The best expert versus the smartest algorithm
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1 Theoretical Computer Science The best expert versus the smartest algorithm Peter Chen a, Guoli Ding b; a Department of Computer Science, Louisiana State University, Baton Rouge, LA, 70803, USA b Department of Mathematics, Louisiana State University, Baton Rouge, LA, 70803, USA Abstract In this paper, we consider the problem of online prediction using expert advice. Under dierent assumptions, we give tight lower bounds on the gap between the best expert and any online algorithm that solves the problem. c 004 Elsevier B.V. All rights reserved. Keywords: Online algorithm; Online prediction; Expert advice 1. Introduction The problem of online prediction using expert advice is for a predictor to predict, along with n other experts, a sequence 1 ; ;:::; {0; 1}. Here, the only assumptions we make are the following. Assumption 1.1. Before predicting each j, the predictor knows the predictions of the experts on this term. Also, right after predicting each j, the predictor is given the true value of this term. Notice that we do not assume anything on possible patterns of either or the sequences of the predictions of the experts. The goal of the predictor is to score as close to the best expert as possible. We point out that this is dierent from the goal that tries to predict as accurate as possible, which is a problem studied in [1]. Next, we make the problem more precise. Suppose x x 1 ;x ;:::;x is a sequence of predictions, made by the predictor or by the experts. It is worth mentioning that, Corresponding author. address: gding1@lsu.edu G. Ding /$ - see front matter c 004 Elsevier B.V. All rights reserved. doi: /j.tcs
2 36 P. Chen, G. Ding / TheoreticalComputer Science sometimes, terms of x might be allowed to take any value in the interval [0; 1]. The loss of x is dened to be Lx; x j j : j1 Let i be the sequence of predictions made by expert i and let Then { i :16 i 6 n}: L ; min{l i ;:16 i 6 n} is the loss of the best expert. For any strategy A of the predictor, 1 let A ; bethe sequence of predictions generated according to A. We measure the performance of A by the worst gap between the losses of the predictor and the best expert, i.e., by G A n; sup L A ; ; L ; : ; Clearly, the goal of the predictor is to minimize G A n; over all strategies A. In this paper, we analyze upper and lower bounds of G A n;. To get a lower bound, let us make the following assumption, which is more in favor of the predictor, and thus will make the result stronger: Assumption 1.. The predictions of the predictor can be any real number in the interval [0; 1], while the predictions of the experts can only be 0 or 1. In addition, before predicting 1, the predictor knows not only, but also, the entire prediction sequence of each expert. In other words, other than the actual value of each j, the predictor knows everything else before predicting 1. Under Assumption 1., it is proved in [] that, for all strategies A, lim inf n lim inf G A n; ln n 1: 1.1 We should point out that, because of the order the two limits are taken, it is assumed, implicitly, in the above inequality that is signicantly larger than n. We will see later that the situation is quite dierent if is smaller than n. For upper bounds, let us make a dierent assumption, which is less in favor of the predictor, and thus will make the result stronger. 1 Here we assume that, when predicting two sequences and,if and turn out to be the same, and the predictions of the experts are also the same, then the strategy should generate two identical sequences of predictions. For a strategy that generates two dierent sequences of predictions, due to randomization or similar reasons, we will consider it as several strategies under our term and this change does not aect our discussions.
