Expectations
Expectations
Definition
Let X be a discrete rv with set of possible values D and pmf p(x).
The expected value or mean value of X , denoted by E (X ) or
µX , is
X
E (X ) = µX =
x · p(x)
x∈D
Expectations
Expectations
Proposition
If the rv X has a set of possible values D and pmf p(x), then the
expected value of any function h(X ), denoted by E [h(X )] or µhX ,
is computed by
X
E [h(X )] =
h(x) · p(x)
D
Expectations
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X ) − 900 = 700
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X ) − 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).
2. For any constant b, E (X + b) = E (X ) + b.
Expectations
Expectations
Definition
Let X have pmf p(x) and expected value µ. Then the variance of
X, denoted by V (X ) or σX2 , or just σX2 , is
V (X ) =
X
(x − µ)2 · p(x) = E [(X − µ)2 ]
D
The stand deviation (SD) of X is
q
σX = σX2
Expectations
Proposition
X
V (X ) = σ 2 = [
x 2 · p(x)] − µ2 = E (X 2 ) − [E (X )]2
D
Expectations
Expectations
Proposition
If h(X ) is a function of a rv X , then
X
2
V [h(X )] = σh(X
{h(x)−E [h(X )]}2 ·p(x) = E [h(X )2 ]−{E [h(X )]}2
) =
D
If h(X ) is linear, i.e. h(X ) = aX + b for some nonrandom constant
a and b, then
2
2
2
V (aX + b) = σaX
+b = a · σX and σaX +b =| a | ·σX
In particular,
σaX =| a | ·σX , σX +b = σX
Binomial Distribution
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Possible values for X in an n-trial experiment are x = 0, 1, 2, . . . , n.
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Possible values for X in an n-trial experiment are x = 0, 1, 2, . . . , n.
Notation
We use X ∼ Bin(n, p) to indicate that X is a binomial rv based on
n trials with success probability p.
We use b(x; n, p) to denote the pmf of X , and B(x; n, p) to denote
the cdf of X , where
B(x; n, p) = P(X ≤ x) =
x
X
y =0
b(x; n, p)
Binomial Distribution
Theorem
( n
b(x; n, p) =
x
0
p x (1 − p)n−x
x = 0, 1, 2, . . . , n
otherwise
Binomial Distribution
Binomial Distribution
Mean and Variance
Theorem
If X ∼ Bin(n, p), then E (X ) = np, V (X ) = np(1 − p) = npq, and
√
σX = npq (where q = 1 − p).
Hypergeometric Distribution
Hypergeometric Distribution
Assume we are drawing cards from a deck of well-shulffed cards
with replacement, one card per each draw. We do this 5 times and
record whether the outcome is ♠ or not. Then this is a binomial
experiment.
Hypergeometric Distribution
Assume we are drawing cards from a deck of well-shulffed cards
with replacement, one card per each draw. We do this 5 times and
record whether the outcome is ♠ or not. Then this is a binomial
experiment.
If we do the same thing without replacement, then it is NO
LONGER a binomial experiment.
Hypergeometric Distribution
Assume we are drawing cards from a deck of well-shulffed cards
with replacement, one card per each draw. We do this 5 times and
record whether the outcome is ♠ or not. Then this is a binomial
experiment.
If we do the same thing without replacement, then it is NO
LONGER a binomial experiment.
However, if we are drawing from 100 decks of cards without
replacement and record only the first 5 outcomes, then it is
approximately a binomial experiment.
Hypergeometric Distribution
Assume we are drawing cards from a deck of well-shulffed cards
with replacement, one card per each draw. We do this 5 times and
record whether the outcome is ♠ or not. Then this is a binomial
experiment.
If we do the same thing without replacement, then it is NO
LONGER a binomial experiment.
However, if we are drawing from 100 decks of cards without
replacement and record only the first 5 outcomes, then it is
approximately a binomial experiment.
What is the exact model for drawing cards without replacement?
Hypergeometric Distribution
Hypergeometric Distribution
1. The population or set to be sampled consists of N individuals,
objects, or elements (a finite population).
Hypergeometric Distribution
1. The population or set to be sampled consists of N individuals,
objects, or elements (a finite population).
2. Each individual can be characterized as a success (S) or a
failure (F), and there are M successes in the population.
Hypergeometric Distribution
1. The population or set to be sampled consists of N individuals,
objects, or elements (a finite population).
