Download Exercise 2 – Probability

Exercise 2 – Probability
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November 2, 2015
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Independence
a. Let X1 , . . . , Xn be independent discrete random variables. Show using the definition
of independence that:
E[X1 X2 ] = E[X1 ]E[X2 ],
b. Let (Ω1 , P 1), . . . , (Ωk , Pk ) be distributions, and for every i ∈ [k] let Xi : Ωi → U be
a random variable.
Let (Ω, P ) = (Ω1 , P 1) × · · · × (ω, Pk ). For every i ∈ [k] we define Xi0 : Ω → U to by
∀ω = (ω1 , . . . , ωk ) ∈ Ω,
Xi0 (ω) = Xi (ωi )
Show that {Xi0 }ki=1 are independent.
c. Let {Ai }ni=1 be a set of independent events. Show that if we replace a subset I ⊆ [n]
with their complements
the set is still independent. Namely, that for every I ⊆ [n]
the set {Ai }i∈I
/ ∪ Āi i∈I is independent.
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Bayes’ theorem
Let A, B ∈ Ω be two events. Assuming P r(B) 6= 0, define the probability of A conditioned
on B as
P r(A ∩ B)
P r(A|B) =
.
P r(B)
The intuitive interpretation of this number is the probability that A occurs given that B
has occurred.
a. Proof Bayes’ theorem, namely that
P r(A|B) =
P r(A)P r(B|A)
,
P r(A)P r(B|A) + P r(A)P r(B|A)
where A is the complement of A.
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b. A card is drawn from a standard deck of 52 cards and discarded (i.e. not replaced).
A second card is drawn from the remaining deck of 51 cards. Given that the second
card was a spade, what is the probability that the first card was also a spade?
c. A medical company touts its new test for a certain genetic disorder. The false
negative rate is small: if you have the disorder, the probability that the test returns
a positive result is 0.98. The false positive rate is also small: if you do not have
the disorder, the probability that the test returns a positive result is only 0.04.
Assume that 2% of the population has the disorder. If a person chosen uniformly
at random from the population is tested and the result comes back positive, what
is the probability that the person has the disorder?
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Phase transitions in random graphs
In this question we consider a few examples of a phase transition in a random graphs.
Roughly speaking, phase transition is a phenomenon of abrupt change in the behavior of
an object when some parameter passes a certain threshold.
We look at a random graph G ∼ G(n, p).
Definition 3.1. Let p : N → (0, 1) be a function. We say that a property P of a graph
has a threshold at p if for any function p0 : N → (0, 1) it holds that
n→∞
• If p0 = cp with c > 1 then PrG∼G(n,p0 (n)) [G has the proerty P ] −→ 1.
n→∞
• If p0 = cp with c < 1 then PrG∼G(n,p0 (n)) [G has the proerty P ] −→ 0
Show that the property of having no isolated vertexes has a threshold at
p=
ln n
n
Namely:
n→∞
a. if p0 = c lnnn , c > 1 then Pr [G has no isolated vertexes] −→ 1
n→∞
b. if p0 = c lnnn , c < 1 then Pr [G has no isolated vertexes] −→ 0
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Stock Pricing
The stock price of googface behaves the following way: on each day its price either
gains 60% or looses 40% compared to its price of the previous day, and the two events
happen with probability 1/2 each, independently from historical price movements. Let
the random variable Xi denote the price of the stock on i-th day, where X0 = 1.
a. Compute E[Xi ] for all i.
b. Compute E[ln(Xi )] for all i.
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c. Compute var[ln(Xi )] for all i (hint: write ln Xi as a sum of independent variables).
d. Suppose you buy the stock at day 0 and sell it on day 10, 000. Use Chebyshev’s
inequality to bound the probability that you do not lose.
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The Binomial distribution
Recall the binomial distribution from the tirgul - We flip n independent coins, where
each comes out heads with probability p. Let X be the total number of heads. The
distribution of X is called a binomial distribution with parameters n and p, and denoted
X ∼ B(n, p).
) ≤ n4 . Here we prove a for a similar case
We saw in the tirgul that for p = 12 - Pr(X ≥ 3n
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that the bound is actually exponentially small in n.
a. Use Chebyshev’s inequality to show that if p = c for some constant c ∈ (0, 1/2),
then the probability that X > n2 goes to zero as n goes to infinity.
b. Prove by an elementary direct computation (without using Stirling’s formula or
any other heavy machinery) that the above probability is exponentially small in n.
(Hint: start by bounding the probability that X = n2 , and show that any higher
1
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value
isn even less probable. Also, remember that ∀p ∈ (0, 2 ) p(1 − p) < 4 and
n
< 2 ).
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