Download Econ 162a: Introduction to Probability and Statistics Midterm Exam

Econ 162a: Introduction to Probability and Statistics
Midterm Exam, Fall 2001
Instructor: Hanming Fang
October 25, 2001
Instructions: You have 80 minutes. Answer every question. Allocate your time as
suggested by the points. The total points of this exam is 100. Be sure to answer the
questions directly and precisely, and present your argument clearly. Good luck!
1. [Bayes’ Theorem, 15 points] Three plants, C1 , C2 and C3 , produce respectively,
10, 50 and 40 percent of a company’s output. Although plants C1 is a small plant, its
manager believes in high quality and only 1 percent of its products are defective. The
other two, C2 and C3 , are worse and produce items that are 3 and 4 percent defective,
respectively. All plants are sent to a central warehourse. One item is selected and
observed to be defective. What is the conditional probability that it comes from plant
C1 ? [Hint: Write Ci as the event that a product is produced at firm i; and write D
as the event that a product is defective]
Suggested Answer: Write Ci as the event that the product is from plant i. From
the description of the problem, we know that
P (C1 ) = 0.1, P (C2 ) = 0.5, P (C3 ) = 0.4.
Denote D as the event that a product is defective. We know that
P (D|C1 ) = 0.01, P (D|C2 ) = 0.03, P (D|C3 ) = 0.04.
Hence
P (C1 |D) =
=
P (D|C1 ) P (C1 )
P (D|C1 ) P (C1 ) + P (D|C2 ) P (C2 ) + P (D|C3 ) P (C3 )
0.1 × 0.01
1
= .
0.1 × 0.01 + 0.5 × 0.03 + 0.4 × 0.04
32
2. [Marginal and Conditional Probabilities, 20 points] A national survey of couples showed that 30% of the wives watched the superbowl, and 50% of the husbands.
Also, if the wife watched, the probability that the husband watched increased to 60%.
For a couple drawn at random, what is the probability that:
a. the couple both watch;
b. at least one watches;
c. neither watches;
d. if the husband watches, the wife watches.
1
Suggested Answer: Let H (respectively, W ) be the random variable which equals
to 1 if the husband (respectively, the wife) watches the superbowl, and 0 otherwise.
We are told that PW (1) = 0.3, PH (1) = 0.5, and furthermore PH|W (1|1) = 0.6. Now
let us denote P (h, w) be the joint probability of W = w, H = h. We have
PW (1) = P (1, 1) + P (0, 1) = 0.3
PH (1) = P (1, 1) + P (1, 0) = 0.5
P (1, 1)
PH|W (1|1) =
= 0.6.
PW (1)
Hence P (1, 1) = 0.6 × 0.3 = 0.18, P (0, 1) = 0.3 − 0.18 = 0.12, and P (1, 0) =
0.5 − 0.18 = 0.32. Finally P (0, 0) = 1 − P (1, 1) − P (1, 0) − P (0, 1) = 0.38. Therefore,
the answers are: (a). P (1, 1) = 0.18; (b). 1 − P (0, 0) = 0.62; (c). P (0, 0) = 0.38; (d).
PW |H (1|1) = P (1, 1) /PH (1) = 0.18/0.5 = 0.36.
3. [Discrete Distribution, 15 points] A manufacturer of X-ray machines has fitted
all machines with an excess radiation warning device, yet experience shows that 10
percent of the machines tested fail to issue the warning. Assume that 10 machines
coming off the assembly line during a given month are selected at random.
a. What is the probability mass function for the number of faulty machines, denoted
by X, found in the sample? [you only need to write down the formulas]
b. Calculate the expected value and standard deviation for the random variable X.
Suggested Answer: (a). It has a binomial distribution as follows:
fX (x) =
µ
10
x
¶
0.1x 0.910−x if x = 0, 1, ..., 10; and fX (x) = 0 otherwise.
(b). EX = 0.1 × 10 = 1, VarX = 10 × 0.1 × 0.9 = 0.9.
4. [Distribution of Functions of Random Variables, 15 points] Let X1 , X2 , ..., Xn
be independent and identically distributed random variables from a Uniform distribution on [0, 1] . Let Y = max {X1 , ..., Xn } and Z = min {X1 , ..., Xn } .
