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ECO 317 – Economics of Uncertainty – Fall Term 2009
Problem Set 1 – Answer Key
The distribution was as follows:
100 90-99 80-89 70-79 60-69 < 60
2
9
0
3
2
1
Generally a very good start. Here are a few points worth bringing to your attention:
Q.1. Mostly well done. A few people stated the answer without any explanation.
Q.2. Sometimes explanation that events have to be mutually exclusive was missing. Some
people made mistakes in algebra. E.g. they haven’t expressed correctly the probability
of getting a double six.
Q.3. Often people tried to show that conditional probabilities are the same as unconditional,
and in the process omitted some of conditional probabilities. It is easier to use the plain
product definition of independent events.
Q.4. Mostly well done. Here though some people used the formula with a ratio of positive
outcomes to all outcomes. This assumes that all elementary outcomes are equally likely.
In this case it happens to produce the right results since p = 0.5 But in other cases it
would not; you should be careful.
Q.5. Mostly well done if tried. Do your reading; the EGS textbook does essentially this
calculation on p.21.
Q.6. Produced some confusion sometimes. Some people have not quite grasped the abstract
notions of the CDF. Review your ECO 202 or ORF 245 work; if still confused, ask.
Question 1:
The reasoning is not correct. The error is a failure to identify events that are truly elementary
in the process by which the data are generated. Most of the listed events are actually
combinations of two or more elementary events in which the same three numbers appear in
different orders on the three dice. The order does not matter for the sum, but it does matter
when one determines the probabilities of compound events by recognizing their composition
in terms of elementary event of which we know the probabilities for fair dice. (2 points for
getting the idea right)
When two of the numbers are equal, that event consists of three elementary events; when
all three numbers are unequal, this consists of six elementary events. Therefore the six
different ways in which the number of spots can sum to 9,
1, 2, 6;
1, 3, 5;
1, 4, 4;
2, 3, 4;
2, 2, 5;
3, 3, 3
correspond to
6 + 6 + 3 + 6 + 3 + 1 = 25
elementary events, and the six ways in which the number of spots can sum to 10,
1, 3, 6;
1, 4, 5;
2, 2, 6;
2, 3, 5;
1
2, 4, 4;
3, 3, 4.
correspond to
6 + 6 + 3 + 6 + 3 + 3 = 27
elementary events. (6 points for getting the calculations right)
Since there are 63 = 216 elementary events in all, the probabilities are 25/216 = 0.1157
and 27/216 = 0.125 respectively. (2 points for getting the probabilities right; OK if the
fractions are not converted to decimals)
Question 2:
He is equating the probability of a union of events to the sum of the probabilities of the
separate events even though the events are not disjoint (mutually exclusive). (2 points)
The probability of not getting six in any of the four rolls is (5/6)4 = 0.4823. Therefore
the probability of getting at least one six is 1 − (5/6)4 = 0.5177. (3 points)
The probability of not getting a double-six in one throw of a pair of dice is 35/36.
Therefore the probability of not getting even one double-six in 25 throws is (35/36)25 =
0.4945. Therefore the probability of getting at least one double-six is 0.5055. (2 points)
For 26 throws, the numbers become 0.4807 and 0.5193 respectively. (2 points)
Thus the event “at least one six in four throws of one die” is more likely than “at least
one double-six” in the 25-throws case but less likely in the 26-throws case.
Question 3:
Check the definition. For example,
P r(Blue and 1) = 2/6 = 1/3;
P r(Blue) P r(1) =
1 4
2 6
= 1/3.
So the answer is yes.
Question 4:
(a) A tie vote results if 5 of the 10 vote one way and 5 the other way. The probability is
10
5
!
5 5
1
2
1
2
1
10!
5! ∗ 5! 210
10 ∗ 9 ∗ 8 ∗ 7 ∗ 6 −10
=
2
5∗4∗3∗2∗1
= 0.2461
=
(b) Similar calculation yields the probability
20!
2−20 = 0.1762.
10! ∗ 10!
2
Question 5:
(a) (5 points) The standard formula for the normal density is
1
(t − µ)2
√ exp −
2 σ2
σ 2π
"
#
(b) (25 points) We have
E[exp(kX)] =
σ
1
√
Z ∞
2π
−∞
(t − µ)2
exp(k t) exp −
2 σ2
"
#
dt
Change the variable to z = t − µ. Then
"
#
exp(k µ) Z ∞
z2
√
E[exp(kX)] =
exp k z −
dz
2 σ2
σ 2π −∞
Next complete the square inside the square brackets:
kz −
(z − k σ 2 )2
z2
1 2 2
=
k
σ
−
2
2 σ2
2 σ2
Therefore
exp(k µ + 12 k 2 σ 2 ) Z ∞
(z − k σ 2 )2
√
exp −
E[exp(kX)] =
2 σ2
−∞
σ 2π
Change the variable of integration to w = z − k σ 2 . Then
"
#
dz
"
#
exp(k µ + 12 k 2 σ 2 ) Z ∞
w2
√
E[exp(kX)] =
exp −
dz
2 σ2
−∞
σ 2π
But
w2
exp −
dz = 1
2 σ2
σ 2π −∞
being the integral of the density function of a standard normal variable. Therefore
1
√
Z ∞
"
#
E[exp(kX)] = exp(k µ + 12 k 2 σ 2 )
Remember this formula well. We will use it to find expected utility in the case of constant
absolute risk aversion and normally distributed uncertainty. It also lies behind the famous
Ito’s Formula in finance.
Question 6:
For any t ∈ [a, b], we have
G(t) =
=
=
=
=
P rob(Y (s) ≤ t)
by definition of G as the CDF of Y
P rob(F (Y (s) ) ≤ F (t))
since F is strictly increasing
P rob(X(s) ≤ F (t))
by definition of Y
H(F (t))
by definition of H as the CDF of X
F (t)
since X has a uniform distribution
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Thus the random variable Y has the desired CDF F . So, to get a sample of random
numbers with this distribution, take a sample of uniformly distributed random numbers x1 ,
x2 , . . . generated by your computer program, and then calculate yi = F −1 (xi ). The numbers
y1 , y2 , . . . follow the distribution we want.
Extra information not required in your answer: If F is only weakly increasing (has flat
portions), we need to use
Y (t) = inf{ x | F (x) ≥ X(t) }.
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