Download Proof Assignment #7 1. You and the bank play the following game

Proof Assignment #7
1. You and the bank play the following game: You flip n coins. If X of them come up
heads, you receive 2X dollars.
(a)[1 point] You have to buy a ticket to play this game. What is the fair price of a
ticket?
The fair price of a ticket is its expected value. We can think of this as a Bernoulli trials
process, where “success” is when the coin lands heads. Let Yi = 2i (i.e. the amount that we
expect to receive if the coin lands heads i times). Then, if Y = 2X ,
n n−i
n
n i n−i
X
X
X
1
n i 1
n i 1
=
1
.
E(Y ) =
E(Yi ) =
2
2
2
i
2
i
i=0
i=0
i=0
According to the Binomial Theorem, this last sum is equivalent to (1 + 12 )n . Hence, a fair
price for the ticket would be ( 23 )n dollars.
(b)[1 point] Prove: The probability that you break even (i.e. receive at least your
ticket’s worth) is exponentially small. (Hint: At least how many “heads” do you need to
break even?)
In order to break even, we need to have the number of times i that the coin lands heads
satisfy 2i ≥ ( 32 )n (i.e. “the payoff is at least as large as the amount that we pay for the ticket”).
Taking logs of both sides of the equation yields: log 2i ≥ log( 32 )n , so i log 2 ≥ n log 3 − n log 2.
Hence, i ≥ n(log 3/ log 2 − 1). Using a calculator, we see that this implies that i ≥ .585n.
We can use the Central Limit Theorem to determine the probability of getting at least .585n
heads. We note that p = 1/2 and µ = np = n/2. Hence,
.585n − n/2
.085n
∗
∗
√
√
P (Sn ≥ .585n) = P Sn ≥
= P Sn ≥
.
1/2 n
1/2 n
It’s clear that
n → ∞.
.058n
√
1/2 n
approaches infinity as n → ∞, so P (Sn∗ ≥
.085n
√ )
1/2 n
must tend to 0 as
Another approach that many people took in solving this problem used Chebyshev’s Inequality in lieu of the Central Limit Theorem. Both are valid approaches.
(c)[1 point] Calculate the standard deviation of the variable 2X . Your answer should
be a simple formula.
To find the standard deviation of 2X , we first must find its variance. Recall that, if
Y = 2X , then V (Y ) = E(Y 2 ) − (E(Y ))2 . We have already computed E(Y ) in part (a), so
we just need to find E(Y 2 ). Using the same idea as in part (a), we have
i n−i X
n−i
n
n
X
1
1
1
2
2i
i
E(Y ) =
2
=
2
.
2
2
2
i=0
i=0
According to the Binomial Theorem, this last sum is equivalent to (2 + 21 )n = ( 52 )n . Hence,
q
V (Y ) = ( 25 )n − ( 32 )2n = ( 25 )n − ( 94 )n . Therefore, SD= ( 52 )n − ( 49 )n .
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Proof Assignment #7
(d)[1 point] Show that your answer to (c) is asymptotically equal to an even simpler
formula. Make it as simple as possible.
q
The simplest choice is ( 52 )n . It is easy to see that this is asymptotically equal to the
expression that we obtained in part (c). Namely,
q
s
s
s
n
( 52 )n − ( 94 )n
( 25 )n − ( 94 )n
( 52 )n ( 94 )n
9
n
q
lim
=
= lim
− 5 n = lim 1 −
.
5
5
n
n
n→∞
n→∞
n→∞
10
(
)
(
)
(
)
5 n
2
2
2
( )
2
Observing that (9/10)n approaches 0 as n → ∞ allows us to conclude that the limit above
is 1, i.e. our solution to part (d) is, in fact, asymptotically equal to our solution to part (c).
(e)[1 point] State what the (Weak) Law of Large Numbers would say about the variable
2X .
The Weak Law of Large Numbers would say that, as n tends to infinity, the average
winnings (i.e. 2X ) get closer and closer to the price of the ticket (i.e. ( 23 )n ). Of course, as
we’ll see in part (f), the Weak Law of Large Numbers is not applicable in this situation.
(f )[Bonus! +2 points] Prove that the (Weak) Law of Large Numbers does NOT hold
for this variable.
There were a number of reasons that the Weak Law of Large Numbers should fail. Here is
one possible answer: The Weak Law of Large Numbers says that, as n → ∞, the probability
of winning back the price that you paid for a ticket should approach 1. However, we showed
in part (b) that, as n → ∞, the probability of winning back the price that you paid for the
ticket actually approaches 0. Thus, the Weak Law of Large Numbers cannot hold in this
scenario!
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