COMPSCI 230 Discrete Mathematics for Computer Science Probability

COMPSCI 230 Discrete Mathematics for Computer Science

Probability Refresher What’s a Random Variable? A Random Variable is a real-valued function on a sample space S E[X+Y] = E[X] + E[Y]

Random Walks

How to walk home drunk

Abstraction of Student Life Eat No new ideas Wait Hungry Work 0. 3 0. 4 0. 3 probability 0. 01 Solve HW problem 0. 99 Work

Abstraction of Student Life No new ideas Eat Wait Hungry Work Like finite automata, but 0. 3 instead 0. 4 of a determinisic or 0. 99 0. 01 non-deterministic action, 0. 3 we have a probabilistic action Work Solve HW Example questions: “Whatproblem is the probability of reaching goal on string Work, Eat, Work? ”

Simpler: Random Walks on Graphs - At any node, go to one of the neighbors of the node with equal probability

Random Walk on a Line You go into a casino with $k, and at each time step, you bet $1 on a fair game You leave when you are broke or have $n 0 n k Question 1: what is your expected amount of money at time t? Let Xt be a R. V. for the amount of $$$ at time t

Random Walk on a Line You go into a casino with $k, and at each time step, you bet $1 on a fair game You leave when you are broke or have $n 0 n k Xt = k + d 1 + d 2 +. . . + dt, (di is RV for change in your money at time i) E[di] = 0 So, E[Xt] = k

Random Walk on a Line You go into a casino with $k, and at each time step, you bet $1 on a fair game You leave when you are broke or have $n 0 n k Question 2: what is the probability that you leave with $n?

Random Walk on a Line Question 2: what is the probability that you leave with $n? E[Xt] = k E[Xt] = E[Xt| Xt = 0] × Pr(Xt = 0) + E[Xt | Xt = n] × Pr(Xt = n) + E[ Xt | Xt is neither] × Pr(Xt is neither) = n × Pr(Xt = n) + (somethingt) × Pr(neither) Allow play to infinity, but di = 0 after reaching 0 or n. As t ∞, Pr(neither) 0, and somethingt < n Hence Pr(Xt = n) E[Xt]/n = k/n

Another Way To Look At It You go into a casino with $k, and at each time step, you bet $1 on a fair game You leave when you are broke or have $n 0 n k Question 2: what is the probability that you leave with $n? = probability that I hit green before I hit red

Random Walks and Electrical Networks What is chance I reach green before red? - Same as voltage if edges are identical resistors and we put 1 -volt battery between green and red

Random Walks and Electrical Networks px = Pr(reach green first starting- from x) pgreen= 1, pred = 0 And for the rest px = Averagey Nbr(x)(py) Same as equations for voltage if edges all have same resistance!

Another Way To Look At It You go into a casino with $k, and at each time step, you bet $1 on a fair game You leave when you are broke or have $n 0 n k Question 2: what is the probability that you leave with $n? voltage(k) = k/n = Pr[ hitting n before 0 starting at k] !!!

Getting Back Home - Lost in a city, you want to get back to your hotel How should you do this? Depth First Search! Requires a good memory and a piece of chalk

Getting Back Home - How about walking randomly?

Will this work? Is Pr[ reach home ] = 1? When will I get home? What is E[ time to reach home ]?

Pr[ will reach home ] = 1

We Will Eventually Get Home Look at the first n steps There is a non-zero chance p 1 that we get home Also, p 1 ≥ (1/n)n Suppose we fail Then, wherever we are, there is a chance p 2 ≥ (1/n)n that we hit home in the next n steps from there Probability of failing to reach home by time kn = (1 – p 1)(1 – p 2) … (1 – pk) 0 as k ∞

Furthermore (no proof provided): If the graph has n nodes and m edges, then E[ time to visit all nodes ] ≤ 2 m × (n-1) E[ time to reach home ] is at most this

Cover Times Cover time (from u) Cu = E [ time to visit all vertices | start at u ] Cover time of the graph C(G) = maxu { Cu } (worst case expected time to see all vertices)

Cover Time Theorem If the graph G has n nodes and m edges, then the cover time of G is C(G) ≤ 2 m (n – 1) Any graph on n vertices has < n 2/2 edges Hence C(G) < n 3 for all graphs G

Actually, we get home pretty fast… Chance that we don’t hit home by (2 k)2 m(n-1) steps is (½)k

A Simple Calculation True or False: If the average income of people is $100 then more than 50% of the people can be earning more than $200 each False! else the average would be higher!!!

