Download HW_1 _AMS_570 1.5 Approximately one

HW_1 _AMS_570
1.5 Approximately one-third of all human twins are identical (one-egg) and twothirds are fraternal (two egg) twins. Identical twins are necessarily the same sex,
with male and female being equally likely. Among fraternal twins, approximately
one-fourth are both female, one-fourth are both male, and half are one male and one
female. Finally, among all U.S. births, approximately 1 in 90 is a twin birth. Define
the following events:
𝑨 = {𝒂 𝑼. 𝑺. 𝒃𝒊𝒓𝒕𝒉 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 𝒊𝒏 𝒕𝒘𝒊𝒏 𝒇𝒆𝒎𝒂𝒍𝒆𝒔}
𝑩 = {𝒂 𝑼. 𝑺. 𝒃𝒊𝒓𝒕𝒉 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 𝒊𝒏 𝒊𝒅𝒆𝒏𝒕𝒊𝒄𝒂𝒍 𝒕𝒘𝒊𝒏𝒔}
𝑪 = {𝒂 𝑼. 𝑺. 𝒃𝒊𝒓𝒕𝒉 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 𝒊𝒏 𝒕𝒘𝒊𝒏𝒔}
(a) State, in words, the event𝑨 ∩ 𝑩 ∩ 𝑪.
(b) Find𝑷(𝑨 ∩ 𝑩 ∩ 𝑪).
a. A ∩ B ∩ C = {a U.S. birth results in identical twins that are female}
b. P (A ∩ B ∩ C) = 1/90 * 1/3 * 1/2 = 1/540
1.33 Suppose that 5% of men and 0.25% of women are color-blind. A person is
chosen at random and that person is color-blind. What is the probability that the
person is male? (Assume males and females to be in equal numbers.)
Using Bayes rule
𝑃(𝐶𝐵|𝑀)𝑃(𝑀)
P (M|CB) = 𝑃(𝐶𝐵|𝑀)𝑃(𝑀) + 𝑃(𝐶𝐵|𝐹)𝑃(𝐹) =
1
2
0.05∗
1
2
1
2
0.05∗ +0.0025∗
= 0.9524.
1.41 As in example 1.3.6, consider telegraph signals “dot” and “dash” sent in the
proportion 3:4, where erratic transmissions cause a dot to become a dash with
𝟏
𝟏
probability 𝟒 and a dash to become a dot with probability𝟑.
(a) If a dash is received, what is the probability that a dash has been sent?
(b) Assuming independence between signals, if the message dot-dot was received,
what is the probability distribution of the four possible messages that could have
been sent?
a. P (dash sent | dash rec) =
𝑃 ( dash rec | dash sent)𝑃 ( dash sent)
𝑃 ( dash rec | dash sent)𝑃 ( dash sent) + 𝑃 ( dash rec | dot sent)𝑃 ( dot sent)
32
.
41
(2/3)(4/7)
= (2/3)(4/7) + (1/4)(3/7) =
b. By a similar calculation as the one in (a) P (dot sent|dot rec) = 27/43. Then we have P
(dash sent|dot rec) =16/43.Given that dot-dot was received, the distribution of the four
possibilities of what was sent are:
Event
dash-dash
dash-dot
dot-dash
dot-dot
Probability
(16/43)2
(16/43)*(27/43)
(27/43)*(16/43)
(27/43)2
1.51 An appliance store receives a shipment of 30 microwave ovens, 5 of which are
(unknown to the manager) defective. The store manager selects 4 ovens at random,
without replacement, and tests to see if they are defective. Let X = number of
defectives found. Calculate the pmf and cdf of X and plot the cdf.
This kind of random variable is called hypergeometric in Chapter 3. The probabilities are
obtained by counting arguments, as follows.
The c.d.f is a step function with jumps at x=0, 1, 2, 3, and 4.
2.12 A random right triangle can be constructed in the following manner. Let X be a
𝝅
random angle whose distribution is uniform on(𝟎, 𝟐 ). For each X, construct a
triangle as pictured below. Here, Y = height of the random triangle. For a fixed
constant d, find the distribution of Y and EY.