2. Each individual can be characterized as a success (S) or a
failure (F), and there are M successes in the population.
3. A sample of n individuals is selected without replacement in
such a way that each subset of size n is equally likely to be chosen.
Hypergeometric Distribution
1. The population or set to be sampled consists of N individuals,
objects, or elements (a finite population).
2. Each individual can be characterized as a success (S) or a
failure (F), and there are M successes in the population.
3. A sample of n individuals is selected without replacement in
such a way that each subset of size n is equally likely to be chosen.
Definition
For any experiment which satisfies the above 3 conditions, let X =
the number of S’s in the sample. Then X is a hypergeometric
random variable and we use h(x; n, M, N) to denote the pmf
p(x) = P(X = x).
Hypergeometric Distribution
Hypergeometric Distribution
Examples:
Hypergeometric Distribution
Examples:
In the second cards drawing example (without replacement and
totally 52 cards), if we let X = the number of ♠’s in the first 5
draws, then X is a hypergeometric random variable with n = 5,
M = 13 and N = 52.
Hypergeometric Distribution
Examples:
In the second cards drawing example (without replacement and
totally 52 cards), if we let X = the number of ♠’s in the first 5
draws, then X is a hypergeometric random variable with n = 5,
M = 13 and N = 52.
For the pmf, the probability for getting exactly x (x = 0, 1, 2, 3, 4,
or 5) ♠’s is calculated as following:
13
39
x · 5−x
p(x) = P(X = x) =
52
5
Hypergeometric Distribution
Examples:
In the second cards drawing example (without replacement and
totally 52 cards), if we let X = the number of ♠’s in the first 5
draws, then X is a hypergeometric random variable with n = 5,
M = 13 and N = 52.
For the pmf, the probability for getting exactly x (x = 0, 1, 2, 3, 4,
or 5) ♠’s is calculated as following:
13
39
x · 5−x
p(x) = P(X = x) =
52
5
13
39
where x is the number of choices for getting x ♠’s, 5−x
is the
number
of choices for getting the remaining 5 − x non-♠ cards and
52
is
the
total number of choices for selecting 5 cards from 52
5
cards.
Hypergeometric Distribution
Hypergeometric Distribution
Examples:
Hypergeometric Distribution
Examples:
For the same experiment (without replacement and totally 52
cards), if we let X = the number of ♠’s in the first 20 draws, then
X is still a hypergeometric random variable, but with n = 20,
M = 13 and N = 52.
Hypergeometric Distribution
Examples:
For the same experiment (without replacement and totally 52
cards), if we let X = the number of ♠’s in the first 20 draws, then
X is still a hypergeometric random variable, but with n = 20,
M = 13 and N = 52.
However, in this case, all the possible values for X is 0, 1, 2, . . . , 13
and the pmf is
13
39
x · 20−x
p(x) = P(X = x) =
52
20
where 0 ≤ x ≤ 13.
Hypergeometric Distribution
Hypergeometric Distribution
Proposition
If X is the number of S’s in a completely random sample of size n
drawn from a population consisting of M S’s and (N − M) F ’s,
then the probability distribution of X , called the hypergeometric
distribution, is given by
N−M M
x · n−x
P(X = x) = h(x; n, M, N) =
N
n
for x an integer satisfying max(0, n − N + M) ≤ x ≤ min(n, M).
Hypergeometric Distribution
Proposition
If X is the number of S’s in a completely random sample of size n
drawn from a population consisting of M S’s and (N − M) F ’s,
then the probability distribution of X , called the hypergeometric
distribution, is given by
N−M M
x · n−x
P(X = x) = h(x; n, M, N) =
N
n
for x an integer satisfying max(0, n − N + M) ≤ x ≤ min(n, M).
Remark:
If n < M, then the largest x is n. However, if n > M, then the
largest x is M. Therefore we require x ≤ min(n, M).
Hypergeometric Distribution
Proposition
If X is the number of S’s in a completely random sample of size n
drawn from a population consisting of M S’s and (N − M) F ’s,
then the probability distribution of X , called the hypergeometric
distribution, is given by
N−M M
x · n−x
P(X = x) = h(x; n, M, N) =
N
n
for x an integer satisfying max(0, n − N + M) ≤ x ≤ min(n, M).
Remark:
If n < M, then the largest x is n. However, if n > M, then the
largest x is M. Therefore we require x ≤ min(n, M).