(a) Find the PDF of the random variables Y and Z;
(b) Use the results of (a) to calculate EY and EZ.
Suggested Answer: (a). First find the CDF of Y and Z. The support of both Y
and Z are [0, 1] .
FY (y) = Pr (Y ≤ y) = Pr (max {X1 , ..., Xn } ≤ y)
= Pr (X1 ≤ y, ..., Xn ≤ y) = Pr (X1 ≤ y) × ... × Pr (Xn ≤ y)
= yn ,
Hence
fY (y) = ny n−1 , 0 ≤ y ≤ 1.
2
FZ (z) = Pr (Z ≤ z) = Pr (min {X1 , ..., Xn } ≤ z)
= 1 − Pr (X1 > z, ..., Xn > z)
= 1 − (1 − z)n
Hence
fZ (z) = n (1 − z)n−1 , 0 ≤ z ≤ 1.
(b).
¯1
ny n+1 ¯¯
n
yny
dy =
=
.
EY =
¯
n+1 0 n+1
0
Z 1
Z 1
Z 1
n−1
n−1
zn (1 − z)
dz =
n (1 − z)
dz −
(1 − z) n (1 − z)n−1 dz
EZ =
0
0
0
¯
n+1 ¯1
n (1 − z)
¯
= 1+
¯
¯
n+1
Z
1
n−1
0
n
1
= 1−
=
.
n+1
n+1
5. [Conditional Distributions, 15 points] Define a continuous random vector (X, Y )
by the joint p.d.f
fXY (x, y) = e−y , 0 < x < y < +∞.
(a) Verify that fXY (x, y) defined above indeed is a valid p.d.f.
(b) Compute the conditional p.d.f of Y given X = x.
Suggested Answer: (a). Notice that f (x, y) ≥ 0 everywhere and
Z ∞Z ∞
Z ∞
£ −y ¤¯∞
−e ¯x dx
e−y dydx =
0
x
0
Z ∞
£
¤¯∞
e−x dx = −e−x ¯0 = 1.
=
0
(b). To find the conditional p.d.f of Y giveng X = x, we first find the marginal density
of X.
Z ∞
£
¤¯∞
fX (x) =
e−y dy = −e−y ¯x = e−x , 0 < x < ∞
x
Hence
fY |X (y|x) =
fXY (x, y)
=
fX (x)
½
e−y
e−x
0 if y ≤ x
= ex−y if y > x.
[It is important to write out the domain of the density]
6. [Moment Generating Function and Expectations, 20 points] The p.d.f of random variable X is given by
fX (x) = θxθ−1 , 0 < x < 1
where θ > 0 is a parameter. (caution: notice the support of r.v. X)
3
(a) Find the c.d.f of X;
(b) Let Y = −2θ ln X. Find the p.d.f of Y ;
(c) Calculate the moment generating function of Y ;
(d) Calculate EY and V arY.
Suggested Answer: (a). For x ∈ (0, 1)
Z x
FX (x) =
θz θ−1 dz = z θ |x0 = xθ .
0
[This should be clear enough.]
(b). The support of Y is (0, +∞) . The CDF of Y is
Hence the pdf is
FY (y) = Pr (Y ≤ y) = Pr (−2θ ln X ≤ y)
³
y´
= Pr ln X ≥ −
³
³2θ y ´´
= Pr X ≥ exp −
2θ
h
³ y ´iθ
= 1 − exp −
´
³ y2θ
= 1 − exp −
2
fY (y) =
(c). The m.g.f of Y is
³ y´
1
exp − , y > 0.
2
2
Z
³ y´
∞
1
ety exp − dy
MY (t) = Ee =
2
2
0
µ
¶
Z ∞
1
(1 − 2t) y
exp −
dy
=
2
2
0
·
µ
¶¸¯
(1 − 2t) y ¯¯∞
1
× − exp −
=
¯
1 − 2t
2
0
1
1
=
,t < .
1 − 2t
2
tY
[It is important to specify the domain of t on which MY (t) is defined.
(d).
EY
= M 0 (0) = 2
EY 2 = M 00 (0) = 8
Hence V arY = EY 2 − [EY ]2 = 4. [Alternatively one can recognize that MY (t) above
is the m.g.f of exponential with parameter λ = 2 and use the formula for exponential:
EY = λ, V arY = λ2 ].
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