1
Sample Final Problems
1. (25 points) The number of daily trac accidents at a certain intersection is a random
variable that has a Poisson distribution. The mean of the Poisson distribution depends on
whether it is a week day (Mon.-Fri.) or weekend day (
131B HW#6 solution
8.7 Unbiased Estimators
12. (a) Let X denote the value of the characteristic for a person chosen at random from
the total population, and let Ai denote the event that the person belongs to stratum i(i =
1, . . . , k). Then
= E(X) =
k
E
Lecture 2: Measuring the Macroeconomy
Last time
Income per person in the United States
$2,800 in 1870
$44,000 in 2012
Many countries have not experienced similar increases
in living standards.
The analysis of economic growth helps explain
these long
Supplementary notes for STA131A spring 2017
Dr. Susan Alber
These notes are not meant to either replace the book or lecture notes. They are
notes I prepared for a previously taught intro probability theory class that you may
find useful. This is my attemp
STAIBl/I 4/7?
ovemle
_. relafionskgf between the Size,
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uMou/rtably imcinlte) 01]: the son/lg
5/4? and the Sula/oft
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n. : n(nI)/Y\) [ 0) A Fad'o/m)
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a: )( Two t/fes DF (hula
131B HW#5 solution
7.8 Jointly Sucient Statistics
10. The p.d.f. of the uniform distribution is f (x|) = 1 1cfw_0x , so the likelihood is
fn (x|) =
n
f (xi |) =
i=1
1
1cfw_maxcfw_x1 ,.,xn 1cfw_0mincfw_x1 ,.,xn
n
It implies that the M.L.E. = maxcfw_X1 ,
Handout 6
STA 131B
Let X1 , . . . , Xn be a random sample from f (x|) and T = r(X1 , . . . , Xn ) be a statistic.
Definition of Sufficiency:
T is sufficient for if the conditional distribution of
(X1 , . . . , Xn ) given T = t does not involve (or is the
Final Review Topics:
1. Combinatorial Analysis
- Basic techniques for counting
- Equally likely outcomes
- Permutation
- Combination
2. Axioms of Probability
- Probability rules
3. Conditional Probability and Independence
- Conditional probability
- Rule
STA 131A - Sample Final Solutions
Problem 1
a)
X is the number of accidents in the chosen day. P (weekday) = 5/7, P (weekend) = 2/7.
P (X > 1 | weekday)
=
=
P (X > 1 | weekend)
1 P (X = 0 | weekday) P (X = 1 | weekday)
0.06155
=
1 P (X = 0 | weekend) P (X
Do not start until instructed to do so. You can detach this page for scratch paper.
Midterm II
STA 131B
Winter 2016
University of California, Davis
Exam Rules: This exam is closed book and closed notes. You can use one page (two-sided)
of notes. Use of ca
Handout 1: Calculus Review
STA 131B
For more details, please see a calculus text.
1. Series
1an+1
if a 6= 1.
1a
P
1an+1
i
1 + a + a2 + =
i=0 a = limn 1a
P i1 P d i
d a
1
= i=1 da a = da
= (1a)
2,
i=1 ia
1a
a) 1 + a + + an =
b)
c)
P
i(i 1)ai2 =
P
ai
i=2
d
Handout 3: Method of Moments
STA 131B
Definition of Method of Moments Estimator
This is another quick method for getting estimates that relies on a substitution idea. Let
X1 , . . . , Xn be a random sample from a population with parameter indexed by . Sup
STA 131B
Handout 2
Winter 2016
Let cfw_Zn be a sequence of random variables.
Definition: Modes of Convergence
1. Zn Z in probability Zn Z 0 in probability.
For any > 0, P (|Zn Z| > ) 0, as n .
2. Zn Z in distribution Zn Z 0 in distribution.
FZn (t) FZ
STA 131B HW2
7.5
8. The likelihood function is
(
exp(n
0
fn (x|) =
Pn
i=1 xi )
for min(x1 , . . . , xn ) >
otherwise.
(a) For each value of x, fn (x|) will be a maximum when is made as large as possible subject
to the strict inequality < min(x1 , . . .
131B HW#1 solution
3.7 Multivariate Distributions
1. (a) We have
1
1
1
f (x1 , x2 , x3 )dx1 dx2 dx3 = 3c
0
0
0
Since the value of this integral must be equal to 1, it follows that c = 1/3.
(b) For 0 x1 1 and 0 x3 1,
1
1
f13 (x1 , x3 ) =
f (x1 , x2 , x3 )