Stochastic Processes - lesson 3

Bo Friis Nielsen

Institute of Mathematical Modelling

Technical University of Denmark

2800 Kgs. Lyngby – Denmark

Email: bfn@imm.dtu.dk

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

? Lack of memory property

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

? Lack of memory property

• Continuous random variables

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

? Lack of memory property

• Continuous random variables

� Exponential distribution

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

? Lack of memory property

• Continuous random variables

� Exponential distribution

� Moments for continuous random variables

Bo Friis Nielsen – 12/9-2000 2C04141

OutlineOutline

• Discrete random variables (from last lesson)

� Geometric distribution

? Lack of memory property

• Continuous random variables

� Exponential distribution

� Moments for continuous random variables

• Reading recommendations

Bo Friis Nielsen – 12/9-2000 3C04141

• Bernoulli distribution

� Two possibilities succes(X = 1)/failure(X = 0)

• Binomial distribution

� Number of succeses in a sequence of independent

Bernoulli trials

� pdf (Probability Density Function)

? f(x) = P{X = x} =

n

x

px(1− p)n−x

� Mean and variance

? E(X) = np V (X) = np(1− p)

Bo Friis Nielsen – 12/9-2000 4C04141

Poisson distributionPoisson distribution

• Number of faults/accidents/occurrences

• X ∈ P(λ) ⇔ P{X = x} = f(x) = λx

x!e−λ

• E(X) = V (X) = λ

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

• Possible values

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

• Possible values

� If we count all trials

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

• Possible values

� If we count all trials 1, . . . ,∞

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

• Possible values

� If we count all trials 1, . . . ,∞

� If we only count the failures

Bo Friis Nielsen – 12/9-2000 5C04141

Geometric distributionGeometric distribution

• How many items produced to get one passing the quality

control

• Number of days to get rain

• Sequence of failures until the first success - sequence of

Bernoulli trials

• Possible values

� If we count all trials 1, . . . ,∞

� If we only count the failures 0, . . . ,∞

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

• The probability of having x− 1 failures before the first

succes is

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

• The probability of having x− 1 failures before the first

succes is

P{x trials} =

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

• The probability of having x− 1 failures before the first

succes is

P{x trials} = P{x− 1 failures and then a succes} =

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

• The probability of having x− 1 failures before the first

succes is

P{x trials} = P{x− 1 failures and then a succes} =

(1− p)x−1p

Bo Friis Nielsen – 12/9-2000 6C04141

Derivation of probability density function -

counting all trials

Derivation of probability density function -

counting all trials

• Let’s look at the sequence FFFFS with probability

(1− p)4p

• A general sequence will be like FFF. . .FFS

• The probability of having x− 1 failures before the first

succes is

P{x trials} = P{x− 1 failures and then a succes} =

(1− p)x−1p

• The cumulative distribution can be found to be

� F (x) =∑x

t=1(1− p)t−1p = 1− (1− p)x

Bo Friis Nielsen – 12/9-2000 7C04141

• E(X) = 1p

V (X) = 1−p

p2

Bo Friis Nielsen – 12/9-2000 7C04141

• E(X) = 1p

V (X) = 1−p

p2

The expressions when counting only failuresThe expressions when counting only failures

• f(x) = (1− p)xp

• F (x) = 1− (1− p)x+1

• E(X) = 1−p

p

• V (X) = 1−p

p2

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

= (1−p)x+n

(1−p)n

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

= (1−p)x+n

(1−p)n= (1− p)x

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

= (1−p)x+n

(1−p)n= (1− p)x = P{X > x}

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

= (1−p)x+n

(1−p)n= (1− p)x = P{X > x}

• That is, the probability of exceeding x + n having reached

n is the same as the property of exceeding x starting from

the beginning.

Bo Friis Nielsen – 12/9-2000 8C04141

The memoryless propertyThe memoryless property

• What will happen to the distribution knowing that n

failures already occured?

