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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. π’ƒπ’Šπ’“π’•π’‰ 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 π’Šπ’ π’•π’˜π’Šπ’ π’‡π’†π’Žπ’‚π’π’†π’”}
𝑩 = {𝒂 𝑼. 𝑺. π’ƒπ’Šπ’“π’•π’‰ 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 π’Šπ’ π’Šπ’…π’†π’π’•π’Šπ’„π’‚π’ π’•π’˜π’Šπ’π’”}
π‘ͺ = {𝒂 𝑼. 𝑺. π’ƒπ’Šπ’“π’•π’‰ 𝒓𝒆𝒔𝒖𝒍𝒕𝒔 π’Šπ’ π’•π’˜π’Šπ’π’”}
(a) State, in words, the event𝑨 ∩ 𝑩 ∩ π‘ͺ.
(b) Find𝑷(𝑨 ∩ 𝑩 ∩ π‘ͺ).
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(𝐢𝐡|𝑀)𝑃(𝑀)
P (M|CB) = 𝑃(𝐢𝐡|𝑀)𝑃(𝑀) + 𝑃(𝐢𝐡|𝐹)𝑃(𝐹) =
1
2
0.05βˆ—
1
2
1
2
0.05βˆ— +0.0025βˆ—
= 0.9524.
1.41 As in example 1.3.6, consider telegraph signals β€œdot” and β€œdash” sent in the
proportion 3:4, where erratic transmissions cause a dot to become a dash with
𝟏
𝟏
probability πŸ’ and a dash to become a dot with probabilityπŸ‘.
(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
π‘“π‘Œ (𝑦) =
2
βˆ—
πœ‹π‘‘
1
𝑦 2
1+( )
𝑑
𝑦
𝑑
𝑑 tanβˆ’1 ( )
𝑑𝑦
=
1
𝑦 2
1+( )
𝑑
1
𝑑π‘₯
βˆ— 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 = π’Žπ’Šπ’(𝑿, 𝒀) 𝒂𝒏𝒅 𝑿 ∨ 𝒀 = π’Žπ’‚π’™(𝑿, 𝒀).
Analogous to the probability law 𝑷(𝑨 βˆͺ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) βˆ’ 𝑷(𝑨 ∩ 𝑩), show that
𝑬(𝑿 ∨ 𝒀) = 𝑬𝑿 + 𝑬𝒀 βˆ’ 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 βˆ’ E(X ∧ Y).
𝟏
2.17 A median of a distribution is a value m such that 𝑷(𝑿 ≀ π’Ž) β‰₯ 𝟐 and
π’Ž
𝟏
∞
𝟏
𝑷(𝑿 β‰₯ π’Ž) β‰₯ 𝟐. (if X is continuous, m satisfiesβˆ«βˆ’βˆž 𝒇(𝒙)𝒅𝒙 = βˆ«π’Ž 𝒇(𝒙)𝒅𝒙 = 𝟐.) Find
the median of the following distribution.
(a) 𝒇(𝒙) = πŸ‘π’™πŸ , 𝟎 < 𝒙 < 𝟏
(b) 𝒇(𝒙) =
𝟏
𝝅(𝟏+π’™πŸ )
, βˆ’βˆž < 𝒙 < ∞
1
a.
π‘š
∫0 3
2
3
1
βˆ— π‘₯ 𝑑π‘₯ = π‘š = 2 , 𝑔𝑖𝑣𝑒𝑠 π‘š =
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 andβˆ«βˆ’βˆž 𝑓(π‘₯)𝑑π‘₯ = 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 = π‘˜) = (1 βˆ’ 𝑝)π‘˜βˆ’1 𝑝, where p = probability of a daughter. Thus, X is geometric
random variable, and
∞
𝐸𝑋 = βˆ‘ π‘˜ βˆ— (1 βˆ’ 𝑝)
π‘˜=1
π‘˜βˆ’1
∞
∞
π‘˜=1
π‘˜=1
𝑑
𝑑
(1 βˆ’ 𝑝)π‘˜ = βˆ’π‘ βˆ—
βˆ— 𝑝 = βˆ’π‘ βˆ— βˆ‘
[βˆ‘(1 βˆ’ 𝑝)π‘˜ ]
𝑑𝑝
𝑑𝑝
𝑑 1
1
= βˆ’π‘ βˆ—
[ βˆ’ 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
πœΆπŸ‘ =
ππŸ‘
ππŸ’
𝒂𝒏𝒅
𝜢
=
πŸ’
(𝝁𝟐 )πŸ‘/𝟐
𝝁𝟐𝟐
The value πœΆπŸ‘ is called the skewness and πœΆπŸ’ 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 point𝜢, then πœΆπŸ‘ = 𝟎.
