Lecture 5 Linearity Of Expectation

CSE 713: Random Graphs and Applications SUNY at Buffalo, Fall 2003

Lecturer: Hung Q. Ngo Scribe: Dazhen Pan

Lecture 5: Linearity of Expectation

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Linearity of Expectation

Theorem 1.1 (Linearity of Expectation Theorem). E[c1 x1 + · · · + cn xn ] = c1 E[x1 ] + · · · + cn E[xn ]

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Applications

Proposition 2.1. Let G be any graph with n vertices, m edges, then G contains a bipartite subgraph with at least m 2 edges. Proof. Place each vertex v ∈ V on either side of a bipartite graph with probability 12 . Let ( 1 if e is not crossing the two sides Ie = 0 if e is crossing Then we have E[#edges crossing] = E[

X

Ie ] =

e∈E

X

E[Ie ] =

e∈E

X e∈E

m 1 P r[Ie = 1] = |E| = 2 2

Remark 2.2. To show the existence of ≥ k(or ≤ k) objects with certain property, we can show E[# of such objects] = k Proposition 2.3. For complete graph Kn , there exists a 2-coloring with at least Ka .

n a




21−a monochromatic

Proof. Let S be the set of all Ka in Kn , for all s ∈ S, let ( 0 if s is not mono. Is = 1 if s is mono. Consequently,   X X X 2 n 1−a E[# mono. Ka ] = E[ Is ] = E[Is ] = = 2 a 2 a s∈S

s∈S

e∈E

Proposition 2.4 (Balancing Vectors). Given n vectors v1 , . . . , vn , where |vi | = 1 for all i. We have √ 1. ∃1 , . . . , n ∈ {−1, 1}, such that, |1 v1 , . . . , n vn | ≤ n 1

2. ∃1 , . . . , n ∈ {−1, 1}, such that, |1 v1 , . . . , n vn | ≥

√

n

Proof. Pick i ∈ {−1, 1} with probability 21 . X E[|1 v1 , . . . , n vn |2 ] = E[ i j vi vj ] i,j

X X i j vi vj ] = E[ 2i vi2 ] + E[ i

=n+

i6=j

X

E[vi vj ]E[i j ]

i6=j

=n+0 =n

Definition 2.5 (List Coloring). Given a graph G = (V, E) and, for every v ∈ V , a list L(v) of colors, construct a valid vertex coloring of G such that each vertex v receives a color from its color list L(v). Proposition 2.6. Given bipartite graph G with n vertices, and assign every v ∈ V a color list L(v). Then G is list-colorable if |L(v)| > log2 n for every v. S Proof. Let C = v L(v) the set of all colors. ∀v, put c on either side of G with probability 12 . A vertex v is a “bad vertex” if all the colors in L(v) are assigned to the other side. If there is no “bad vertex”, we can always find a valid coloring by coloring v with a color c ∈ L(v) which is not assigned to the other side. X X 1 n E[# “bad vertices”] = P r[v is a bad vertex] = < log n = 1 |L(v)| 2 2 2 v v So there is no “bad vertex” when |L(v)| > log2 n for every v.

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