Download Solutions 4 - Institut für Mathematik

Universität Zürich
Institut für Mathematik
Y23G04
STA110
Dr. Candia Riga (lectures)
Antonio De Rosa (exercise classes)
Solutions 4
1. Suppose that a die is rolled two times. What are the possible values that the following
random variables can take on?
(a) W = the maximum value that appears in the two rolls
possible values are 1, . . . , 6
(b) X = the minimum value that appears in the two rolls
possible values are 1, . . . , 6
(c) Y = the sum of the two rolls
possible values are 2, . . . , 12
(d) Z = the value of the first roll minus the value of the second roll
possible values are −5, . . . , 5
What are the probabilities for each of values of the random variable. . .
(a) W ?
We denote by Di the r.v. representing the outcome of the ith die, for i = 1, 2. Observe
that Di ∈ {1, . . . , 6} and Pr(Di = d) = 16 for each d ∈ {1, . . . , 6}. For w ∈ {1, . . . , 6},
we compute
p(w) = Pr(W = w) = Pr(W ≤ w) − Pr(W ≤ w − 1)
= Pr(D1 ≤ w, D2 ≤ w) − Pr(D1 ≤ w − 1, D2 ≤ w − 1)
w 2 w − 1 2 2w − 1
2
= Pr(D1 ≤ w) − Pr(D1 ≤ w − 1) =
−
=
.
6
6
36
We can therefore complete the table as follows.
w
p(w)
1
1/36
2
3/36
3
5/36
4
7/36
5
9/36
6
11/36
(b) X?
Arguing as above, we compute for x ∈ {1, . . . , 6}:
p(x) = Pr(X = x) = Pr(X ≥ x) − Pr(X ≥ x + 1)
= Pr(D1 ≥ x, D2 ≥ x) − Pr(D1 ≥ x + 1, D2 ≥ x + 1)
= Pr(D1 ≥ x)2 − Pr(D1 ≥ x + 1)2
= (1 − Pr(D1 ≤ x − 1))2 − (1 − Pr(D1 ≤ x))2
2 x−1
x 2 13 − 2x
= 1−
− 1−
=
.
6
6
36
We complete the table as follows.
x
p(x)
1
11/36
2
9/36
3
7/36
4
5/36
5
3/36
6
1/36
(c) Y ?
In the previous notations, Y = D1 + D2 . Then, we clearly have:
p(y) = Pr(Y = y) = Pr(D1 + D2 = y)
∞
X
=
Pr(D1 = r) Pr(D1 + D2 = y|D1 = r)
=
r=1
∞
X
Pr(D1 = r) Pr(D2 = y − r),
r=1
where the sum is clearly a finite sum, since the summands are non-zero only for
those values of r such that r, y − r ∈ {1, . . . , 6}. Since for these values of r one has
Pr(D1 = r) = Pr(D2 = y − r) = 61 , it suffices to compute the number of such non-zero
values to conclude.
y
p(y)
2
3
4
5
6
7
8 9
10
11
12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
(d) Z?
Arguing as above:
p(z) = Pr(Z = z) = Pr(D1 − D2 = z) =
∞
X
Pr(D2 = r) Pr(D1 = z + r),
r=1
where the sum is extended to the values of r for which r, z + r ∈ {1, . . . , 6}. We can
complete the table as follows.
z
p(z)
−5
1/36
−4
2/36
−3
3/36
−2
4/36
−1
0
1 2
3
4
5
5/36 6/36 5/36 4/36 3/36 2/36 1/36
2. During 2012 there were 42, 5001 couples married in Switzerland.
(a) Estimate the probability that for at least one of the couples both people have birthdays on March 14th.
Let X = the number of couples with both people having birthday on March 14th
1
X ∼ Bin(42500, 365
2 ) can be approximated with a Poisson distribution with
1
λ = (42500)( 3652 ) ' 0.3190, and therefore
Pr(X ≥ 1) = 1 − Pr(X = 0) ' 1 − exp(−0.319) ' 0.2731
(b) Estimate the expected number of couples where both have their birthday on March
14th.
Using λ = 0.319 would suggest an estimate of 0.
1
Source: www.bfs.admin.ch
(c) Estimate the probability that for at least one of the couples both people have the
same birthday.