(b) Calculate πœΆπŸ‘ for 𝒇(𝒙) = π’†βˆ’π’™ , 𝒙 β‰₯ 𝟎, a pdf that is skewed to the right.
(c) Calculate πœΆπŸ’ for each of the following pdfs and comment on the peakedness of
each.
𝒇(𝒙) =
𝟏
βˆšπŸπ…
𝒇(𝒙) =
𝒇(𝒙) =
π’™πŸ
βˆ— π’†βˆ’ 𝟐 , βˆ’βˆž < 𝒙 < ∞
𝟏
, βˆ’πŸ < 𝒙 < 𝟏
𝟐
𝟏 βˆ’|𝒙|
𝒆 , βˆ’βˆž < 𝒙 < ∞
𝟐
a.
∞
π‘Ž
3
∞
3
πœ‡3 = ∫ (π‘₯ βˆ’ π‘Ž) 𝑓(π‘₯)𝑑π‘₯ = ∫ (π‘₯ βˆ’ π‘Ž) 𝑓(π‘₯)𝑑π‘₯ + ∫ (π‘₯ βˆ’ π‘Ž)3 𝑓(π‘₯)𝑑π‘₯
βˆ’βˆž
0
βˆ’βˆž
π‘Ž
∞
= ∫ 𝑦 3 𝑓(𝑦 + π‘Ž)𝑑𝑦 + ∫ 𝑦 3 𝑓(𝑦 + π‘Ž)𝑑𝑦 (π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘π‘™π‘’)
βˆ’βˆž
0
∞
∞
= ∫ βˆ’π‘¦ 3 𝑓(βˆ’π‘¦ + π‘Ž)𝑑𝑦 + ∫ 𝑦 3 𝑓(𝑦 + π‘Ž)𝑑𝑦
0
(𝑓(βˆ’π‘¦ + π‘Ž)
0
= 𝑓(𝑦 + π‘Ž), 𝑑𝑒𝑒 π‘‘π‘œ π‘ π‘¦π‘šπ‘’π‘‘π‘Ÿπ‘–π‘ 𝑝. 𝑑. 𝑓)
=0
b.
For 𝑓(π‘₯) = 𝑒 βˆ’π‘₯ , πœ‡1 = πœ‡2 = 1, therefore, 𝛼3 = πœ‡3 .
∞
∞
πœ‡3 = ∫0 (π‘₯ βˆ’ 1)3 𝑒 βˆ’π‘₯ 𝑑π‘₯ = ∫0 (π‘₯ 3 βˆ’ 3π‘₯ 2 + 3π‘₯ βˆ’ 1)𝑒 βˆ’π‘₯ 𝑑π‘₯ = 3! βˆ’ 3 βˆ— 2! + 3 βˆ’ 1 = 2.
c.
Each distribution hasπœ‡1 = 0, therefore we must calculate πœ‡2 = 𝐸𝑋 2 π‘Žπ‘›π‘‘ πœ‡4 = 𝐸𝑋 4 .
πœ‡
𝛼4 = πœ‡42 = 3.
πœ‡2 = 1, πœ‡4 = 3,
For (i),
1
2
1
πœ‡
2
πœ‡4
πœ‡2 = 2, πœ‡4 = 24,
For (iii),
9
𝛼4 = πœ‡42 = 5.
πœ‡2 = 3 , πœ‡4 = 5 ,
For (ii),
𝛼4 = πœ‡2 = 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) =
π’†βˆ’π€ 𝝀𝒙
𝒙!
𝒕
, 𝑴𝑿 (𝒕) = 𝒆𝝀(𝒆 βˆ’πŸ) , 𝒙 = 𝟎, 𝟏, … ; 𝝀 > 𝟎
𝒑
(b) 𝑷(𝑿 = 𝒙) = 𝒑(𝟏 βˆ’ 𝒑)𝒙 , 𝑴𝑿 (𝒕) = πŸβˆ’(πŸβˆ’π’‘)𝒆𝒕 , 𝒙 = 𝟎, 𝟏, … ; 𝟎 < 𝒑 < 𝟏
(c) 𝒇𝑿 (𝒙) =
𝟏
βˆšπŸπ…πˆ
𝒆
βˆ’
(π’™βˆ’π)𝟐
𝟐𝝈𝟐
𝑑π‘₯
a. 𝑀𝑋 (𝑑) = βˆ‘βˆž
βˆ—
π‘₯=0 𝑒
𝐸𝑋 =
𝟏 𝟐 𝟐
𝒕
, 𝑴𝑿 (𝒕) = 𝒆𝝁𝒕+𝟐𝝈
𝑒 βˆ’πœ† πœ†π‘₯
π‘₯!