We have tan(x) = y/d, therefore tan-1 (y/d) = x, and
𝑓𝑌 (𝑦) =
2
∗
𝜋𝑑
1
𝑦 2
1+( )
𝑑
𝑦
𝑑
𝑑 tan−1 ( )
𝑑𝑦
=
1
𝑦 2
1+( )
𝑑
1
𝑑𝑥
∗ 𝑑 = 𝑑𝑦. Thus,
, 0 < 𝑦 < ∞.
This is a Cauchy distribution restricted to(0, ∞), and the mean is infinite.
2.15 Betteley provides an interesting addition law for expectations. Let X and Y be
any two random variables and define
𝑿 ∧ 𝒀 = 𝒎𝒊𝒏(𝑿, 𝒀) 𝒂𝒏𝒅 𝑿 ∨ 𝒀 = 𝒎𝒂𝒙(𝑿, 𝒀).
Analogous to the probability law 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩), show that
𝑬(𝑿 ∨ 𝒀) = 𝑬𝑿 + 𝑬𝒀 − 𝑬(𝑿 ∧ 𝒀)
(Hint: Establish that𝑿 + 𝒀 = (𝑿 ∧ 𝒀) + (𝑿 ∨ 𝒀).)
Assume without loss of generality that X ≤ Y. Then X ∨ Y = Y and X ∧ Y = X. Thus, X + Y
= (X ∧ Y) + (X ∨ Y). Taking expectations:
E[X] +E[Y] =E[X + Y] = E [(X ∧ Y) + (X ∨ Y)] = E(X ∧ Y) + E(X ∨ Y).
Therefore, E(X ∨ Y) = EX + EY − E(X ∧ Y).
𝟏
2.17 A median of a distribution is a value m such that 𝑷(𝑿 ≤ 𝒎) ≥ 𝟐 and
𝒎
𝟏
∞
𝟏
𝑷(𝑿 ≥ 𝒎) ≥ 𝟐. (if X is continuous, m satisfies∫−∞ 𝒇(𝒙)𝒅𝒙 = ∫𝒎 𝒇(𝒙)𝒅𝒙 = 𝟐.) Find
the median of the following distribution.
(a) 𝒇(𝒙) = 𝟑𝒙𝟐 , 𝟎 < 𝒙 < 𝟏
(b) 𝒇(𝒙) =
𝟏
𝝅(𝟏+𝒙𝟐 )
, −∞ < 𝒙 < ∞
1
a.
𝑚
∫0 3
2
3
1
∗ 𝑥 𝑑𝑥 = 𝑚 = 2 , 𝑔𝑖𝑣𝑒𝑠 𝑚 =
1 3
(2)
= 0.794.
b. The function is symmetric about zero, therefore m=0 as long as the integral is finite.
+∞
We know it’s a Cauchy p.d.f and∫−∞ 𝑓(𝑥)𝑑𝑥 = 1.
2.20 A couple decides to continue to have children until a daughter is born. What is
the expected number of children of this couple?
From Example 1.5.4, if X = number of children until the first daughter, then
𝑃(𝑋 = 𝑘) = (1 − 𝑝)𝑘−1 𝑝, where p = probability of a daughter. Thus, X is geometric
random variable, and
∞
𝐸𝑋 = ∑ 𝑘 ∗ (1 − 𝑝)
𝑘=1
𝑘−1
∞
∞
𝑘=1
𝑘=1
𝑑
𝑑
(1 − 𝑝)𝑘 = −𝑝 ∗
∗ 𝑝 = −𝑝 ∗ ∑
[∑(1 − 𝑝)𝑘 ]
𝑑𝑝
𝑑𝑝
𝑑 1
1
= −𝑝 ∗
[ − 1] = .
𝑑𝑝 𝑝
𝑝
Therefore, if p=1/2, the expected number of children is two.
2.28 Let 𝝁𝒏 denote the nth central moment of a random variable X. Two quantities
of interest, in addition to the mean and variance, are
𝜶𝟑 =
𝝁𝟑
𝝁𝟒
𝒂𝒏𝒅
𝜶
=
𝟒
(𝝁𝟐 )𝟑/𝟐
𝝁𝟐𝟐
The value 𝜶𝟑 is called the skewness and 𝜶𝟒 is called the kurtosis. The skewness
measures the lack of symmetry in the pdf. The kurtosis, although harder to
interpret, measures the peakedness or flatness of the pdf.