Similarly, if n < N − M, then the smallest x is 0. However, if
n > N − M, then the smallest x is n − (N − M). Thus
x ≥ min(0, n − N + M).
Hypergeometric Distribution
Hypergeometric Distribution
Example: (Problem 70)
Hypergeometric Distribution
Example: (Problem 70)
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
Hypergeometric Distribution
Example: (Problem 70)
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
a. What is the probability that exactly 10 of these are from the
second section?
Hypergeometric Distribution
Example: (Problem 70)
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
a. What is the probability that exactly 10 of these are from the
second section?
b. What is the probability that at least 10 of these are from the
second section?
Hypergeometric Distribution
Example: (Problem 70)
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
a. What is the probability that exactly 10 of these are from the
second section?
b. What is the probability that at least 10 of these are from the
second section?
c. What is the probability that at least 10 of these are from the
same section?
Hypergeometric Distribution
Hypergeometric Distribution
Proposition
The mean and variance of the hypergeometric rv X having pmf
h(x; n, M, N) are
N −n
M
M
M
V (X ) =
·n·
· 1−
E (X ) = n ·
N
N −1
N
N
Hypergeometric Distribution
Proposition
The mean and variance of the hypergeometric rv X having pmf
h(x; n, M, N) are
N −n
M
M
M
V (X ) =
·n·
· 1−
E (X ) = n ·
N
N −1
N
N
Remark:
The ratio M
N is the proportion of S’s in the population. If we
replace M
N by p, then we get
Hypergeometric Distribution
Proposition
The mean and variance of the hypergeometric rv X having pmf
h(x; n, M, N) are
N −n
M
M
M
V (X ) =
·n·
· 1−
E (X ) = n ·
N
N −1
N
N
Remark:
The ratio M
N is the proportion of S’s in the population. If we
replace M
N by p,
then we get E (X ) = np and
V (X ) =
N−n
N−1
· np(1 − p).
Hypergeometric Distribution
Proposition
The mean and variance of the hypergeometric rv X having pmf
h(x; n, M, N) are
N −n
M
M
M
V (X ) =
·n·
· 1−
E (X ) = n ·
N
N −1
N
N
Remark:
The ratio M
N is the proportion of S’s in the population. If we
replace M
N by p,
then we get E (X ) = np and
V (X ) = N−n
N−1 · np(1 − p).
Recall the mean and variance for a binomial rv is np and np(1 − p).
Hypergeometric Distribution
Proposition
The mean and variance of the hypergeometric rv X having pmf
h(x; n, M, N) are
N −n
M
M
M
V (X ) =
·n·
· 1−
E (X ) = n ·
N
N −1
N
N
Remark:
The ratio M
N is the proportion of S’s in the population. If we
replace M
N by p,
then we get E (X ) = np and
V (X ) = N−n
N−1 · np(1 − p).
Recall the mean and variance for a binomial rv is np and np(1 − p).
We see that the mean for binomial and hypergeometric rv’s are
equal, while the variances differ by the factor (N − n)/(N − 1).
Hypergeometric Distribution
Hypergeometric Distribution
Example (Problem 70) continued:
Hypergeometric Distribution
Example (Problem 70) continued:
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
Hypergeometric Distribution
Example (Problem 70) continued:
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
d. What are the mean value and standard deviation of the number
of projects among these 15 that are from the second section?
Hypergeometric Distribution
Example (Problem 70) continued:
An instructor who taught two sections of engineering statistics last
term, the first with 20 students and the second with 30, decided to
assign a term project. After all projects had been turned in, the
instructor randomly ordered them before grading. Consider the
first 15 graded projects.
d. What are the mean value and standard deviation of the number
of projects among these 15 that are from the second section?
e. What are the mean value and standard deviation of the number
of projects not among these 15 that are from the second section?
Negative Binomial Distribution
Negative Binomial Distribution
Consider the card drawing example again. This time, we still draw
cards from a deck of well-shulffed cards with replacement, one card
per each draw. However, we keep drawing until we get 5 ♠’s. Let
X = the number of draws which do not give us a ♠, then X is NO
LONGER a binomial random variable, but a negative binomial
random variable.
Negative Binomial Distribution
Negative Binomial Distribution
1. The experiment consists of a sequence of independent trials.
Negative Binomial Distribution
1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either s success (S) or a failure (F).
Negative Binomial Distribution
1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either s success (S) or a failure (F).
