Math 340
Piryatinska
minutes)
Name:
Fall 2013
Midterm (90
KEY
SHOW ALL WORK
1)An investor is monitoring stock from company A and company B, where each either increases
or decreases each day. On a given day, suppose that there is a probability of 0.38 that
3.5 Expected Values
141
Comment We assume that both the sum and the integral in Definition 3.5.1
converge absolutely:
!
all k
"
|k| p X (k) <
|y| f Y (y) dy <
If not, we say that the random variable has no finite expected value. One immediate
reason for
3.5 Expected Values
145
Therefore,
E(X ) =
!
all k
2k p X (2k ) =
!
k=1
2k
1
=1+1+1+
2k
which is a divergent sum. That is, X does not have a finite expected value, so in order
for this game to be fair, our ante would have to be an infinite amount of mone
3.5 Expected Values
Figure 3.5.4
147
Resultant
Wave 1
Wave 2
A Second Measure of Central Tendency: The Median
While the expected value is the most frequently used measure of a random variables central tendency, it does have a weakness that sometimes makes
140 Chapter 3 Random Variables
and
p X (1) = P(X = 1) =
20 10
=
38 19
9
Then 19
of the time we will win $1 and 10
of the time we will lose $1. Intuitively,
19
then, if we persist in this foolishness, we stand to lose, on the average, a little more
than 5
3.5 Expected Values
3.5.9. Recall Question 3.4.4, where the length of time Y
(in years) that a malaria patient spends in remission has
pdf f Y (y) = 19 y 2 , 0 y 3. What is the average length of
time that such a patient spends in remission?
3.5.10. Let th
142 Chapter 3 Random Variables
Now, let j = k 1. It follows that
E(X ) = np
Finally, letting m = n 1 gives
#
n1 "
!
n1
j=0
E(X ) = np
j
m " #
!
m
j
j=0
p j (1 p)n j1
p j (1 p)m j
and, since the value of the sum is 1 (why?),
(3.5.3)
E(X ) = np
!
Comment Th
144 Chapter 3 Random Variables
negative for the disease, all fifty individuals must necessarily be free of the infection,
and no further testing would need to be done. If the pooled sample gave a positive
reading, of course, all fifty B samples would have
3.5 Expected Values
143
r
Comment Let p represent the proportion of red balls in an urnthat is, p = r +w
.
The formula, then, for the expected value of a hypergeometric random variable
has the same structure as the formula for the expected value of a bino
146 Chapter 3 Random Variables
!
Let w = y/, so that dw = 1/ dy. Then E(Y ) = 0 wew dw. Setting u = w and
dv = ew dw and integrating by parts gives
"
(3.5.4)
E(Y ) = [wew ew ]" =
0
Equation 3.5.4 shows that is aptly namedit does, in fact, represent the a
148 Chapter 3 Random Variables
More specifically, the median lifetime for these bulbsaccording to Definition 3.5.2is the value m for which
! m
0.001e0.001y dy = 0.5
But
"m
0
0
0.001e
0.001y
dy = 1 e
0.001m
. Setting the latter equal to 0.5 implies that
m
1. Suppose that the three inspectors at a lm factory are supposed
to stamp the expiration data on each package of lm at the end of
assembly line. John , who stamps 25% of the packages, fails to stamp
the expiration date once in every 200 packages, Tom , w
Math 340
Piryatinska
Spring 2011
Practice test, Midterm
1. Given that P (A) = P (B ) = 1 , and P (A B ) = 0.1, nd:
3
P (A B ), P (A B ), and P (A B ).
2. Players A, B, and C toss a fair coin in order. The rst to throw a head
wins. What is the probability
MATH 340
Syllabus
Spring 2011
Instructor: Dr. Alexandra Piryatinska
Office: TH 942
Office hours: T 1.00-2.00 pm, TH 8.30-9.30 or by appointment
Email: alpiryat@sfsu.edu
Class: TTH 9:35-10:50, TH 409
Prerequisites: MATH 228 with a grade of C or better (may
MATH 340
Syllabus
Spring 2012
Instructor: Dr. Alexandra Piryatinska
Office: TH 942
Office hours: T, TH 9.50-10.50 or by appointment
Email: alpiryat@sfsu.edu
Class: TTH 11:00-12:15, T-R 4
Prerequisites: MATH 228 with a grade of C or better (may be taken co
Table 1: Special Discrete Distributions
Notation and Parameters
Discrete pdf f (x)
Mean
Variance
MGF MX (t)
np
npq
(pet + q )n
p
pq
pet + q
r/p
rq/p2
1/p
q/p2
nM/N
nM 1
N
e(e
N +1
2
N 2 1
12
1 et e(N +1)
N
1et
Binomial
X BIN (n, p)
0<p<1
n
x
px q nx
x =