3 P. Chen, G. Ding / TheoreticalComputer Science Assumption 1.3. The predictions of the predictor and the experts can be any real number in [0; 1]. The predictor also knows before predicting 1. Under Assumption 1.3, an online algorithm a strategy for the predictor A is given in [] for which lim inf n lim inf G A n; ln n 6 1: 1. In fact, what has been proved is that, for all positive integers n and, the algorithm A satises lnn +1 G A n; 6 + log n +1 : 1.3 Notice that 1.3 is much stronger than 1. since it upper bounds G A n; for all n and. Having such a bound is important because very often, in various applications, n and are not arbitrarily large. In this paper, we improve lower bound 1.1 inthe same way. Theorem 1.1. Under Assumption 1., for any algorithm A, any integer n, and any [0; 1], if n : 8 ln n + 8 lnn +1n 1 ; 1.4 then 1 ln n G A n; 1 1 n 1 n ; 1.5 where 1 lnn ln n +8 : +1 n 1 4 First, as easily shown below, 1.1 is a consequence of Theorem 1.1, and thus our theorem is indeed an improvement of 1.1 in the sense that our result implies 1.1 yet it is not implied by 1.1. Corollary 1.1. Inequality 1.1 holds for all online prediction algorithms A. Proof. By setting lnnn 1, it is straightforward to verify that lim inf n lim inf G A n; ln n
4 364 P. Chen, G. Ding / TheoreticalComputer Science lim inf n lim inf n 1 ; lim inf 1 1 ln n 1 n 1 n 1 1 n n holds for all 0; 1, and so 1.1 follows. Further remarks on Theorem 1.1. a If n 1, it is easy to see that the algorithm that copies the only expert will perform exactly the same as the best expert and thus G A n; 0, for all. Because of this, we can say that the assumption n in the theorem does not miss any interesting cases. b Inequality 1.5 still holds if we set 0. We introduce this extra parameter because we need it in proving Corollary 1.1. c For any 0, it is clear that nn 0, as n. Therefore, the requirement n is more or less the same as n. c One may wonder if the requirement n can be dropped completely. For instance, one may ask if there could exist a constant c 0, a function dn; with lim n dn; 0, and such that ln n G A n; c + dn; 1.6 holds for all n,, and A. Unfortunately, the answer is negative. Consider the strategy A that predicts 1 all the time. Then L A; ;, for all and. It follows that G A n; sup L A ; ; L ; ; inf ; L ; 6 : 1.7 Clearly, this inequality contradicts 1.6, for every 0, when n is suciently large. This contradiction indicates that a condition similar to n is required to prove any lower bound of the form 1.6. So far we have discussed the situation when is bigger than n. When is smaller than n, we have seen from remark d that lower bounds 1.1 and 1.5 no longer hold. Moreover, as indicated by our next result, that upper bounds 1. and 1.3 are not very close to the truth either. Theorem 1.. For all n,, and A, under Assumption 1., we have G A n; 1 1 n n :
5 P. Chen, G. Ding / TheoreticalComputer Science By combining this result with 1.7 we obviously have the following. Corollary 1.. For all, under Assumption 1., lim n inf A G An; : This result suggests that, if n is signicantly larger than, then the predictor cannot catch up with the best expert. The only thing the predictor can do is to predict 1 all the time so that it would not be left too far behind the best expert. Remaining questions. We have seen the behavior of G A n; when n and are far apart. But the situation is not very clear when n and are very close. For instance, it would be very interesting to know the asymptotic behavior of inf A G A ;, or the exactly value of inf A G A n; n. We closed this section by giving an outline of the rest of the paper. In Section, we present all necessary mathematical formulas. Then we prove Theorem 1. in Section 3. The proof of Theorem 1.1 will be given in the last section.. Preliminaries In this section, we present all necessary mathematical tools needed for the rest of the paper. Our rst lemma is well known and can be found in many Calculus books. Lemma.1 Stirling formula. For each positive integer m, m! m m m e m ; e where m satises 0 m 1 1m : The next lemma consists of some basic facts about normal distribution, which can also be found in most probability books. Lemma.. Let z 1 z e x dx. Then i 0 1 ; ii z1 z, for all z 0; and iii z 1 z+1 e z, for all z 0. Lemma.3. If n 1, then xx n +1 x n is decreasing over the interval [ 0; 1 ].
