CS 1951-k and CS 2951-z: Homework 9
Professors Greenwald and Oyakawa
2017-04-19
Due Date: Tuesday, May 2, 2017. 9:00 PM.
We encourage you to work in groups of size two. Each group
need only submit one solution. Your submission must be typeset
using LATEX. You must submit both the LATEX source and the rendered
PDF. LATEX files without an accompanying PDF will not be graded.
Likewise, PDF files without accompanying LATEX source will not be
graded. Use the handin script to submit.
For 1000-level and 2000-level credit, you must solve all the problems.
1
Lemonade Equilibria
In class we presented the pure Nash equilibria for the Lemonade
game, but we1 were wrong, so it is up to you to fix our mistake!
Describe all pure Nash equilibria for the one-shot Lemonade game
and prove your answer. See the lab directory for the description of
the Lemonade game. For clarification, if exactly two people select
the same number, then those two people receive a payoff of 6 and
the third player receives a payoff of 12. If all three select the same
number, then they all receive a payoff of 8.
2
1
Gabe
The Randomized Halving Mistake Bound
In this problem, you will prove that the randomized halving algorithm2 mistake bound is
E [m] ≤ ln n,
Algorithm 3 in the Halving Algorithm
lecture notes.
2
(1)
when there exists one “perfect” expert that never makes any mistakes.
At the start of the algorithm, we have a set of experts X1 = N.
Some members of X1 will be wrong. Let W1 ⊂ X1 be the set of experts in X1 that are wrong at timestep 1. This means the probability
of making a mistake at round 1 is
α1 =
|W1 |
.
| X1 |
(2)
Generalizing, let Xt be the set of experts we have available at timestep
t. Let Wt ⊂ Xt be the set of experts in Xt that are wrong at timestep t.
cs 1951-k and cs 2951-z: homework 9
The probability of making a mistake at round t is
αt =
|Wt |
.
| Xt |
(3)
In expectation, the number of mistakes made up to timestep T is
E [m] =
T
∑ Probability of making a mistake at timestep t
(4)
∑ αt .
(5)
t =1
T
=
t =1
1. Prove that the fraction of experts remaining at timestep T is
T
1
.
n
(6)
ln (1 − x ) ≤ − x.
(7)
∏ (1 − α t ) ≥
t =1
2. Prove analytically that, for 0 ≤ x < 1,
Hint: Take a Maclaurin series expansion.
3. Prove the mistake bound, Equation (1).
3
The Multiplicative Weights Algorithm
In our story for the multiplicative weights algorithm so far, we say
that on each round some “event” either does or does not occur. The
experts who made the correct prediction maintain the same weight,
while the experts who did not get their weight reduced by the same
fraction. In this problem, we generalize this story to allow for some
experts to be “more wrong” than others. If, for instance, the experts
have some measure of confidence associated with their prediction,
then we can punish them proportional to their confidence level. Or
if there are more than two possible outcomes, the experts could give
probability distributions across the outcome space and we could
judge them based on their given probability of the event that actually
occurred.
See section 3 of the lecture notes for a formalization of this new
setting. In this section, we state Theorem 3.1, which is a generalization of the bound we proved in class (Theorem 2.1 in the notes).
1. Show that if we restrict mt ∈ {0, 1}n , then Theorem 3.1 follows
immediately from Theorem 2.1.
2. Prove Theorem 3.1. Hint: Your proof should closely resemble the
proof of Theorem 2.1. Remember that Theorem 2.1 made use of
2
cs 1951-k and cs 2951-z: homework 9
two log tricks: log(1 − x ) < − x for 0 ≤ x ≤ 1 and log( 1−1 x ) ≤ x +
x2 for 0 ≤ x ≤ 1/2, both of which you are free to assume without
proof. You may find an additional trick useful: (1 − x )n ≤ 1 − nx
for 0 ≤ n, x ≤ 1, which are you also free to assume without proof.
3. We3 pulled a fast one on you in class by using this more general
version of the theorem for the Zero-Sum Game application. Briefly
explain why our method of generating p̃ relies on Theorem 3.1 in
its full generality.
3
Luke
3