MATH 5707 HOMEWORK 4 SOLUTIONS CİHAN BAHRAN 1. Let v 1,..., v n R m, all lengths v are not larger than 1. Let p 1,..., p n [0, 1] be arbtrary and set w = p 1 v 1 + + p n v n. Then there exst ε 1,..., ε n each equal to 0 or 1, so that f v = ε 1 v 1 + + ε n v n, then n w v. Fx v 1,..., v n R m wth v 1 and p 1,..., p n [0, 1], and set w = p v as stated n the queston. For each = 1,..., n, let X be the projecton / ndcator functon X : {0, 1} n {0, 1} ε = (ε 1,..., ε n ) ε. Next, let V = n X v R n and D = w V [0, ). Note that we haven t put a probablty space structure on {0, 1} n yet. Once we do so, X s and n turn V and D wll become random varables. Snce our am s to show the exstence of an ε {0, 1} n wth D(ε) n 4, t suffces to fnd a probablty dstrbuton on {0, 1}n whch satsfes E(D) n 4. Wrtng, for the Eucldean nner product, we have n D = w V, w V = (p X )v, (p j X j )v j j=1 = (p X )(p j X j ) v, v j,j=1 = X ) (p v + (p X )(p j X j ) v, v j j X ) (p + (p X )(p j X j ) v, v j. j Now we mpose the probablty space structure on {0, 1} n such that each ε s 1 wth probablty p and 0 wth probablty 1. Then X has the Bernoull probablty dstrbuton B(p ), so has expectaton E(X ) = p. Moreover, for j the random varables X and X j are ndependent, so E ((p X )(p j X j )) = E(p X )E(p j X j ) = 0. 1
Therefore, by the lnearty of the expectaton, MATH 5707 HOMEWORK 4 SOLUTIONS E(D) E((p X ) Ä ) = p p E(X ) + E(X ) ä Ä = E(X ) p ä ÄÄ = p 1 + (1 p ) 0 ä p ä = (p p ). Fnally, by elementary calculus the global maxmum of the functon f(x) = x x on the nterval [0, 1] occurs at 1, hence f(x) 1 for every x [0, 1]. As a result, 4 1 E(D) f(p ) 4 = n 4.. Suppose n and let H = (V, E) be an n-unform hypergraph wth E = 4 n 1 edges. Show that there s a colorng of V by four colors so that no edge s monochromatc. Let C to be a set of four colors. Put the unform probablty dstrbuton on C, and the product dstrbuton on C V. Snce the set C V can be dentfed wth the set of colorngs of the vertces wth four colors (functons from V to C), nformally ths means that gven c C, a vertex has a 1 -chance to be colored wth c. 4 Note that every element of E s a subset of V wth e = n. In partcular, the vertex colorng f C V makng e monochromatc means that f(e) s a sngleton. For every e E, consder the random varable X e : C V {0, 1} 1 f f(e) s a sngleton, So X e s 1 f e s monochromatc and 0 otherwse. Settng X = e E X e, for f C V, X(f) s the number of monochromatc edges nduced by the colorng f. We want to show that there exsts f such that X(f) = 0 Frst, ( ) E(X) = E X e = E(X e ) e E e E = e E P(f(e) s a sngleton) = P(f(e) = {c}) e E c C = Ç å 1 n = E C 4 n e E c C 4 = 4 n 1 4 4 n = 1.
MATH 5707 HOMEWORK 4 SOLUTIONS 3 Second, choosng f 0 C V to be a constant functon (every vertex s colored the same), we have X(f 0 ) = E = 4 n 1 > 1 as n. Snce E(X) = 1, we deduce that there exsts an f C V such that X(f) < 1, that s, X(f) = 0. 3. Let F be a famly of subsets of N = {1,..., n} and suppose there are no A, B F satsfyng A B. Let σ S n be a random permutaton of elements of N and consder the random varable X whch s the number of -s such that {σ(1),..., σ()} F. By consderng the expectaton of X prove that ( ) n F (here [x] denotes the nteger part of x). Let X be the ndcator functon defned by 1 f {σ(1),..., σ()} F, X (σ) = Observe that by the condton on F, at most one of X 1 (σ),..., X n (σ) s an element of F for each σ. Thus X = n X and n partcular X 1. Thus, no matter what probablty space structure we mpose on the set of permutatons S n, usng the addtvty of the expectaton we get ( n ) 1 E(X) = E X = E(X ). Now we choose unform dstrbuton on S n, that s, every permutaton s equally probable. Then for each, wrtng N = {A N : A = } and F = F N (so F conssts of -element subsets of N contaned n F ), we have By usng the fact Hence so F ( ) n. E(X ) = P({σ(1),..., σ(n)} F ) = F N = F ä. ( ) ( ) n n for any = 1,..., n, we conclude that E(X ) F ä. 1 E(X) n F ä = F ä, 4. Let G = (V, E) be a bpartte graph wth n vertces and a lst S(v) of more than log n for each vertex v V. Prove that there s a proper colorng of G assgnng to each vertex v V a color from ts lst S(v). Let S = v V S(v). Put the unform probablty dstrbuton on the two-element set {C, D}, and then the product dstrbuton on {C, D} S.
MATH 5707 HOMEWORK 4 SOLUTIONS 4 Snce G s bpartte, there exsts A V such that every edge n G s n between A and B := V A. For each a A, defne a random varable Y a : {C, D} S {0, 1} 1 f f(s(a)) = {D}, Smlarly, for each b B, defne a random varable Note that for every a A, Z b : {C, D} S {0, 1} 1 f f(s(b)) = {C}, E(Y a ) = P(f(S(a)) = {D}) = Ç å 1 S(a) < Ç å 1 log n = 1 n. Smlarly, E(Z b ) < 1 n for each b B. Thus lettng X = Y a + Z b, we have a A b B E(X) = E(Y a ) + E(Z b ) a A b B < A 1 n + B 1 n = V n = 1. Therefore, there exsts f {C, D} S such that X(f) = 0. Ths means that Y a (f) = 0 for every a A and Z b (f) = 0 for every b B. In turn, ths means C f(s(a)) and D f(s(b)) for every a A and b B, respectvely. Thus, for every v V, there exsts a color c v S(v) such that C f v A, f(c v ) = D f v B. In partcular, for every a A and b B, c a c b. Snce every edge n G s between A and B, we conclude that the choce of c v s s a a proper colorng. 7. Let F be a fnte collecton of sequences of 0-s and 1-s of fnte length. Assume no member of F s a begnnng of a dfferent member of F. For example, F cannot contan both 101 and 10. Let N denote the number of sequences of length n F. Prove that N 1. Let l be the length of the longest sequence n F. So f we let Q to be the set of all sequences of length, we have the decomposton l F = (F Q ) where N = F Q. By our choce of l, we have N k = 0 when k > l. Gven a sequence q of 0-s and 1-s, we wrte T (q) for the -th truncaton of q. So for nstance T 3 (00101) = 001.
MATH 5707 HOMEWORK 4 SOLUTIONS 5 Now, put the unform dstrbuton on the set Q l. For each {1,..., l} consder the random varable X : Q l {0, 1} 1 T (q) F, q 0 T (q) / F. Gven q Q l, by the assumpton on F, at most one of X 1 (q),..., X l (q) can be equal to 1. Thus defnng X = l X, we have 1 E(X) = E(X ) = F Q Q = N = N.