6 366 P. Chen, G. Ding / TheoreticalComputer Science Proof. Since n 1 and x 1 0;, it is clear that x nx n 1 1 x n 1 0 and thus the lemma follows. Lemma.4. For every x in [ 1 0; ], we have i ln1 + x x x + x3 3 x4 4, and ii ln1 x x x x3 3 x4. Proof. Notice that both inequalities are equalities when x 0. In addition, ln1 + x x + x x3 3 + x x 1+x x + x 3 x4 1+x 0 holds for all x 0, and thus i follows. Similarly, ln1 x+x + x + x3 3 + x4 1 1 x +1+x + x +x 3 x3 1 x 1 x holds for all x 0; 1, and thus ii follows. 0 Lemma.5. Let x and y be nonnegative numbers. Then e x y e y x. Proof. Notice that the inequality is an equality when x 0. Moreover, when taking derivative with respect to x we have e x y e y + x e x y +1 0: Therefore, the result follows. Lemma.6. Let fx be a function dened on the interval [c; c + ]. i If f x 0 on c; c +, then c+ c fx fc + 1 dx 0; ii If f x 0 on c; c +, then c+1 c fx fc dx + c+ c+1 Proof. If f x 0, then fx is convex and thus 1 fx 1+ 1 fx x1 + x f fx fc + dx 0:
7 P. Chen, G. Ding / TheoreticalComputer Science holds for all x 1 ;x in [c; c + ]. Consequently, c+ c fx fc + 1 dx c+1 c c+1 c c+1 c 0; fx fc + 1 dx + fx fc + 1 dx + c+ c+1 c+1 fx+fc + x fc + 1 dx c fx fc + 1 dx fc + x fc + 1 dx as required. On the other hand, if f x 0, then fx is concave and thus, for all x 1 ;x [c; c + ] and all [0; 1], we have and fx 1 +1 x fx 1 +1 fx f1 x 1 + x 1 fx 1 +fx : Let x x 1 +1 x. Then, adding the last two inequalities gives us fx+fx 1 + x x fx 1 +fx : Now, it is clear that c+1 c fx fc dx + c+1 c 0; c+ fx fc dx + and thus the lemma is proved. fx fc + dx c+1 c+1 c fc + x fc + dx Lemma.7. The inequality z 4 + z max{50z ; 48z 4 +1} holds for all z 0. Proof. It is easy to see that, for z 0, 48z z6 1 ; max{50z ; 48z 4 +1} 50z 1 6 z 6 ; 48z z ; where :908 and :157 are the only two positive solutions to the equation 50z 48z Therefore, to prove the Lemma, we only need to prove the following three claims.
8 368 P. Chen, G. Ding / TheoreticalComputer Science Claim 1. z z 4 +1 z, for z 1;. Claim. z z 4 +1 z, for z 0; 0:3. Claim 3. z z z, for z 0;. Since z is an increasing function and 0:075, it follows that: 48z 4 +1 z 6 48z :050:8z 4 +0:1 z 4 + and thus Claim 1 is proved. To prove Claim, let Then fz z4 + 48z 4 +1 z: 1+8z f 4 15z 3 e z z 1+8z 4 e z and f z z 4 1+8z z3 40z 4 e z 1+8z 4 3 e z : For each z 0; 0:3, it is easy to see that 3 40z 4 0, and so we have f z60. Consequently, f z is a decreasing function, which implies that fz min{f0;f0:3} f00 for all z 0; 0:3, which proves Claim. We prove Claim 3 by considering three intervals, 0; 0:73, [0:73; 0:8], and 0:8;. If z 0:8;, we deduce from 0:8 0:0548 0:056 that z z z z 4 + :8z z 1:4 +0:4 0; and thus the desired inequality holds. Next, let gz z4 + 50z : Then gz is decreasing over [0:73; 0:8], since g z 0:04 z 4 z 3. Recall that z is an increasing function. Therefore, for z [0:73; 0:8], gz g0:80:0753 0:071 0:73 z; which is what we want. It remains to prove z z z, over 0; 0:73. In the remainder of this proof, we assume that z belongs to 0; 0:73. Let h 1 z z +z 50 z and h z z 5lnz:
