It is seen from Chapter 3 that the probability of block error with ML decoding is upper-bounded by the Gallager bound P.(elm)s∑Pv(ylkn)】 0,p20 (4.62) Let(y)be an arbitrary probability measure of y (which may depend on thex).Then from (4.62)we have F(elm)s()(广EIs2 0,p心0(4.63) 4w「N(yXw） By invoking the Jensen's inequality (the expectatin is taken with respect to(y)),(4.63) becomes Pa(elm)s (4.64) B(ylx) For the binary-input AWGN channels,we select (y)=().Applying(4.64)to every subcode Cyields Pds{Σi)p,)p.w (4.65) wherex equals to"1"or 0"for d and N-d indexes of n.respectively,and p(y)=Prx(Y=y=x)denotes the conditional probability density function (pdf) given by 1 (y-E,)2】 0naon0） 1 The right-hand side(R.H.S.)of(4.65)can be written as (4.66 4444-44 It is seen from Chapter 3 that the probability of block error with ML decoding is upper-bounded by the Gallager bound ' ' (| ) (| ) (| ) (| ) N m B Nm m m N m P P em P P                     y y x y x y x , >0, 0 (4.62) Let ( ) m  N y be an arbitrary probability measure of y (which may depend on the xm). Then from (4.62) we have 1 1 ' ' (| ) (| ) () () ( | ) (| ) m m N m B N N Nm m m N m P P em P P                            y y x y y yx y x , >0, 0 (4.63) By invoking the Jensen’s inequality (the expectatin is taken with respect to ( ) m  N y ), (4.63) becomes 1 1 1 ' ' (| ) (| ) ( | ) () (| ) m N m B N mN m m N m P P em P P                           y y x yx y y x (4.64) For the binary-input AWGN channels, we select 1 () ( ) N N n n   y  y   . Applying (4.64) to every subcode d yields 1 1 1 |0 0 1 ( ) . ( ) ( ) ( ) n d N e n n xn n Pd y py p y d                               x x 0 y  (4.65) where xn equals to “1” or “0” for d and N-d indexes of n, respectively, and | () ( | ) nx n YX n n p y p Y yX x   denotes the conditional probability density function (pdf) given by 2 0 2 1 ( ) ( ) exp 2 2 s y E p y           , 2 1 2 1 ( ) ( ) exp 2 2 s y E p y           . The right-hand side (R.H.S.) of (4.65) can be written as 11 1 1 1 1 01 0 () () () () () d Nd A y p y p y dy y p y dy d                                       (4.66)