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Math 402 Study Guide

Problem 1
In a “four number bet” in roulette, you win if any of the numbers 00, 0, 1, or 2 comes up
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To compute the fair payout that yields an expected loss of − 19
, consider the
following: There are 38 equally likely outcomes in American roulette
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winning the four-number bet is 38
38

Let x be the payout upon winning
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(b) The actual payout for this bet is 8:1
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With an 8:1 payout, the net gain is 8 dollars when you win, and the loss is 1
dollar when you lose
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0526
38
19
Therefore, the expected loss is −

1
or approximately $0
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19
1

Problem 2
In a dice game, you bet $2 on a number from 1 to 6
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If your
number appears on k ∈ {1, 2, 3} dice, you win 2k dollars (and keep your original wager)
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What is your expected net
profit or loss per round?
Solution
...

We first compute the probabilities P (k) for k = 0, 1, 2, 3 matches, using the binomial distribution:
   k  3−k
5
3
1
, k = 0, 1, 2, 3
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216

A $2 stake yields a net profit of 2k dollars when k ≥ 1, and a loss of $2 when k = 0:
(
2k, k = 1, 2, 3,
X=
−2, k = 0
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1574 dollars
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16 per round
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One contains an unknown amount of money, and the
other contains 1
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You pick one envelope at random and find $120 inside
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(a) Compute the expected value if you switch
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We identify the two possibilities for the unknown envelope:
80 =

120
1
...
5×120 (if 120 is the smaller amount)
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2
2
2
Because you already hold $120, this suggests an expected gain of $10 by switching
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Solution
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Let X denote the smaller of the two amounts
prepared
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However, the actual probability distribution over X determines which scenario is more likely,
and the equal probability assumption generally does not follow from a coherent prior
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This leads to a contradiction of the conservation of
expected value and thus reveals the logical flaw
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, n, and mean µ = E[X]
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(a) Show that the conditional distribution for X is:
pˆk =

kpk
µ

Solution
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” By Bayes’
theorem,
Pr(A | X = k) Pr(X = k)
Pr(X = k | A) =

...
Since the
proportionality constant cancels during normalization, we may write:
Pr(A | X = k) ∝ k
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j=1

Combining the components yields the conditional probability:
Pr(X = k | A) =

kpk
= pˆk ,
µ

k = 1,
...


(b) Explain why this is referred to as a “size-biased” distribution
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The term “size-biased” arises because the original mass pk is weighted by the
“size” factor k before renormalization
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The observation that a genotype is present biases the distribution in favor of larger
group sizes, hence the name size-biased
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Of these, nk have k children, where k ∈ {1, 2, 3, 4}
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You observe one child
playing and note the number of children in their household
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µ

Solution
...
”
We apply Bayes’ theorem:
Pr(family size = k | A) =

Pr(A | k) Pr(family size = k)

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Since the total number of children is N = j jnj , we
may write:
Pr(A | k) ∝ k
...

j

j

Hence the conditional probability becomes:
Pr(family size = k | A) =
4

kpk

...
This result is algebraically identical to the genotype problem in Question 4
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The structure is the same,
creating a size-biased distribution; the only difference lies in the interpretation—household
size versus genotype carrier count
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bution with mean 12 seconds
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Solution
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λ2

(b) Find Pr(T ≤ 6)
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Using the exponential CDF F (t) = 1 − e−λt :
Pr(T ≤ 6) = 1 − e−λ·6 = 1 − e−6/12 = 1 − e−0
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60653 ≈ 0
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(c) What is the probability that the next job arrives within 4 seconds, given that the last
job arrived 20 seconds ago?
Solution
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Hence,
Pr(T ≤ 4 | T > 20) = Pr(T ≤ 4) = 1 − e−λ·4 = 1 − e−4/12 = 1 − e−1/3 ≈ 1 − 0
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283
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Solution
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368
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393, 0
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368, respectively
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The shelf life T
(in hours) of a mango at this facility follows an exponential distribution with a mean of 120
5

hours
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Suppose a shipment contains 6 mangoes
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Solution
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05 ≈ 0
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Hence, the probability of passing is q = 1 − p ≈ 0
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For a crate of n = 6 mangoes, the number Y of spoiled mangoes follows a binomial distribution:
 
6 k 6−k
Y ∼ Binomial(n = 6, p ≈ 0
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, 6
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Solution
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04877 ≈ 0
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(c) Compute the variance of the number of spoiled mangoes per crate
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The variance of a binomial distribution is:
Var[Y ] = npq = 6 · 0
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95123 ≈ 0
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Thus, on average, roughly 0
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28
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(a) Compute Pr(X > 90)
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Using the exponential survival function Pr(X > t) = e−λt , we determine λ from
the half-life:
ln 2
λ=
≈ 0
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70
Then,
Pr(X > 90) = e−λ·90 = exp(−0
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410
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6

Solution
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Then,
Pr(X ≤ 40) = 1 − e−λ·40 = 1 − exp(−0
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327
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Solution
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01 × 102 min,
λ
ln 2
2

1
70
Var[X] = 2 =
≈ 1
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λ
ln 2
E[X] =

Problem 9
Calls arrive at a hotline according to a Poisson process with mean rate 3 calls per minute
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(a) Derive the PMF of Z, the number of calls conditioned on the minute being busy
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Since Y ∼ Poisson(λ = 3), the conditional PMF of Z given that the minute is
busy (i
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, Y ≥ 1) is:
k

e−3 · 3k!
Pr(Y = k)
=
,
Pr(Z = k) = Pr(Y = k | Y ≥ 1) =
Pr(Y ≥ 1)
1 − e−3

k ≥ 1
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Solution
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−λ
1
−
e
k=1
k=1

Since

P∞

k=1

k · Pr(Y = k) = E[Y ] = λ = 3, it follows that:
E[Z] =

λ
3
3
=
≈
≈ 3
...