3. The probability of success is constant from trial to trial, so
P(S on trial i) = p for i = 1, 2, 3, . . . .
Negative Binomial Distribution
1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either s success (S) or a failure (F).
3. The probability of success is constant from trial to trial, so
P(S on trial i) = p for i = 1, 2, 3, . . . .
4. The experiment continues (trials are performed) until a total of
r successes have been observed, where r is a specified positive
integer.
Negative Binomial Distribution
1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either s success (S) or a failure (F).
3. The probability of success is constant from trial to trial, so
P(S on trial i) = p for i = 1, 2, 3, . . . .
4. The experiment continues (trials are performed) until a total of
r successes have been observed, where r is a specified positive
integer.
Definition
For any experiment which satisfies the above 4 conditions, let X =
the number of failures that precede thr r th success. Then X is a
negative binomial random variable and we use nb(x; r , p) to
denote the pmf p(x) = P(X = x).
Negative Binomial Distribution
Negative Binomial Distribution
Remark:
1. In some sources, the negative binomial rv is taken to be the
number of trials X + r rather than the number of failures.
Negative Binomial Distribution
Remark:
1. In some sources, the negative binomial rv is taken to be the
number of trials X + r rather than the number of failures.
2. If r = 1, we call X a geometric random variable. The pmf for
X is then the familiar one
nb(x; 1, p) = (1 − p)x p
x = 0, 1, 2, . . .
Negative Binomial Distribution
Negative Binomial Distribution
Proposition
The pmf of the negative binomial rv X with parameters r =
number of S’s and p = P(S) is
x +r −1
nb(x; r , p) =
· p r (1 − p)x
r −1
Then mean and variance for X are
E (X ) =
respectively
r (1 − p)
r (1 − p)
and V (X ) =
,
p
p2
Negative Binomial Distribution
Negative Binomial Distribution
Example: (Problem 78)
Individual A has a red die and B has a green die (both fair). If
they each roll until they obtain five “doubles”
(1 − 1, 2 − 2, . . . , 6 − 6), what is the pmf of X = the total number
of times a die is rolled? What are E (X ) and V (X )?
Poisson Distribution
Poisson Distribution
Consider the following random variables:
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
2. The number of drivers who travel between Salt Lake City and
Sandy during each day.
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
2. The number of drivers who travel between Salt Lake City and
Sandy during each day.
3. The number of trees in each square mile in a forest.
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
2. The number of drivers who travel between Salt Lake City and
Sandy during each day.
3. The number of trees in each square mile in a forest.
None of them are binomial, hypergeometric or negative binomial
random variables.
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
2. The number of drivers who travel between Salt Lake City and
Sandy during each day.
3. The number of trees in each square mile in a forest.
None of them are binomial, hypergeometric or negative binomial
random variables.
In fact, the experiments associated with above random variables
DO NOT involve trials.
Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency
room in each hour.
2. The number of drivers who travel between Salt Lake City and
Sandy during each day.
3. The number of trees in each square mile in a forest.
None of them are binomial, hypergeometric or negative binomial
random variables.
In fact, the experiments associated with above random variables
DO NOT involve trials. We use Poisson distribution to model the
experiment for occurence of events of some type over time or area.
Poisson Distribution
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
1. The value λ is frequently a rate per unit time or per unit area.
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
1. The value λ is frequently a rate per unit time or per unit area.
2. e is the base of the natural logarithm system.
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
1. The value λ is frequently a rate per unit time or per unit area.
2. e is the base of the natural
logarithm system.
P
3. It is guaranteed that ∞
p(x;
λ) = 1.
x=0
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
1. The value λ is frequently a rate per unit time or per unit area.
2. e is the base of the natural
logarithm system.
P
3. It is guaranteed that ∞
p(x;
λ) = 1.
x=0
∞
X λx
λ2 λ3
e =1+λ+
+
+ ··· =
2!
3!
x!
λ
x=0
Poisson Distribution
Poisson Distribution
Example:
The red blood cell (RBC) density in blood is estimated by means of
a hematometer. A blood sample is thoroughly mixed with a saline
solution, and then pipetted onto a slide. The RBC’s are counted
under a microscope through a square grid. Because the solution is
throughly mixed, the RBC’s have an equal chance of being in a
particular square in the grid. It is known that the number of cells
counted in a given square follows a Poisson distribution and the
parameter λ for certain blood sample is believed to be 1.5.
