0.125, 0.125).Solution:(a) Since , we have iipl1og1log1liipWhich implies that .)()(XHlLXHiBy the choice of , we have . Thus , differs from il )1(2ii lil jFiby at least , and will therefore differ from is
17、 at least one place jFil2 iin the first bits of the binary expansion of . Thus the codeword for il i, , which has length , differs from the codeword for at jijliFleast once in the first places. Thus no codeword is a prefix of any ilother codeword.(b)We build the following tableSymbol Probability in
18、iFdecimalin ibinaryilCodeword1 0.5 0.0 0.0 1 02 0.25 0.5 0.10 2 103 0.125 0.75 0.110 3 1104 0.125 0.875 0.111 3 1113.5 AEP. Let be independent identically distributed random ,21Xvariables drawn according to the probability mass function. Thus . We know that mxp,21),(niixpxp121)(),(in probability. Le
19、t , )(),(log21XHXnn niixqq121)(),(where q is another probability mass function on .m,(a) Evaluate , where are i.i.d. .),(logim21nqn21X)(xpSolution: Since the are i.i.d., so are nX21, ,， ，and hence we can apply the strong law of large )(1Xq2)(nqnumbers to obtain )(log1lim),(log1im21 inXqXn.(pwE)(lx)(
20、log)(logxpqp|(HD8.1 Preprocessing the output. One is given a communication channel with transition probabilities and channel capacity . A )|(xyp );(max)YXICphelpful statistician preprocesses the output by forming . He _gclaims that this will strictly improve the capacity.(a) Show that he is wrong.(b
21、) Under what condition does he not strictly decrease the capacity?Solution:(a) The statistician calculates . Since forms a )(_Yg_YXMarkov chain, we can apply the data processing inequality. Hence for every distribution on ,x.);();(_YXIILet be the distribution on that maximizes . Then )(_xpx);(_YXI._
22、)()(_)()( max;(;ma _ CIYXIIC pxpxpxp 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 , we have forming a Markov chain.);(_YXI YX_8.3 An addition noise channel. Find the channel capacity of the following discrete memoryless channel:
