Big Sale in Shopping Paradise Japan | JAPAN Monthly Web MagazineSTAT/MATH 395 & MATH/STAT 395
STAT/MATH 395
STAT/MATH 395, Probability II, continues STAT/MATH 394.
topic will be modeling interactions between random variables.
(Essentially the first part of this page).
Instructor
For better or worse, I, Hoyt Koepke, am your instructor.
I’m a PhD
student in statistics.
I did my masters in Computer Science and my
undergrad in physics.
M I’m not
on email all the time, but I will try to get back to you ASAP.
Prerequisites
The official prerequisite is MATH/STAT 394, the prequel to this
The website to that one is stull up at
The book for the course is A First Course in Probability, 7th
edition, by Sheldon Ross.
If you have another edition, it covers the
same material and will likely work, but it’s your responsibility to
get the homework problems from a copy of this edition.
There will be one midterm and one final.
The final exam will be held
on the last day of class, and I’ll decide when to have the midterm
The exams will all be closed book.
I’ll decide as the course
progresses whether to all you to write up a cheat sheet, provide one
myself, or make you memorize everything.
The last option has low
probability.
Also, at least half of the exam material will come directly from the
recommended problems (see below) with little or no modification.
NOTE: You do not need to bring a calculator to the exams.
Because this course moves very quickly – we have only about four
weeks! – homework will be assigned in almost every class and then
collected the next class.
This translates into 8-10 homework sets,
and I’ll drop the lowest one in calculating the final grade. Having
homeworks due this frequently makes sure you stay on top of the
Generally, the assignments will be short, only 2-4 graded
problems, to also make sure you can have a life.
The exercises in the book are divided into two types, regular problems
and self-assessment problems.
The answers to all of the
self-assessment problems and some of the regular problems are in the
back of the book.
In addition to the graded problems, I’ll also recommend a few ungraded
exercises with each homework.
These will usually have answers in the
back of the book, and I’ll generally take half or more of the exam
questions from these.
Understanding and working these problems is a
great way to prepare for the exams.
Discussion Board
The course discussion board is for discussion of homeworks, concepts,
and pretty much anything related directly to the material of the
It is online at
requires a UW net ID to log on.
It is restr if
you have trouble, send me an email.
The grade distribution will be as follows:
25% Homework.
30% (or 20%) Midterm.
40% (or 50%) Final Exam.
5% Class Participation / Extra Credit (may give additional credit).
For the exams, if you do better on the final than the midterm, it will
count as 50% of your final grade and the midterm will count as 20%.
The class participation / extra credit grade will be based on how much
above the “minimum” you participate in the course and how much I see
you try to engage probability theory.
Asking questions in class / on the course discussion board.
Answering and explaining questions other students ask on the
discussion board.
Sending me a link to a news article that deals with probabilities,
along with a 2-5 sentence discussion/assessment of how the article
handles them.
You can talk about how well it is explained, whether
they interpreted it correctly, if there’s room for people to hide
something within their assessment, etc.
Other things I haven’t thought of yet...
Office Hours
Office hours will be from 10:45 - 11:20 MWF (right after class), and
12:30 - 2:00 on Tue/Thu (note change), and by appointment.
office is Padelford B-312 (Note that Padelford has 3 sections, A, B,
and C; I’m in B).
Schedule (In Progress)
Here is the (tentative) schedule for the course.
Anything that hasn’t
yet happened may change.
If I need to change anything that may
potentially affect your grade (e.g. dropping/changing/modifying a
homework problem), I’ll either email it out or announce it in class.
Otherwise, it is your responsibility to keep track of the content
Friday, July 23
Introduce Jointly Distributed Random Variables, start Chapter 6.
Monday, July 26
Discuss final exam from 394, continue with examples of joint random
variables.
HW 1 + Solutions
Due Wednesday, July 28:
Problems 6.10 and 6.17.
Recommended Problems: Self-Test 6.4, 6.11, Problem 6.22.
Problem 6.10:
The joint probability desnity function of
This amounts to choosing an integral with the correct bounds:
The interesting thing about this problem is that it illustrates
one thing we’ since the two variables are independent
and continuous, the event that one is greater than the other is
exactly as likely as the other one being greater due to the
symmetry of the problem.
Thus the two probabilities should be
since there are two choices,
As in part (a) the condition that $X&a$ defines a set of points
over which
can be integrated:
Problem 6.17:
Three points , , and
are selected
at random on a line .
What is the probability that
lies between
assume that all three come from the same distribution.)
exploit the symmetry of the problem to get your final answer.
Assume that the 3 points are drawn from the same distribution
(i.e. the three R.V.s have the same distribution).
symmetry, all such order i.e. the problem is
the same if we relabel any of the three variables.
Thus of the
possible orderings, 2 of them have
middle. Thus there is a
probability
being between the other two points.
Wednesday, July 28
Review of continuous random variables, independence of random
variables having a joint distribution.
HW 2 + Solutions
Due Friday, July 30: Problems 6.7, 6.16, 6.24; Theory Exercise 6.30.
Bonus Problem: Theory Exercise 6.9.
Recommended Problems: Problem 6.15, 6.28; Self Test 6.6, 6.16.
Problem 6.7:
Consider a sequence of independent Bernoulli trials, each of which
is a success with probability .
number of failures preceding the first success, and let
be the number of failures between the first two successes.
joint mass function of
First recall that when ; so
models the number of Bernoulli trials before a
success occurs.
This experiment “restarts” after finding the first
success and stops after the second.
However, we’re interested in
the number of failures before the first success, and between the
first success and the second success, so this isn’t the negative
binomial distribution.
We’re interested here in the joint distribution of
In this case,
the probability that there were
failures, then a
success, then
failures followed by another success, i.e.
Now we can see from the first line above that this factors into two
terms multiplied together, one containing only
other containing only .
are independent.
Problem 6.16:
Suppose that
points are independently chosen at random on
the perimeter of a circle, and we want the probability that they all
lie in some semicircle.
(That is, we want the probability that
there is a line passing throught the center of the circle such that
all the points are on one side of that line.)
denote the event that all the points are contained in some
semicircle, and let
be the event that ll the points lie
in the semicircle beginning at the point
clockwise for , .
in terms of the .
mutually exclusive?
If any of the
occurs, then
occurs, at least one of the
Yes, the events are mutually exclusive.
To see this assume that
some particular
occured, say .
line drawn through the center of the circle and point
would contain all points, by definition. Note that this implies
that for an arbitrary point on the semicircle, call it
, the angle (clockwise)
to is constrained to
. Therefore the angular,
clockwise distance from
. Or in other words, the
semicircle drawn from
through the center will not
include , so by definition
occur. Thus if
occurs no other
can occur,
and thus the events
are mutually exclusive.
For each , it’s a given that point
that hemisphere.
Now, there is a probability of
that each of the other points is contained in this hemisphere.
Furthermore, each of these points is placed independently of the
other points.
since each of the
is mutually
exclusive, we can just sum up the
probabilities to get
Problem 6.24:
Consider independent trials, each of which results in outcome
with respective probabilities
the number of trials needed to obtain an outcome that is not equal
to , and let
be that outcome.
Show that .
Is it intuitive to you that
is independent of ?
Is it intuitive to you that
is independent of ?
can be determined by noting that
a geometric random variable with probability of success – not
getting 0 as the outcome – being .
The condition on
is simply that it doesn’t equal 0 (see
Problem 3.76 in ). If we define a new random varaible to be the outcome of the trial, say , it would have a distribution such that .
The condition that
requires that
trials have 0 as the outcome, and the
as the outcome.
trials are independent, the probability of this happening is just
Now here we can see that the terms containing
containing
thus they are independent.
also be seen by noting that
is really when the outcome differs from
, it would make sense that
doesn’t depend on
which of the other
just that 0 did
not occur.
simply is the outcome when a 0 thus it
makes sense that knowing which experiment resulted in a non-zero
value would not give you information about that value.
Theoretical Exercise 6.30:
be a set of independent and identically
distributed continuous random variables having distribution function
denote their ordered values.
, independent of the , , also
has distribution , determine:
This probl it is extra credit.
(a) is just the
probability that
is the maximum when
the same rando since all orderings are equally
likely, this is just the probability that a specific one is the
(b) is the probability that it is
by the same reasoning, the probability that
is the minimum is , so (b) is .
For (c), we can look at the probability that
falls between
This probability is the same
as it being the maximum or minimum b
Now extending this to the probability
considered in (c) means that it can be in any of the
intervals between the values ,
each of which has probability .
Friday, July 30
Conditionals and Marginals.
HW 3 + Solutions
Due Monday, August 2: Problem 6.12; Theory Exercises 6.14 and 6.22.
Recomended Problems: Theory Exercise 6.18, Self Test problems 6.10 and 6.14.
Problem 6.12:
The number of people that enter a drugstore in a given hour is a
poisson random variable with parameter .
Compute the conditional probability that at most 3 men entered the
drugstore, given that 10 women entered in that hour.
assumptions have you made?
Solution: There are several ways of approaching this problem. One
can see the number of men entering and women entering as two
separate Poisson random variables, with the total number of people
entering being another Poisson with rates being the sum of the two.
Then the number of men entering, say , is Poisson
distributed with rate , and the number of woment
entering is also Poisson with rate .
If we assume
that men and women come in at the same rate (which is a probably a
bad assumption, but we’ll have to run with it here), we have that
From the problem, we have that
, since the sum of two Poissons is
also a Poisson with the rates adding.
Thus , and
Note that knowing the number of women coming in doesn’t affect this
Theory Exercise 6.14:
Suppose that
are independent geometric
random variables with the same parameter .
Without any computations, what do you think is the value of
Hint: Imagine that you continually flip a coin having
probability
of coming up heads.
If the second head
occurs on the
th flip, what is the probability mass
function of the time of the first head?
Prove your conjecture in part (a).
Hint: Understand example 6.4b.
Conditioning on
is conditioning on the second
success happening at trial .
Now the individual
probability of any trial in the first
trials being a
success is equal.
Conditioning on there being only one success
trials does not change this, so we would
expect each of the trials being that one success to be equally
Thus we would expect the probability to be
Proceding using standard conditional probabilities:
Now probability that
is the probability that
the second success occurs at , which is the pmf at
of a NegativeBinomial distribution with parameters 2
Proceding:
Theory Exercise 6.22:
be a random variable with rate .
is defined as the largest integer less than
or equal to .
Can you conclude that
are independent?
Hint: Use conditional probabilities to get this into a form where
you can use the memoryless property of the exponential.
This problem wa Here are two ways of doing it:
Since this example is separable into the product of two terms, one
containing
and one containing , we have that
are actually
independent.
The other way of doing it is perhaps simpler:
Monday, August 2
Sums of independ tie-up of sections 6.1-6.5.
HW 4 + Solutions
Due Wednesday, August 4:
Problems 6.31, 6.33, and 6.45.
Recommended Problems: Problems 6.30, 6.40, and Self-Test 6.18(a).
Problem 6.31:
The monthly worldwide average number of airplane crashes of
commercial airlines is 2.2.
What is the probability that there will
More than 2 such acciden
More than 4 such accidents in the next 2
More than 5 such accidents in the next 3 months?
Explain your reasoning.
The number of crashes over a period of time is simply a random
variable with a Poisson distribution.
In this case, the sum of two
Poisson random variables is just a new random variable with the new
rate being the sum of the old rates.
We have that ; this
be the number of accidents that happen in a two
month period.
By additivity of the Poisson,
be the number of accidents that happen in a
three month period.
By additivity of the Poisson,
Problem 6.33:
Jill’s bowling scores are approximately normally distributed with
mean 170 and standard deviation 20, while Jack’s scores are
approximately normally distributed with mean 160 and standard
deviation 15.
If Jack and Jill each bowl one game, and assuming
that their scores are independent random variables, what is the
approximate probability that
Jack’s score is higher?
The total of their scores is above 350?
For this problem note that the sum of two independent normal random
variables is again a normal random variable, with the mean and
variances summed. In this particular case let
Jill’s score, and let
denote Jack’ then
Note that since the distribution of the normal is symmetric
around its mean, .
Now, to compare which is higher (since bowling scores are
integers), it’s appropriate to put the continuity correction in
thus the probability that Jack’s score is higher than
Jill’s is approximately
Similarly,
Problem 6.45:
If , , and
are independent random
variables that are uniformly distributed over , compute
the probability that the largest of the three is greater than the
sum of the other two.
In this problem, it turns out it’s easiest to condition on each
being the maximum, then integrate over the appropriate
Looking at one of the conditionals,
is larger than the sum of
, it will be larger than them individually as they are
all positive.
Now we also have the cases where we condition on
being the largest.
Thus, since all the conditionals
will be the same, we have that
Wednesday, August 4
A little more on combinations of random variables, transformations
between density functions, Gamma distribution, introduce expectations
of sums of random variables.
HW 5 + Solutions
Due on Friday, August 6: Problems 6.44, 6.54, 6.57, 7.5.
Recommended Problems: Problems 5.40, 6.55, 7.6; Self-Test 6.13.
Problem 6.44:
An insurance company supposes that each person has an accident
parameter and that the yearly number of accidents of someone whose
accident parameter is
is Poisson distributed with
They also suppose that the parameter value of
a newly insured person can be assumed to be the value of a gamma
random variable with parameters
newly insured person has
accidents in her first year, find
the conditional density of her accident parameter.
Also, determine
the expected number of accidents that she will have in the following
Hint: You don’t actually have to evaluate any integrals to do this
recognize the form of the conditional density as one of the
distributions we’ve looked at, then plug in the proper constants.
denote the number of accidents a person has in a year,
denote the (random) value of their accident
parameter.
Here, we have a joint distribution between
, given in terms of the conditionals.
Together, these define a full joint distribution for the two random
variables.
First, note that the resulting joint distribution is a mix between a
discrete and an continuous distribution, with the distribution of
the discrete random variable, , being specified as a
conditional distribution dependent on the value of .
Second, what we are interested in is not just the distribution of
, which is given in the problem, but the distribution of
knowing that the person had n accidents. In
other words, we want .
This can be
readily expanded out.
Third, the book is stupid as it specifies the rate parameter as
and the shape parameter as ; typically,
is the shape parameter.
Either answer will be
accepted in the homework.
are constants that depend only on
Since we are interested
in a density over , then this constant is determined
entirely by normalizing, i.e. ensuring that
is set such
In this case, we have one m this density is
simply that of a Gamma distribution with parameters
and ; this density has the form
Notice that the terms containing
exactly match up, thus
the value for the constant
in order for the density to integrate to 1.
Thus the expected number of accidents is going to be
Problem 6.54:
have the joint density function
Compute the joint density function of .
What are the marginal densities? (i.e. what are
The form of the joint density function is easy to find using the
density transformation the bounds on the
input variables – i.e. where the density function is non-zero –
is a bit more tricky.
First, we need to find functions
and . This amounts to
solving a system of equations:
This system of equations gives us
This gives us what we need to
know for the relationships between the index variables to plug
into the density transformation.
Now, to calculate the Jacobian, we need to get the partial
derivatives:
The Jacobian is then
Thus . We can get this in terms of
using the functions
Putting these together:
However, this expression for the new joint distribution is not
The original function had
; if these were satisfied, the function took the
value ; outside of
the value 0.
to be non-zero, the
values that these translate back to must
also satisfy those relations.
Specifically,
Now, note that
immediately gives us a lower bound
for ; the smallest possible value of
is , so the smallest value their product can
take is also .
Furthermore, note that ,
is positive. Now, we derive the relationship between
and ; this is trickier.
From this we have the final bounds:
Thus we end up with
To calculate the marginal densities, we integrate out the
appropriate variables.
is a little more tricky, since the bounds on
integrating out
are a little more subtle.
case, we have that, given , ;
is positive, there are really just two cases, one
being the lower bound, and one with
being the lower bound.
If , this bound is
, and we have:
If , the other bound kicks in:
Combining these gives us:
Problem 6.57:
are independent exponenential random
variables each having rate parameter , find the joint
density function of
First, we try to find
in terms of
As before, this gives us a way to express the index variables
in terms of
Next, we need to calculate the Jacobian.
To do this, first
calculate the partial derivatives.
Thus the Jacobian is:
is ; in terms of
, this is .
Now we’re ready to calculate
the form of our final density function:
To refine the region where this is positive, note that
As , we also have that .
Problem 7.5:
The county hospital is located at the center of a square whose sides
are 3 miles wide.
If an accident occurs within this square, then
the hospital sends out an ambulance.
The road network is
rectangular, so the travel distance from the hospital, whose
coordinates are , to the point , is just
If an accident occurs at a point
that is uniformly distributed in the square, find the expected
travel distance of the ambulance.
Because expectation is a linear operator,
Friday, August 6
A little more on conditional probabilities, review for exam.
Monday, August 9
Some more on expectations of problems + Midterm.
Homework 6 (one
problem) below.
Midterm Exam
We are having the midterm exam this Monday.
You’ll be allowed a
single page reference sheet of your own making, which you have to turn
in with the exam (though you’ll get it back with the graded exam); you
make this yourself.
In addition, I’ll include the ,
with the addition of the Gamma and Beta densities.
This exam will cover sections 6.1-6.5 and 6.7. You won’t be
responsible for the following examples/topics, however: 2e, 2g, 2i,
2j, lognormals & 3d, 5a (silly), and 7c and subsequent material in
section 7.
Again, many of the questions will be based heavily on reference
problems, though don’t count on merely being able to copy down
solutions from
know and understand how to work
these problems.
More information may appear the closer we get to Monday.
HW 6 + Solutions
Due Wednesday, August 11: Problem 7.13.
Recommended Problems: Self Test 7.2, 7.4.
Problem 7.13:
A set of 1000 cards, numbered 1 through 1000, are distributed among
1000 people.
Compute the expected number of cards that are given to
people whose age matches the number on the card.
be the card distributed to person .
are dependent, as the possible
values for each card depends on the previously distributed values.
However, using expectation, we can get around this.
be the age of person
(assume ), and let
We are then interested in
Wednesday, August 11
Expectations and marginals, Variance, Covariance, Correlation, Skew,
and Kurtosis.
Recommended Reading:
Sections 7.1-7.4.
However, here’s a few comments:
Skip examples 7.2m and subsequent material (though the bonus
problem references 7.2m).
I don’t find the discussion of higher moments to be very well
the way I presented in class is more commonly what you
Thus just skim section 7.3.
In 7.4, just read the discussion on example 4.c; the math is a bit
involved and not really needed.
HW 7 + Solutions
Due Friday, August 13: Problems 7.14, 7.17, 7.19, 7.36.
Bonus Problem: Problem 7.24.
Recommended Problems: Problems 7.12, 7.37, 7.45, Self Test 7.3.
Problem 7.14:
An urn has
black balls.
At each stage a black ball is
removed and a new ball, that is black with probability
white with probability , is put in its place.
expected number of stages needed until there are no more black balls
in the urn.
Hint: Connect this to a previously seen distribution.
Each stage decreases the number of black balls in the urn by 1 with
probability
and leaves it the same with probability
Thus the number of stages needed to decrease the number
of black balls to 0 is a negative binomial random variable with
parameters
and ; a “success” here is decreasing
the number of black balls, and
successes are required
before stopping.
Thus the expected number of stages before there
are no more black balls is .
Problem 7.17:
cards, numbered 1 through , is
thoroughly shuffled so that all possible
orderings are
equally likely.
Suppose you are to make
sequntially, where the
one is a guess of the card
in position .
denote the number of correct
If you are not given any information about the correctness of a
guess after you make it, show that, for any strategy, .
Suppose that after each guess you are shown the card that was in
the position in question.
What do you think is the best
Show that under this strategy,
Suppose that you are told after each guess whether you are right
In this case, it can be shown that the strategy that
is one which keeps on guessing the same
card until you are told you are correct and then changes to a new
For this strategy, show that
equal 1 if guess
is correct and 0
otherwise. Since any guess will be correct with probability
, so , it follows that
The best strategy in this case is to simply guess a card that has
not yet appeared. For this strategy, the
will be correct with probability
cards left to choose from.
Suppose that you intend to guess in the order ;
because these labels are arbitrary, choosing this ordering will
yield probabilities that are the same as any other ordering.
Thus we guess 1 until we hit 1, then switch to 2, if/when we hit
2, we switch to 3 and so on.
Now, how many will we get right?
be 1 if you eventually guess card
correctly and 0 otherwise.
That is, if your guesses for 1, then
2, and then 3 are correct, but you hit the end before guessing 4
correct, then
With this definition, the number
of cards you get correct is .
Now, the event that you make it through the first
is equivalent to the event that those
cards appear
within the random sample in same order you guess them.
words, if 1 appears before 2, which appears before 3, you’ll get
all three correct since you switch to 2 after guessing 1 and
switch to 3 after guessing 2.
However, if they appear in the
order 2,1,3, you’ll have passed over 2 while looking for 1, thus
you’ll never guess 2 correctly after guessing 1 correctly and
switching to 2.
Putting this together gives us:
Problem 7.19:
A certain region is inhabited by
distinct types of a
certain kind of insect species, and each insect caught will,
independently of the types of the previous catches, be of type
with probability
Compute the mean number of insects that are caught before the
first type
Compute the mean number of types of insects that are caught
before the first type 1 catch.
The number of catches before an insect of type 1 is caught is
just a geometric random variable with the probability of success
Thus the number of insects caught before this
is , where , so
the expected number caught is just .
The the number of types of insects caught before a type 1 insect
is caught is just .
is just the probability that an insect of type
is caught before one of type 1.
Since we are dealing
with expectations, we can consider this independently of any
we simply keep catching insects, stopping
when either a type
is caught, and
look at the probability that it’s a type .
Problem 7.36:
be the number of 1’s and
the number of 2’s
that occur in
rolls of a fair die. Compute
be the values of each of the
dice rolls.
Then define
if , as separate dice
rolls are independent.
can only be
nonzero if both
and , which is
thus . Thus
It makes sense that the quantities would be negatively correlated as
we would expect the number of rolls of 2 to go down as the number of
rolls of 1 goes up.
Friday, August 13
Discuss exam, continue with another discussion / example of
covariance, introduce conditional expectation.
HW 8 + Solutions
Due Monday, August 16: Problems 7.47, 7.51.
Recommended Problems: Problems 7.45, 7.57, Self Test 7.18, Theory Exercise 7.32.
Bonus Problem: Problem 7.43.
Problem 7.47:
Consider a graph having
vertices labeled
, and suppose that between each of the
pairs of distinct viertices an edge is,
independently, present with probability .
The degree of
vertex , designated as , is the number of edges
that have vertex
as one of its vertices.
What is the distribution of ?
Find , the correlation between
Since each edge occurs independently with probability ,
the number of edges connected to a given node is a binomial with
parameters
( is the number
hence the number of possible edges).
if there is an edge between vertices
and 0 otherwise.
Where the step on the last line follows from noting that
are the only possible values of
are not independent.
Thus the correlation is
Problem 7.51:
The joint density of
is given by
Monday, August 16
Conditional Expectation, moment
mention central
limit theorems and other laws of large numbers.
HW 9 + Solutions (Partial)
Full HW Solutions coming ASAP.
Due Wednesday, August 18: Problems 7.56, 7.69, 7.71, and Theory Exercise 7.29.
Recommended Problems: 7.57, 7.68, 7.70, Theory Exercise 7.43, and Self Test 7.18.
Bonus Problem: Theory Exercise 7.44.
Problem 7.56:
The number of people who enter an elevator on the ground floor is a
Poisson random variable with mean 10.
If there are
above the ground floor and if each person is equally likely to get
off at any one of these
floors, independently of where the
others get off, compute the expected number of stops that the
elevator will make before discharging all of its passengers.
denote the number of people that enter the elevator in
the first place, and let
equal 1 if the elevator stops
and 0 otherwise.
Then we are interested in the
quantity .
Where the probability that no one gets off on floor
probability that all
people get off on a different floor.
where the sum on the left becomes one as it is summing over all the
values of a Poisson distribution with rate .
Plugging in
final answer, .
Another way of seeing this problem is to note that since the rate of
people entering the elevator is a Poisson, the rate of people
getting off at each floor will also be Poisson but with rate
since with probability , each person
will also get off on a given floor.
This method will yield the same
Problem 7.68: (Recommended problem):
The number of accidents that a person has in a given year is a
Poisson random variable with mean .
However, suppose
that the value of
changes from person to person,
being eqaul to 2 for 60 percent of the population and 3 for the
other 40 percent.
If a person is chosen at random, what is the
probability that he will have
0 Accidents?
Exactly 3 accidents?
Exactly 3 accidents, given that he had no accidents the preceding
Problem 7.69 (Homework):
Repeat problem 7.68 when the proportion of the population having a
is equal to .
We have that
(The problem gives a weird
way of s the
cumulative distribution function of an
random variable.)
as the integral over the exponential density is equal to 1.
Similarly to part (a),
Where the last step follows from simplification and noting that
the integral over the gamma distribution is exactly 1.
be the number of accidents the first year, and
be the number of accidents the second year.
Here, we need the condi we can do these as
Where the final value is calculated the same way as in part b.
Problem 7.70 (Recommended problem):
Consider an urn containing a large number of coins and suppose that
each of the coins has some probability
of turning up heads
when it is flipped. However, this value of
varies from
coin to coin.
Suppose that the composition of the urn is such that
if a coin is selected at random from the urn, then its
value can be regarded as being the value of a random variable that
is uniformly distributed over .
If a coin is
selected at random from the urn and flipped twice, compute the
probability that
Both flips are heads.
Problem 7.71: (Homework):
In problem 70, suppose that the coin is tossed
denote the number of heads that occur.
Hint: Make use of the fact that
Conditional on , this value becomes a simple binomial
random variable with parameters
Where the evaluation of the integral procedes directly from the hint
Theoretical Exercise 7.29:
Suppose that
are independent and
identically distributed random variables.
Note: Since I misspecified this in class, I’ll also accept
T.E. 7.43 instead of this one.
However, still do the other one, as
one is recommended and the other is homework, however you see it.
This problem is just a clever rearrangement of expectations.
where the last step follows since each of the
same distribution.
Rearranging gives us
Wednesday, August 18
Ways to use conditional probability, review for final.
Friday, August 20
Final Exam.
Final Exam
We’re h been a quick four weeks!
Like the midterm, you’ll be allowed a single page reference sheet of
your own making, which you have to turn in with the exam (though
you’ll get it back with the graded exam); you make this yourself.
addition, I’ll include the .
This has all the relevant
things you’ll need to include on your reference
sheet are any theorems or concepts we’ve covered in class this quarter
not on there. Again, many of the questions will be based heavily on
reference problems, though don’t count on merely being able to copy
down solutions from
know and understand how to
work these problems.
This exam will cover the following:
Chapter 6:
Sections 6.1-6.5 and 6.7. You won’t be responsible for the
following examples/topics, however: 2e, 2g, 2i, 2j, lognormals & 3d,
5a (silly), and 7c and subsequent material in section 7.
Chapter 7:
Sections 7.1-7.5, 7.7.
Skip example 7.2m and subsequent material in section 7.2
I don’t find the discussion of higher moments in 7.3 to be very
the way I presented in class is more commonly what
Thus just skim section 7.3.
In 7.4, just read the discussion on example 4.c; the math is a bit
involved and not really needed.
In 7.5, skip examples 5e-5i (I presented 5i in class, but you
won’t really need that trick on the exam.
The discussion in 7.5.3
is a nice connection back to conditional probabilities, though I
think it’s a bit random to put some
should all be familiar stuff to most of you (with the exception of
5l and 5m, which are they simplify some of
the other things we’ve done.). Skip section 7.5.4.
In 7.7, Know the basic idea behind moment generating functions and
you won’t have to solve any math problems on
them, but you may have to answer a conceptual question.
More information may appear if people ask questions.}