(a) Show that if a pdf is symmetric about a point𝜶, then 𝜶𝟑 = 𝟎.
(b) Calculate 𝜶𝟑 for 𝒇(𝒙) = 𝒆−𝒙 , 𝒙 ≥ 𝟎, a pdf that is skewed to the right.
(c) Calculate 𝜶𝟒 for each of the following pdfs and comment on the peakedness of
each.
𝒇(𝒙) =
𝟏
√𝟐𝝅
𝒇(𝒙) =
𝒇(𝒙) =
𝒙𝟐
∗ 𝒆− 𝟐 , −∞ < 𝒙 < ∞
𝟏
, −𝟏 < 𝒙 < 𝟏
𝟐
𝟏 −|𝒙|
𝒆 , −∞ < 𝒙 < ∞
𝟐
a.
∞
𝑎
3
∞
3
𝜇3 = ∫ (𝑥 − 𝑎) 𝑓(𝑥)𝑑𝑥 = ∫ (𝑥 − 𝑎) 𝑓(𝑥)𝑑𝑥 + ∫ (𝑥 − 𝑎)3 𝑓(𝑥)𝑑𝑥
−∞
0
−∞
𝑎
∞
= ∫ 𝑦 3 𝑓(𝑦 + 𝑎)𝑑𝑦 + ∫ 𝑦 3 𝑓(𝑦 + 𝑎)𝑑𝑦 (𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒)
−∞
0
∞
∞
= ∫ −𝑦 3 𝑓(−𝑦 + 𝑎)𝑑𝑦 + ∫ 𝑦 3 𝑓(𝑦 + 𝑎)𝑑𝑦
0
(𝑓(−𝑦 + 𝑎)
0
= 𝑓(𝑦 + 𝑎), 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑒𝑡𝑟𝑖𝑐 𝑝. 𝑑. 𝑓)
=0
b.
For 𝑓(𝑥) = 𝑒 −𝑥 , 𝜇1 = 𝜇2 = 1, therefore, 𝛼3 = 𝜇3 .
∞
∞
𝜇3 = ∫0 (𝑥 − 1)3 𝑒 −𝑥 𝑑𝑥 = ∫0 (𝑥 3 − 3𝑥 2 + 3𝑥 − 1)𝑒 −𝑥 𝑑𝑥 = 3! − 3 ∗ 2! + 3 − 1 = 2.
c.
Each distribution has𝜇1 = 0, therefore we must calculate 𝜇2 = 𝐸𝑋 2 𝑎𝑛𝑑 𝜇4 = 𝐸𝑋 4 .
𝜇
𝛼4 = 𝜇42 = 3.
𝜇2 = 1, 𝜇4 = 3,
For (i),
1
2
1
𝜇
2
𝜇4
𝜇2 = 2, 𝜇4 = 24,
For (iii),
9
𝛼4 = 𝜇42 = 5.
𝜇2 = 3 , 𝜇4 = 5 ,
For (ii),
𝛼4 = 𝜇2 = 6.
2
Therefore, (iii) is the most peaked one, (i) next, and (ii) is least peaked.
2.33 In each of the following cases verify the expression given for the moment
generating function, and in each case use the mgf to calculate EX and VarX.
(a) 𝑷(𝑿 = 𝒙) =
𝒆−𝝀 𝝀𝒙
𝒙!
𝒕
, 𝑴𝑿 (𝒕) = 𝒆𝝀(𝒆 −𝟏) , 𝒙 = 𝟎, 𝟏, … ; 𝝀 > 𝟎
𝒑
(b) 𝑷(𝑿 = 𝒙) = 𝒑(𝟏 − 𝒑)𝒙 , 𝑴𝑿 (𝒕) = 𝟏−(𝟏−𝒑)𝒆𝒕 , 𝒙 = 𝟎, 𝟏, … ; 𝟎 < 𝒑 < 𝟏
(c) 𝒇𝑿 (𝒙) =
𝟏
√𝟐𝝅𝝈
𝒆
−
(𝒙−𝝁)𝟐
𝟐𝝈𝟐
𝑡𝑥
a. 𝑀𝑋 (𝑡) = ∑∞
∗
𝑥=0 𝑒
𝐸𝑋 =
𝟏 𝟐 𝟐
𝒕
, 𝑴𝑿 (𝒕) = 𝒆𝝁𝒕+𝟐𝝈
𝑒 −𝜆 𝜆𝑥
𝑥!