9 P. Chen, G. Ding / TheoreticalComputer Science Then h 1z z 4z e z ; h 1 z+1z 4 400z e z and h z 5 4z : z Clearly, h z 0, which implies that h z is a decreasing function. Consequently, h z h 0:73 :63935 :5876 ln which implies z ln 400z5 1 : Therefore, h 1 z 1z 4 400z e z 1 z 4 e z ; e z 400z5 1 0; which implies that h 1 z is an increasing function. Notice that h 1 0:5 6:8 0 and h 1 0:6 :1 0. It follows that h 1 z 0 has a unique solution, say z 0, which means that h 1 z achieves its minimum at z 0. To complete our proof, we only need to show that h 1 z 0 0. For this, we apply Taylor s expansion theorem to h 1 z atz 0:56. We have h 1 z h 1 0:56 + h 10:56z 0: h 1 z 0:56 ; where is a number between z and 0:56. Since h 1 0:56 0:35 0, we conclude that z 0 is between 0:56 and 0:6. Now, from h 1 z 0 and h 10:56 0:13 we deduce that h 1 z 0 h 1 0:56 + h 10:56z 0 0:56 0:116 0:40:6 0:560:1 0; which competes our proof of the Lemma. The last is another technical lemma, which we will use in the last section. Lemma.8. Let and be positive numbers and let gx e x 1 1:16 + x: 1 x { Then gx60 if 06x6 min 1 48 ; 5 }.
10 370 P. Chen, G. Ding / TheoreticalComputer Science Proof. Notice that g0 0 and e x g x 1 1 x 3 x+ 1:16 6 e148 0: : Thus, the result follows. 1: Proving Theorem 1. It is clear from the denition of G A n; that, in order to prove a lower bound, we only need to nd one sequence and one set of sequences, for every A, such that L A ; ; L ; is greater than or equal to the lower bound. Since there are too many choices for A, it is dicult or may be impossible to choose and. The way to solve this problem is to use probabilistic method. Recall that, under Assumption 1., the predictions of the experts must be 0 or 1. Suppose each expert makes prediction by tossing, independently, a fair coin. In addition, suppose the outcome is also determined by tossing a fair coin. In the following, we establish a lower bound on the expected gap between the losses of any A and the best expert. Then we prove Theorem 1. using this bound. For each i 1; ;:::;n and j 1; ;:::;, let X i; j be the random variable such that X i; j 0, if the prediction of expert i on j is correct, and X i; j 1, if this prediction is wrong. Then it is not dicult to verify that X {X i;j :16 i 6 n and 1 6 j 6 } is a set of mutually independent random variables with Px i;j 0PX i;j 1 1 : For each i 1; ;:::;n; let S i X i; 1 + X i; + + X i;, which is the number of mistakes made by expert i. It follows that the number of mistakes made by the best expert is S min{s 1 ;S ;:::;S n }. Lemma 3.1. For all j 0; 1;:::;, let p j p j j. Then n E[S] p j : i1 ji Proof. If x is a nonnegative integer not exceeding, it is clear that n PS x PS 1 x; S x; ; S n x n PS i x p j i1 jx
11 and so, Therefore, P. Chen, G. Ding / TheoreticalComputer Science PS x PS x PS x +1 n E[S] x p j x0 x x1 x1 jx jx jx n p j 1 x0 x1 jx+1 x p j n jx+1 p j n n p j jx n n x p j x 1 p j n p j x1 jx and the lemma is proved. jx jx+1 p j n: Lemma 3.. Under Assumption 1., for all n,, and A, we have G A n; Gn;, where n Gn; p j : 3.1 i1 ji Proof. For each j 1; ;:::;, let Y j be the outcome of j. Here, we assume that Y {Y j : j 1; ;:::;} is a set of mutually independent random variables with PY j 0 PY j 1 1. For any algorithm A, let AY; X 1 ; ;:::;. Notice that, for each j, E[ j Y j ] j + 1 j 1 : Therefore, E[L A Y; X ; Y]. Also notice that E[LX ; Y] ES, so, from Lemma 3.1, we deduce that E[L A Y; X ; Y LX ; Y] Gn;. Consequently, for some special values of Y and X, say and, we must have L A ; ; L ; Gn;, and thus the lemma follows. For each i 1; ;:::;+ 1, let P i P i i 1 p j : 3. j0