• That is we have been waiting for an empty cab and have

experienced 7 occupied

• Formally

P{X > x + n|X > n} = P{X>x+n∩X>n}P{X>n}

= P{X>x+n}P{X>n}

=1−(1−(1−p)x+n)1−(1−(1−p)n)

= (1−p)x+n

(1−p)n= (1− p)x = P{X > x}

• That is, the probability of exceeding x + n having reached

n is the same as the property of exceeding x starting from

the beginning. In other words no aging

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

• The probability density function in the continuous case

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

• The probability density function in the continuous case

� f(x) = F ′(x)

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

• The probability density function in the continuous case

� f(x) = F ′(x)

� Probabilistic interpretation

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

• The probability density function in the continuous case

� f(x) = F ′(x)

� Probabilistic interpretation

P{x ≤ X ≤ x + dx} = f(x)dx

Bo Friis Nielsen – 12/9-2000 9C04141

Continuous random variablesContinuous random variables

• Cumulative distribution function, once again

� P{X ≤ x} = F (x)

• The probability density function in the continuous case

� f(x) = F ′(x)

� Probabilistic interpretation

P{x ≤ X ≤ x + dx} = f(x)dx or F (x) =∫ x−∞ f(t)dt

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

• F (x) = 1− e−λx

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

• F (x) = 1− e−λx ⇔ X ∈ exp(λ)

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

• F (x) = 1− e−λx ⇔ X ∈ exp(λ)

• f(x) = λe−λx

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

• F (x) = 1− e−λx ⇔ X ∈ exp(λ)

• f(x) = λe−λx

• The exponential is without memory like the geometric

distribution

Bo Friis Nielsen – 12/9-2000 10C04141

The continuous parallel to the geometric

distribution

the exponential distribution

The continuous parallel to the geometric

distribution

the exponential distribution

• F (x) = 1− e−λx ⇔ X ∈ exp(λ)

• f(x) = λe−λx

• The exponential is without memory like the geometric

distribution

• The geometric/exponential distributions are the unique

memoryless distributions

Bo Friis Nielsen – 12/9-2000 11C04141

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Exponential density with mean=1

’exppdf.lst’

Bo Friis Nielsen – 12/9-2000 11C04141

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Exponential density with mean=1

’exppdf.lst’

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Exponential cumulative distribution with mean=1

’expcdf.lst’

Bo Friis Nielsen – 12/9-2000 12C04141

The normal densityThe normal density

0

0.05

0.1

0.15

0.2

0.25

0.3

-4 -3 -2 -1 0 1 2 3 4

Normal density with mean=0 and variance=1

’normpdf.lst’

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx = 1

λby partial integration

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx = 1

λby partial integration

• The variance of continuous random variables

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx = 1

λby partial integration

• The variance of continuous random variables

� V (X) =∫∞−∞(x− µ)2f(x)dx

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx = 1

λby partial integration

• The variance of continuous random variables

� V (X) =∫∞−∞(x− µ)2f(x)dx = σ2

Bo Friis Nielsen – 12/9-2000 13C04141

Moments revisitedMoments revisited

• The mean for continuous random variables

� E(X) =∫∞−∞ xf(x)dx = µ

�∫∞0 xλe−λxdx = 1

λby partial integration

• The variance of continuous random variables

� V (X) =∫∞−∞(x− µ)2f(x)dx = σ2

� X ∈ exp(λ) V (X) = 1λ2

Bo Friis Nielsen – 12/9-2000 14C04141

Small examples using the exponential

distribution

Small examples using the exponential

distributionThe time between two consecutive attempts access attempts

to a popular web site can be adequately described by an

exponential distribution with mean 5 seconds.

Bo Friis Nielsen – 12/9-2000 14C04141

Small examples using the exponential

distribution

Small examples using the exponential

distributionThe time between two consecutive attempts access attempts

to a popular web site can be adequately described by an

exponential distribution with mean 5 seconds.

• What is the variance of the time between two consecutive

access attempts?

Bo Friis Nielsen – 12/9-2000 14C04141

Small examples using the exponential

distribution

Small examples using the exponential

distributionThe time between two consecutive attempts access attempts

to a popular web site can be adequately described by an

exponential distribution with mean 5 seconds.

• What is the variance of the time between two consecutive

access attempts?