=𝑒
βˆ’πœ†
βˆ— βˆ‘βˆž
π‘₯=0
, βˆ’βˆž < 𝒙 < ∞, βˆ’βˆž < 𝝁 < ∞, 𝝈 > 𝟎
(πœ†π‘’ 𝑑 )
π‘₯!
π‘₯
𝑑
= 𝑒 βˆ’πœ† βˆ— 𝑒 πœ†π‘’ = 𝑒 πœ†(𝑒
𝑑 βˆ’1)
.
𝑑
𝑑2
𝑀𝑋 (𝑑)|𝑑=0 = πœ†; 𝐸𝑋 2 = 2 𝑀𝑋 (𝑑)|𝑑=0 = πœ†2 + πœ†
𝑑𝑑
𝑑𝑑
π‘‰π‘Žπ‘Ÿπ‘‹ = 𝐸𝑋 2 βˆ’ (𝐸𝑋)2 = πœ†
𝑝
∞
𝑑π‘₯
π‘₯
𝑑 π‘₯
b. 𝑀𝑋 (𝑑) = βˆ‘βˆž
π‘₯=0 𝑒 βˆ— 𝑝(1 βˆ’ 𝑝) = 𝑝 βˆ‘π‘₯=0((1 βˆ’ 𝑝)𝑒 ) = 1βˆ’(1βˆ’π‘)𝑒 𝑑
𝐸𝑋 =
𝑑
1βˆ’π‘
𝑑2
𝑝(1 βˆ’ 𝑝) + 2(1 βˆ’ 𝑝)2
𝑀𝑋 (𝑑)|𝑑=0 =
; 𝐸𝑋 2 = 2 𝑀𝑋 (𝑑)|𝑑=0 =
𝑑𝑑
𝑝
𝑑𝑑
𝑝2
π‘‰π‘Žπ‘Ÿπ‘‹ = 𝐸𝑋 2 βˆ’ (𝐸𝑋)2 =
c. 𝑀𝑋 (𝑑) =
∞
βˆ«βˆ’βˆž 𝑒 𝑑π‘₯
βˆ—
1
1βˆ’π‘
𝑝2
1
√2πœ‹
βˆ—πœŽ 𝑒
βˆ’
(π‘₯βˆ’πœ‡)2
2𝜎2
𝑑π‘₯ =
2
2 𝑑π‘₯+πœ‡2
∞ βˆ’π‘₯ βˆ’2πœ‡π‘₯βˆ’2𝜎
2
2𝜎
∫ 𝑒
√2πœ‹πœŽ βˆ’βˆž
1
𝑑π‘₯.
π‘₯ 2 βˆ’ 2πœ‡π‘₯ βˆ’ 2𝜎 2 𝑑π‘₯ + πœ‡ 2 = π‘₯ 2 βˆ’ 2(πœ‡ + 𝜎 2 𝑑)π‘₯ ± (πœ‡ + 𝜎 2 𝑑)2 + πœ‡ 2 = (π‘₯ βˆ’
2
(πœ‡ + 𝜎 2 𝑑)) βˆ’ [2πœ‡πœŽ 2 𝑑 + (𝜎 2 𝑑)2 ],
Then we have𝑀𝑋 (𝑑) = 𝑒
1 2 2
𝑑
𝑒 πœ‡π‘‘+2𝜎
.
1
2
πœ‡π‘‘+ 𝜎2 𝑑 2
βˆ—
∞ βˆ’
∫ 𝑒
√2πœ‹πœŽ βˆ’βˆž
1
(π‘₯βˆ’(πœ‡+𝜎2 𝑑))
2𝜎2
2
1 2 2
𝑑
𝑑π‘₯ = 𝑒 πœ‡π‘‘+2𝜎
βˆ—1=
𝑑
𝑑2
2
𝐸𝑋 = 𝑀𝑋 (𝑑)|𝑑=0 = πœ‡; 𝐸𝑋 = 2 𝑀𝑋 (𝑑)|𝑑=0 = πœ‡ 2 + 𝜎 2
𝑑𝑑
𝑑𝑑
π‘‰π‘Žπ‘Ÿπ‘‹ = 𝐸𝑋 2 βˆ’ (𝐸𝑋)2 = 𝜎 2 .