=𝑒
−𝜆
∗ ∑∞
𝑥=0
, −∞ < 𝒙 < ∞, −∞ < 𝝁 < ∞, 𝝈 > 𝟎
(𝜆𝑒 𝑡 )
𝑥!
𝑥
𝑡
= 𝑒 −𝜆 ∗ 𝑒 𝜆𝑒 = 𝑒 𝜆(𝑒
𝑡 −1)
.
𝑑
𝑑2
𝑀𝑋 (𝑡)|𝑡=0 = 𝜆; 𝐸𝑋 2 = 2 𝑀𝑋 (𝑡)|𝑡=0 = 𝜆2 + 𝜆
𝑑𝑡
𝑑𝑡
𝑉𝑎𝑟𝑋 = 𝐸𝑋 2 − (𝐸𝑋)2 = 𝜆
𝑝
∞
𝑡𝑥
𝑥
𝑡 𝑥
b. 𝑀𝑋 (𝑡) = ∑∞
𝑥=0 𝑒 ∗ 𝑝(1 − 𝑝) = 𝑝 ∑𝑥=0((1 − 𝑝)𝑒 ) = 1−(1−𝑝)𝑒 𝑡
𝐸𝑋 =
𝑑
1−𝑝
𝑑2
𝑝(1 − 𝑝) + 2(1 − 𝑝)2
𝑀𝑋 (𝑡)|𝑡=0 =
; 𝐸𝑋 2 = 2 𝑀𝑋 (𝑡)|𝑡=0 =
𝑑𝑡
𝑝
𝑑𝑡
𝑝2
𝑉𝑎𝑟𝑋 = 𝐸𝑋 2 − (𝐸𝑋)2 =
c. 𝑀𝑋 (𝑡) =
∞
∫−∞ 𝑒 𝑡𝑥
∗
1
1−𝑝
𝑝2
1
√2𝜋
∗𝜎 𝑒
−
(𝑥−𝜇)2
2𝜎2
𝑑𝑥 =
2
2 𝑡𝑥+𝜇2
∞ −𝑥 −2𝜇𝑥−2𝜎
2
2𝜎
∫ 𝑒
√2𝜋𝜎 −∞
1
𝑑𝑥.
𝑥 2 − 2𝜇𝑥 − 2𝜎 2 𝑡𝑥 + 𝜇 2 = 𝑥 2 − 2(𝜇 + 𝜎 2 𝑡)𝑥 ± (𝜇 + 𝜎 2 𝑡)2 + 𝜇 2 = (𝑥 −
2
(𝜇 + 𝜎 2 𝑡)) − [2𝜇𝜎 2 𝑡 + (𝜎 2 𝑡)2 ],
Then we have𝑀𝑋 (𝑡) = 𝑒
1 2 2
𝑡
𝑒 𝜇𝑡+2𝜎
.
1
2
𝜇𝑡+ 𝜎2 𝑡 2
∗
∞ −
∫ 𝑒
√2𝜋𝜎 −∞
1
(𝑥−(𝜇+𝜎2 𝑡))
2𝜎2
2
1 2 2
𝑡
𝑑𝑥 = 𝑒 𝜇𝑡+2𝜎
∗1=
𝑑
𝑑2
2
𝐸𝑋 = 𝑀𝑋 (𝑡)|𝑡=0 = 𝜇; 𝐸𝑋 = 2 𝑀𝑋 (𝑡)|𝑡=0 = 𝜇 2 + 𝜎 2
𝑑𝑡
𝑑𝑡
𝑉𝑎𝑟𝑋 = 𝐸𝑋 2 − (𝐸𝑋)2 = 𝜎 2 .