12 37 P. Chen, G. Ding / TheoreticalComputer Science Lemma 3.3. For all n and, we have where Gn; n; i1 1 P i n + P n i n; ; { 0 if is even; 1 n if is odd: Proof. From 3.1 it is clear that n Gn; p j i1 i1 ji n 1 i 1 p j j0 i +1 n p j n; ji i +1 n p j n; : 3.3 Notice that, by setting i +1 i, j j, and using p k p k, we have n n p j i +1 ji i 1 i 1 i1 p j j+1 i n i 1 p j j 0 i 1 j0 p j n and thus the lemma follows from 3.3 and 3.. Proof of Theorem 1.. In is even, by Lemmas 3., 3.3 and.3, we have G A n; i1 1 P 1 n + P n 1 ji 1 1 n n 1 + and the result follows. If is odd, a little extra eort is needed. When 1,itis straightforward to verify from 3.1 that Gn; 1 1 n n n 1 and thus the theorem holds in this case. When 3, again, using Lemmas 3., 3.3 and.3, we have G A n; 1 1 P 1 n + P1 n n; i1
13 P. Chen, G. Ding / TheoreticalComputer Science The theorem is proved. 1 1 n n n 1 1 n + 1 n : n 1 n n + n 1 8 n 1 4. Proving Theorem 1.1 We prove the theorem by proving a sequence of lemmas. Our starting point is Lemma 3.. In the rest of the paper, let z be a positive number and let 1 a z : 4.1 Lemma 4.1. For every positive integer, we have Gn; a 1 P a+1 ; where is the function dened in Lemma.3. Proof. It follows from Lemmas 3.3 and.3 that Gn; P i n; i1 a P i i1 ia+1 P i n; a ap a+1 n; : 4. The lemma clearly holds if is even. When is odd, using the fact that n; 1, we deduce from 4. and Lemma.3 that Gn; 1 a a P a a 1 P a P a+1 a 1 P a+1 : The lemma is proved.
14 374 P. Chen, G. Ding / TheoreticalComputer Science Next, we need to approximate each p j and then P a+1. Lemma 4.. Let j6 be a nonnegative integer and let h j. If 4h6, then h exp 1 p j 6 exp 1 + h4 3 3 ; 1 h Proof. By Stirling s formula, we have! p j j! j! 6 e e 11 j j e j j j e j 1 j j e 11 : 4.3 j j j j First, it is clear that j j h + h 1 : h Then, when 4h6, we deduce from Lemma.4 that j ln j j ln j 1 h j jln 6 j h h 8h h4 4 h h + h 8h3 + h h + h h h j 3 h h h + 8h h4 4 4h 4 6 h + h and thus the lemma follows from Lemma 4.3. Suppose z 6j6 and 16z. Then p j e j ; 4
15 P. Chen, G. Ding / TheoreticalComputer Science where 8z exp z Proof. Let h j. Then we have 4h64z 16z 6 and so Lemma 4. applies. Notice that h j, and exp h exp h 1 z +1: 1 + z 4 Thus, the result follows from Lemma 4.. Lemma 4.4. If 16z, then P a z +3 3 : Proof. Let x 1 e x and let b a 1. We rst consider the case when is even. Since p j 1 j0 and p j p j, for every j 0; 1;:::;, it follows that j0 p j + 1 p 1 : By 3. and Lemma 4.3, we have P a+1 a j0 p j 1 1 p p j ja b j1 ja+1 j j : 4.6