−λ
−3
1−e
1−e
0
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16 calls
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You want the probability that X falls within c standard deviations of the mean to be at
least 98%
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98
Solution
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Then,

Pr(µ − cσ ≤ X ≤ µ + cσ) = Pr(|Z| ≤ c) = 2Φ(c) − 1,
where Φ is the standard normal cumulative distribution function
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98

⇒

Φ(c) ≥ 0
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Taking the inverse CDF:
c = Φ−1 (0
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326
...
33
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(a) Define the discrete failure rate function r(k) = Pr(X = k | X ≥ k)
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The failure rate function is defined in terms of the PMF and the survival function:
r(k) =

Pr(X = k)
pX (k)
=
,
Pr(X ≥ k)
SX (k)

k = 1, 2,
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(b) Suppose X is geometric with success probability p = 0
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Sketch the PMF and failure
rate
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Solution
...
7k−1 · 0
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The survival function becomes:
SX (k) =

∞
X

0
...
3 = 0
...


j=k

8

Thus, the failure rate is:
r(k) =

0
...
3
= 0
...

0
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3 is constant in k
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• The PMF pX (k) = 0
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7k−1 is a strictly decreasing geometric sequence, dropping
from 0
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• The failure rate r(k) = 0
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The PMF decays because later failures become less likely in absolute terms, whereas the
failure rate reflects a constant conditional chance of failure on the next trial, given survival
so far—demonstrating the memoryless property
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The probability that the machine lasts
more than 4000 hours is 0
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We want to double this survival probability by scaling the
failure rate to cr(t) where c < 1
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The survival function is given by:
 Z t

S(t) = exp −
r(u) du = e−H(t) ,

Z
H(t) =

0

t

r(u) du
...
25

⇒

H = − ln(0
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Modified hazard under scaling:
Sc (4000) = exp(−cH) = e−c ln 4 = (e− ln 4 )c = 0
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Set Sc (4000) = 0
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25c = 0
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25 = ln 0
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50
=
=
...
25
−2 ln 2
2

Conclusion:
c = 0
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9

Problem 13
Let X ∼ Poisson(θ) be a random variable representing the number of defective widgets found
during routine inspection at a factory per hour
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Solution
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k!

Split the expression into two terms:
∞
X

k 2 e−θ

k=0

∞
X
θk
θk
ke−θ
...

k!

Shift indices: k 7→ k + 2 and k 7→ k + 1 respectively
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k

ke−θ θk! = θ
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(b) Use the result from part (a) and the identity X 2 = X(X − 3) + 3X to derive E[X 2 ], and
hence find Var(X)
...
Substitute into the identity:
E[X 2 ] = E[X(X − 3)] + 3E[X] = (θ2 − 2θ) + 3θ = θ2 + θ
...

Thus, the variance of a Poisson-distributed variable is Var(X) = θ
...

Show that the new conditional distribution depends on the second moment E[X 2 ], and
compute the expression for the probability that an individual (Mr
...

E[X 3 ]

(a) Conditional distribution after two positive observations
P
Let pk = P (X = k) and mr = E[X r ] = k≥1 k r pk
...

P (X = k | 2 positives) = P
m2
j P (2 positives | X = j)pj
Thus, the new size-biased PMF is
k 2 pk
,
m2

pˆk =

for k = 1, 2,
...
Y is guilty
Suppose guilt attaches to one of the genotype-positive individuals, and Mr
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Given X = k, Mr
...
Averaging:
P (guilty) =

X 1
1 X k 2 pk
m2

...
Observing two positives favours large k, but larger k implies smaller
individual guilt probabilities; the third moment normalizes this tradeoff
...


11

With the same logic, a unique culprit is chosen uniformly among the k j ordered (j +1)-tuples
(the j positives plus Mr
...

Pj (guilty) =
mj+1
E[X j+1 ]
Thus, each additional observed carrier strengthens the size bias and lowers the guilt probability according to a ratio of successive moments
...
Define the empirical
P
distribution pk = mk /m, and let µ = 5j=1 jpj denote the average number of children per
neighborhood
...

(a) Show that the probability that the randomly encountered child lives in a k-child household is:
kpk
P (household size = k | random child) =

...

µ
j jpj
Hence, the conditional probability is:
P (size = k | random child) =

kpk
µ

(b) Interpretation via biased sampling and weighted-urn analogy
Each household can be viewed as an urn containing k indistinguishable balls (children)
...

Urns with more balls are proportionally more likely to supply the sampled individual, so the
sampling scheme biases toward larger k by a factor of k
...

This mirrors a weighted-urn model, where compartments are drawn according to weights—here,
the household sizes

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