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81∫ ∞−∞|g i (Y i )|f(Y i )dY i =∫ −YA−∞|g i (Y i )|f(Y i )dY i +∫ −∞Y A|g i (Y i )|f(Y i )dY i + C (5.25)Now note that g i (Y i ) is the log of a fraction; making use of equations 5.23 and5.24, the inequality of equation 5.26 will hold when Y i > Y A or Y i < −Y A .K∑K|g i (Y i )| = | log( f(Y i |X i = j, θ 1 )p 1 j ) − log( ∑ Tf(Y i |X i = j, θ 2 )p 2 j )|≤| log(j=1K∑j=1j=1f(Y i |X i = j, θ 1 )p 1 ∑K Tj)| + | log(j=1f(Y i |X i = j, θ 2 )p 2 j )|≤ | log(f(Y i |X i = S 1 , θ 1 ))| + | log(f(Y i |X i = S 2 , θ 2 ))| (5.26)Combining the inequality of equation 5.26 with equation 5.25, we obtain equation5.27, in which C 1 and C 2 are irrelevant constants.≤≤∫ ∞−∞∫ −YA−∞∫ ∞+∫ ∞−∞|g i (Y i )|f(Y i )dY i(| log(f(Y i |X i = S 1 , θ 1 ))| + | log(f(Y i |X i = S 2 , θ 2 ))|)f(Y i )dY iY A(| log(f(Y i |X i = S 1 , θ 1 ))| + | log(f(Y i |X i = S 2 , θ 2 ))|)f(Y i )dY i + C 1(| log(f(Y i |X i = S 1 , θ 1 ))| + | log(f(Y i |X i = S 2 , θ 2 ))|)f(Y i )dY i + C(5.27)2At this point, showing that g i ∈ L 1 reduces to showing that the inequality inequation 5.28 holds.∫ ∞−∞(| log(f(Y i |X i = S 1 , θ 1 ))| + | log(f(Y i |X i = S 2 , θ 2 ))|)f(Y i )dY i < ∞ (5.28)The log terms in the integral can be simplified further, where C 1 , C 2 , and C 3are irrelevant constants and S is either S 1 or S 2 .