Poisson Distribution
Example:
The red blood cell (RBC) density in blood is estimated by means of
a hematometer. A blood sample is thoroughly mixed with a saline
solution, and then pipetted onto a slide. The RBC’s are counted
under a microscope through a square grid. Because the solution is
throughly mixed, the RBC’s have an equal chance of being in a
particular square in the grid. It is known that the number of cells
counted in a given square follows a Poisson distribution and the
parameter λ for certain blood sample is believed to be 1.5.
Then what is the probability that there is no RBC in a given
square?
Poisson Distribution
Example:
The red blood cell (RBC) density in blood is estimated by means of
a hematometer. A blood sample is thoroughly mixed with a saline
solution, and then pipetted onto a slide. The RBC’s are counted
under a microscope through a square grid. Because the solution is
throughly mixed, the RBC’s have an equal chance of being in a
particular square in the grid. It is known that the number of cells
counted in a given square follows a Poisson distribution and the
parameter λ for certain blood sample is believed to be 1.5.
Then what is the probability that there is no RBC in a given
square?
What is the probability for a square containing exactly 2 RBC’s?
Poisson Distribution
Example:
The red blood cell (RBC) density in blood is estimated by means of
a hematometer. A blood sample is thoroughly mixed with a saline
solution, and then pipetted onto a slide. The RBC’s are counted
under a microscope through a square grid. Because the solution is
throughly mixed, the RBC’s have an equal chance of being in a
particular square in the grid. It is known that the number of cells
counted in a given square follows a Poisson distribution and the
parameter λ for certain blood sample is believed to be 1.5.
Then what is the probability that there is no RBC in a given
square?
What is the probability for a square containing exactly 2 RBC’s?
What is the probability for a square containing at most 2 RBC’s?
Poisson Distribution
Poisson Distribution
Proposition
If X has a Poisson distribution with parameter λ, then
E (X ) = V (X ) = λ.
Poisson Distribution
Proposition
If X has a Poisson distribution with parameter λ, then
E (X ) = V (X ) = λ.
We see that the parameter λ equals to the mean and variance of
the Poisson random variable X .
Poisson Distribution
Proposition
If X has a Poisson distribution with parameter λ, then
E (X ) = V (X ) = λ.
We see that the parameter λ equals to the mean and variance of
the Poisson random variable X .
e.g. for the previous example, the expected number of RBC’s per
square is thus 1.5 and the variance is also 1.5.
Poisson Distribution
Proposition
If X has a Poisson distribution with parameter λ, then
E (X ) = V (X ) = λ.
We see that the parameter λ equals to the mean and variance of
the Poisson random variable X .
e.g. for the previous example, the expected number of RBC’s per
square is thus 1.5 and the variance is also 1.5.
In practice, the parameter usually is unknown to us. However, we
can use the sample mean to estimate it. For example, if we
observed 15 RBC’s over 10 squares, then we can use x̄ = 15
10 = 1.5
to estimate λ.
Poisson Distribution
Poisson Distribution
Poisson Process: the occurrence of events over time.
Poisson Distribution
Poisson Process: the occurrence of events over time.
1. There exists a parameter α > 0 such that for any short time
interval of length ∆t, the probability that exactly one event is
received is α · ∆t + o(∆t).
Poisson Distribution
Poisson Process: the occurrence of events over time.
1. There exists a parameter α > 0 such that for any short time
interval of length ∆t, the probability that exactly one event is
received is α · ∆t + o(∆t).
2. The probability of more than one event being received during
∆t is o(∆t) [which, along with Assumption 1, implies that the
probability of no events during ∆t] is 1 − α · ∆t − o(∆t)].
Poisson Distribution
Poisson Process: the occurrence of events over time.
1. There exists a parameter α > 0 such that for any short time
interval of length ∆t, the probability that exactly one event is
received is α · ∆t + o(∆t).
2. The probability of more than one event being received during
∆t is o(∆t) [which, along with Assumption 1, implies that the
probability of no events during ∆t] is 1 − α · ∆t − o(∆t)].
3. The number of events received during the time interval ∆t is
independent of the number received prior to this time interval.
Poisson Distribution
Poisson Distribution
Proposition
Let Pk (t) denote the probability that k events will be observed
during any particular time interval of length t. Then
Pk (t) = e −αt ·
(αt)k
.
k!
In words, the number of events during a time interval of length t is
a Poisson rv with parameter λ = αt. The expected number of
events during any such time interval is then αt, so the expected
number during a unit interval of time is α.