16 376 P. Chen, G. Ding / TheoreticalComputer Science Observe that, by Lemma. and 4.1, 1 0 b j1 j b +1 + z+ b+1 0 b+1 0 xdx 0 b j xdx 0 b j: 4.7 Let k be the largest even integer not exceeding. Then b+1 A : 0 k 1 j0 j k xdx 0 b j j1 j+1 x j dx + k k+3 k+ x k dx + j+ j+1 k+ k+1 x k + dx + b 1 jk+1 j1 j1 x j + dx x k + dx j+3 j+1 x j + dx: 4.8 It is straightforward to verify that x 0 over 0;, and x 0 over ;. Thus, we conclude from Lemma.6 and the fact that x is decreasing on 0; that A k+1 k k+ k+1 k+3 k+ k +1 k dx k + k + dx k +3 k + dx +0 k +1 k+k +3 k + 1 e 4k 4k+1 e 4k + e k+ 4k+5 e k+ : 4.9 By Lemma.5, it follows that A 1 4k +1 4k +5 and thus the lemma follows from k
17 P. Chen, G. Ding / TheoreticalComputer Science The situation for odd is similar. First, P a+1 a p j j0 1 1 ja+1 p j ja+1 j b j 1 j b j 1 j1 b 1 + z+ xdx b Let k + 1 be the largest odd integer not exceeding. Then B : b k xdx b j 1 j1 x 1 dx j+ x j + 1 dx + j0 j+1 k+ k+1 k+4 k+3 x k + 1 dx + 0 k+3 x k + 3 dx + b 1 j1 j+3 j 1 : 4.10 j+ k+ i+ jk+ x j + 3 dx x k + 3 dx k + k +1+k +4 k + 3 4k +5 3 j x j + 1 dx +3 3 : 4.11 Therefore, by 4.10 and 4.11, the lemma also holds for odd t. The proof is complete. Lemma 4.5. Suppose max{5; z 4 + z }. Then P a+1 z z4 +.
18 378 P. Chen, G. Ding / TheoreticalComputer Science Proof. By Lemma.7, we have 16z. Also observe that z z : 3 Thus the lower bound in Lemma 4.4 can be simplied as P a+1 z +3 3 : 4.1 By taking 8z 4 +11, 4z, and x 1, we deduce from Lemmas.8 and.7 that 8z :16 +z 1 : Therefore, and thus the lemma follows from :16 3 z4 +1:16z : z4 + ; Proof of Theorem 1.1. We rst prove that the assumption n implies max {5; z 4 + z}, if we take 1 lnn z : Notice that 8ln + 8 ln+1 11:6 and thus 5 holds. Furthermore, by Lemma., we have z 4 + z 6 z4 + z + 1e z 1 lnn + 1 lnn +1n 1 6 n: Therefore, the assumptions in Lemma 4.5 are satised. Consequently, by Lemma. again, we have P a+1 z z4 + e z z +1 z4 +
19 P. Chen, G. Ding / TheoreticalComputer Science lnn + lnn +1 n 1 : 4.13 In addition, from 1.4 it is clear that 0. Thus, we deduce from Lemma 4.1 and Lemma.3 that Gn; The proof is complete. a 1 z 1 1 : Appendix To illustrate our results, we plot the following curves for n and 6100: The dotted curve is V n; inf A G A n;, which is the function that we are trying to approximate. The top curve is lnn +1 + log n +1 ; an upper bound of V n;, which is obtained in []. The bottom curve is our lower bound given in Theorem 1.1 and the fourth curve is Gn;, which, by our Lemma 3., is another lower bound of V n; Fig Fig. 1. An illustration of our results.
20 380 P. Chen, G. Ding / TheoreticalComputer Science Acknowledgements The authors are grateful to their friend Steve Seiden, who introduced this problem to them. They wish to acknowledge that they have beneted from discussions with Steve. The authors would also like to thank the two anonymous referees for their valuable suggestions. This Research is partially supported by NSF Grant DMS , AFOSR Grant F , and NSF Grant ITR References [1] N. Cesa-Bianchi, Y. Freund, D.P. Helmbold, M. Warmuth, On-line prediction and conversion strategies, Mach. Learning [] N. Cesa-Bianchi, Y. Freund, D. Haussler, D.P. Helmbold, R.E. Schapire, How to use expert advice, J. Assoc. Comput. Mach
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