Markov’s Inequality If X is a non-negative r. v. with mean E[X], then Pr[ X > 2 E[X] ] < ½ Pr[ X > k E[X] ] < 1/k Andrei A. Markov

Markov’s Inequality Non-neg random variable X has expectation A = E[X] = E[X | X > 2 A ] Pr[X > 2 A] + E[X | X ≤ 2 A ] Pr[X ≤ 2 A] ≥ E[X | X > 2 A ] Pr[X > 2 A] (since X is non-neg) Also, E[X | X > 2 A] > 2 A A > 2 A × Pr[X > 2 A] ½ > Pr[X > 2 A] Pr[ X > k × expectation ] < 1/k

Actually, we get home pretty fast… Chance that we don’t hit home by (2 k)2 m(n-1) steps is (½)k

An Averaging Argument Suppose I start at u E[ time to hit all vertices | start at u ] ≤ C(G) Hence, by Markov’s Inequality: Pr[ time to hit all vertices > 2 C(G) | start at u ] ≤ ½

So Let’s Walk Some More! Pr [ time to hit all vertices > 2 C(G) | start at u ] ≤ ½ Suppose at time 2 C(G), I’m at some node with more nodes still to visit Pr [ haven’t hit all vertices in 2 C(G) more time | start at v ] ≤ ½ Chance that you failed both times ≤ ¼ = (½)2 Hence, Pr[ havent hit everyone in time k × 2 C(G) ] ≤ (½)k

Hence, if we know that Expected Cover Time C(G) < 2 m(n-1) then Pr[ home by time 4 k m(n-1) ] ≥ 1 – (½)k

Random walks on infinite graphs

‘To steal a joke from Kakutani (U. C. L. A. colloquium talk): “A drunk man will eventually find his way home but a drunk bird may get lost forever. ”’ R. Durrett, Probability: Theory and Examples, Fourth edition, Cambridge University Press, New York, NY, 2010, p. 191. Shizuo Kakutani

Random Walk On a Line 0 i Flip an unbiased coin and go left/right Let Xt be the position at time t Pr[ Xt = i ] = Pr[ #heads – #tails = i] = Pr[ #heads – (t - #heads) = i] = t (t+i)/2 /2 t

Random Walk On a Line 0 Pr[ X 2 t = 0 ] = 2 t t i /22 t ≤ Θ(1/√t) Y 2 t = indicator for (X 2 t = 0) Sterling’s approx E[ Y 2 t ] = Θ(1/√t) Z 2 n = number of visits to origin in 2 n steps E[ Z 2 n ] = E[ t = 1…n Y 2 t ] ≤ Θ(1/√ 1 + 1/√ 2 +…+ 1/√n) = Θ(√n)

In n steps, you expect to return to the origin Θ(√n) times!

How About a 2 -d Grid? Let us simplify our 2 -d random walk: move in both the x-direction and y-direction…

In The 2 -d Walk Returning to the origin in the grid both “line” random walks return to their origins Pr[ visit origin at time t ] = Θ(1/√t) × Θ(1/√t) = Θ(1/t) E[ # of visits to origin by time n ] = Θ(1/1 + 1/2 + 1/3 + … + 1/n ) = Θ(log n)

But In 3 D Pr[ visit origin at time t ] = Θ(1/√t)3 = Θ(1/t 3/2) limn ∞ E[ # of visits by time n ] < K (constant) E[visits in total ] = Pr[ return to origin ]*(1+ E[visits in total ]) + 0*Pr[ never return to origin ] Hence Pr[ never return to origin ] > 1/(K+1)

A Tale of Two Brothers Carl and Hal (cburch@cs. cmu. edu, hburch@cs. cmu. edu). As competitive as they are similar. They decided to play a fair game once an hour! Each brother wins any event with probability 1/2, independent of any other events. Amazingly, Hal was ahead for the last 50 years of his life! Should we have expected this?

A Random Walk Let xk denote the difference in victories after k events. We can view x 0, x 1, x 2, … , xn as a “random walk” that depends on the outcomes of the individual events.

Theorem: The number of sequences in which xk is never negative is equal to the number that start and end at 0, for even n.

Proof: Put the two sets in one-to-one and onto correspondence.

Lemma: Suppose that Y is an integer random variable in [0, m] such that Then Proof:

Lemma: For odd n, the number of sequences that last touch 0 at position i is equal to the number that last touch 0 at position n-i-1. Proof: Put these sets in one-to-one and onto correspondence using previous theorem to map A to A’ and B to B’.

Theorem: Let Y be the length of the longest suffix that does not touch 0. For odd n, Proof: Y is a random variable in the range [0, n-1]. By previous lemma,

A seemingly contradictory observation: The expected length of the final suffix that does not touch 0 is (n-1)/2, but the expected number of times the sequence touches 0 in the last n/2 steps is

Berthold’s Triangle The number of sequences with k 1’s and n-k 0’s in which the number of 1’s is greater than or equal to the number of 0’s in every prefix. Examples: 111 110 101

Notice that the n’th row sums to 1 0 0 0 0 1 2 2 0 0 0 1 3 5 5 0 1 1 4 9 14 1 5 14 1 6 1

Theorem: Proof: By induction on n. Base cases: Inductive step:

Another strange fact… Let Z be the number of steps before the sequence returns to 0. Let Z’ be the number of steps before the sequence drops below 0. This looks like trouble! E[Z’] isn’t finite!

Random Walk in a Line Cover Time of a Graph Markov’s Inequality Here’s What You Need to Know…