Poisson Distribution
Poisson Distribution
Example: (Problem 92)
Automobiles arrive at a vehicle equipment inspection station
according to a Poisson process with rate α = 10 per hour. Suppose
that with probability 0.5 an arriving vehicle will have no equipemnt
violations.
Poisson Distribution
Example: (Problem 92)
Automobiles arrive at a vehicle equipment inspection station
according to a Poisson process with rate α = 10 per hour. Suppose
that with probability 0.5 an arriving vehicle will have no equipemnt
violations.
a. What is the probability that exactly ten arrive during the hour
and all ten have no violations?
Poisson Distribution
Example: (Problem 92)
Automobiles arrive at a vehicle equipment inspection station
according to a Poisson process with rate α = 10 per hour. Suppose
that with probability 0.5 an arriving vehicle will have no equipemnt
violations.
a. What is the probability that exactly ten arrive during the hour
and all ten have no violations?
b. For any fixed y ≥ 10, what is the probability that y arrive
during the hour, of which ten have no violations?
Poisson Distribution
Example: (Problem 92)
Automobiles arrive at a vehicle equipment inspection station
according to a Poisson process with rate α = 10 per hour. Suppose
that with probability 0.5 an arriving vehicle will have no equipemnt
violations.
a. What is the probability that exactly ten arrive during the hour
and all ten have no violations?
b. For any fixed y ≥ 10, what is the probability that y arrive
during the hour, of which ten have no violations?
c. What is the probability that ten “no-violation” cars arrive during
the next 45 minutes?
Poisson Distribution
Poisson Distribution
In some sense, the Poisson distribution can be recognized as the
limit of a binomial experiment.
Proposition
Suppose that in the binomial pmf b(x; n, p), we let n → ∞ and
p → 0 in such a way that np approaches a value λ > 0. Then
b(x; n, p) → p(x; λ).
Poisson Distribution
In some sense, the Poisson distribution can be recognized as the
limit of a binomial experiment.
Proposition
Suppose that in the binomial pmf b(x; n, p), we let n → ∞ and
p → 0 in such a way that np approaches a value λ > 0. Then
b(x; n, p) → p(x; λ).
This tells us in any binomial experiment in which n is
large and p is small, b(x; n, p) ≈ p(x; λ), where λ = np.
Poisson Distribution
In some sense, the Poisson distribution can be recognized as the
limit of a binomial experiment.
Proposition
Suppose that in the binomial pmf b(x; n, p), we let n → ∞ and
p → 0 in such a way that np approaches a value λ > 0. Then
b(x; n, p) → p(x; λ).
This tells us in any binomial experiment in which n is
large and p is small, b(x; n, p) ≈ p(x; λ), where λ = np.
As a rule of thumb, this approximation can safely be applied if
n > 50 and np < 5.
Poisson Distribution
Poisson Distribution
Example 3.40:
If a publisher of nontechnical books takes great pains to ensure
that its books are free of typographical errors, so that the
probability of any given page containing at least one such error is
0.005 and errors are independent from page to page, what is the
probability that one of its 400-page novels will contain exactly one
page with errors?
Poisson Distribution
Example 3.40:
If a publisher of nontechnical books takes great pains to ensure
that its books are free of typographical errors, so that the
probability of any given page containing at least one such error is
0.005 and errors are independent from page to page, what is the
probability that one of its 400-page novels will contain exactly one
page with errors?
Let S denote a page containing at least one error, F denote an
error-free page and X denote the number of pages containing at
least one error. Then X is a binomial rv, and
P(X = 1) = b(1; 400, 0.005) ≈ p(1; 400 · 0.005) = p(1; 2) =
= 0.270671
e −2 (2)
1!
Poisson Distribution
Poisson Distribution
A proof for b(x; n, p) → p(x; λ) as n → ∞ and p → 0 with
np → λ.
n!
p x (1 − p)n−x
x!(n − x)!
n!
n(n − 1) · · · (n − x + 1) x
lim
p x = lim
p
n→∞ x!(n − x)!
n→∞
x!
(np)[(n − 1)p] · · · [(n − x + 1)p]
= lim
n→∞
x!
λx
=
x!
np n−x
n−x
lim (1 − p)
= lim {1 −
}
n→∞
n→∞
n
λ
= lim {1 − }n−x
n→∞
n
−λ
=e
b(x; n, p) =