Let X = the number of couples that have a common birthday
1
) can be approximated with a Poisson distribution with
X ∼ Bin(42500, 365
1
λ = (42500)( 365 ) ' 116.438, and therefore
Pr(X ≥ 1) = 1 − Pr(X = 0) ' 1 − exp(−116.438) ' 1.
(d) What is an estimate for the actual number of couples in which both people have the
same birthday?
Using λ = 116.4 suggests an estimate of 116.
3. The number of times that a person has a cold each year is a Poisson random variable with
parameter λ = 5. Suppose that a new drug has been developed that reduces this to λ = 3,
but that the drug only works for 75% of the population. For the remaining 25% of the
population there is no effect. A person takes the drug for an entire year. During the year
the person had 2 colds. What is the probability that the drug had an effect on this person?
X ∼ P ois(λ) and let E = event that the drug had the desired effect, then
Pr(X = 2|E) = exp(−3)32 /2! = 0.224, and
Pr(X = 2|E C ) = exp(−5)52 /2! = 0.084. Which leads to
Pr(X = 2|E) · Pr(E)
Pr(X = 2|E) · Pr(E) + Pr(X = 2|E C ) · Pr(E C )
(0.224)(0.75)
= 0.8888.
=
(0.224)(0.75) + (0.084)(0.25)
Pr(E|X = 2) =
4. For a Poisson random variable X ∼ P ois(λ), show that for increasing k, that Pr(X = k)
increases monotonically, then reaches it maximum when k is the largest integer less than
or equal λ, and then decreases monotonically.
Compute Pr(X = k)/ Pr(X = k − 1) and verify that this ratio is equal to λk , which is bigger
than 1 as long as k < λ: in this regime, Pr(X = k) increases monotonically by definition.
Then, the ratio reaches the closest value to 1 that is still less than 1 at k being the largest
integer less than or equal to λ, written k = bλc. Then, for k ≥ bλc + 1 > λ, the ratio is less
than 1, and Pr(X = k) decreases monotonically towards the limit limk→∞ Pr(X = k) = 0.
5. Let X ∼ Bin(n, π), and Y ∼ Bin(n, 1 − π). Show that Pr(X ≤ k) = 1 − Pr(Y ≤ n − k − 1).
Since X = number of successes, and Y = number of failures, then the probability of X = k
successes is equal to the probability of Y = n − k failures, and
Pr(X ≤ k) = Pr(Y ≥ n − k) = 1 − Pr(Y ≤ n − k − 1).
6. If X is a binomial random variable that has expected value 6 and variance 2.4, then what
is Pr(X = 5)?
Use E(X) = nπ = 6, and V ar(X) = nπ(1 − π) = 2.4 to find that
to π = 0.6 and n = 10. Then Pr(X = 5) = 0.20066.
2.4
6
=
nπ(1−π)
nπ
leads
7. Suppose that a certain coin lands on heads with probability p. A trial consists of flipping
the coin two-times in a row. This is repeated until the two flips are either heads or tails.
What is the expected number of trials? And the expected number of total flips?
Let E = event that either 2 heads or 2 tails are flipped when a trial is done. Then
Pr(E) = p2 + (1 − p)2 =: π is the probability of success and suggests that the number
of trials is distributed like a geometric random variable with parameter π and expected
value of π1 . Note that a trial consists of 2 flips, so that the expected number of flips is π2 .
In particular, if the coin is fair then p = 21 , also π = 12 , the expected number of trials is 2
and the expected number of flips is 4.
8. The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random
variable with parameter λ. However, if the insect lays 0 eggs, then it will never be known,
and thus 0 will not be observed. Let X ∼ P ois(λ), and let Y be the number of observed
.
eggs. The probability mass function of Y is Pr(Y = k) = Pr(X = k|X > 0) = Pr(X=k)
Pr(X>0)
Then the expected number of observed eggs is
∞
X
Pr(X = i)
E(Y ) =
i
Pr(X > 0)
i=1
∞
=
X
1
i Pr(X = i)
1 − Pr(X = 0) i=1
1
E(X)
1 − exp(−λ)
λ
=
1 − exp(−λ)
=
solutions04.tex
2017-03-27