• What is the probability that the time between two

consecutive attempts will be less than 1 second?

Bo Friis Nielsen – 12/9-2000 14C04141

Small examples using the exponential

distribution

Small examples using the exponential

distributionThe time between two consecutive attempts access attempts

to a popular web site can be adequately described by an

exponential distribution with mean 5 seconds.

• What is the variance of the time between two consecutive

access attempts?

• What is the probability that the time between two

consecutive attempts will be less than 1 second?

• Knowing that no attempt has been made within the last

minute, what is the probability that there won’t be a new

attempt within the next second?

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

• X : the time between two consecutive attempts

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

• X : the time between two consecutive attempts

• X ∈ exp(λ)

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

• X : the time between two consecutive attempts

• X ∈ exp(λ)

• E(X) = 1λ

= 5s

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

• X : the time between two consecutive attempts

• X ∈ exp(λ)

• E(X) = 1λ

= 5s ⇒ λ = 0.2s−1

Bo Friis Nielsen – 12/9-2000 15C04141

ModelModel

• X : the time between two consecutive attempts

• X ∈ exp(λ)

• E(X) = 1λ

= 5s ⇒ λ = 0.2s−1

• X ∈ exp(0.2)

Bo Friis Nielsen – 12/9-2000 16C04141

What is the variance of the time between

two consecutive access attempts?

What is the variance of the time between

two consecutive access attempts?

Bo Friis Nielsen – 12/9-2000 16C04141

What is the variance of the time between

two consecutive access attempts?

What is the variance of the time between

two consecutive access attempts?

• V (X) = 1λ2

Bo Friis Nielsen – 12/9-2000 16C04141

What is the variance of the time between

two consecutive access attempts?

What is the variance of the time between

two consecutive access attempts?

• V (X) = 1λ2 = 0.04

To exponential distribution

Bo Friis Nielsen – 12/9-2000 17C04141

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

To exponential distribution

Bo Friis Nielsen – 12/9-2000 17C04141

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

• P{X ≤ 1}

To exponential distribution

Bo Friis Nielsen – 12/9-2000 17C04141

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

• P{X ≤ 1} = F (1) = 1− exp−0.2·1

To exponential distribution

Bo Friis Nielsen – 12/9-2000 17C04141

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

What is the probability that the time

between two consecutive attempts will be

less than 1 second?

• P{X ≤ 1} = F (1) = 1− exp−0.2·1 = 0.1813

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60}

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

= P{X>61}P{X>60}

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

= P{X>61}P{X>60}

=1−(1−exp−0.02·61)1−(1−exp−0.02·60)

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

= P{X>61}P{X>60}

=1−(1−exp−0.02·61)1−(1−exp−0.02·60)

= exp−0.02·61

exp−0.02·60

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

= P{X>61}P{X>60}

=1−(1−exp−0.02·61)1−(1−exp−0.02·60)

= exp−0.02·61

exp−0.02·60 = exp−0.02

Bo Friis Nielsen – 12/9-2000 18C04141

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

Knowing that no attempt has been made

within the last minute, what is the

probability that there won’t be a new

attempt within the next second?

• P{X > 61|X > 60} = P{X>61∩X>60}P{X>60}

= P{X>61}P{X>60}

=1−(1−exp−0.02·61)1−(1−exp−0.02·60)

= exp−0.02·61

exp−0.02·60 = exp−0.02 = 0.8187

• Which is exactly the lack of memory property

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions.

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

� E(Y ) = E(g(X)) =∫∞−∞ g(x)f(x)dx

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

� E(Y ) = E(g(X)) =∫∞−∞ g(x)f(x)dx

• One important example

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

� E(Y ) = E(g(X)) =∫∞−∞ g(x)f(x)dx

• One important example

• E(X2) =∫∞−∞ x2f(x)dx

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

� E(Y ) = E(g(X)) =∫∞−∞ g(x)f(x)dx

• One important example

• E(X2) =∫∞−∞ x2f(x)dx = V (X) + (E(X))2

Bo Friis Nielsen – 12/9-2000 19C04141

A unified and more general look at momentsA unified and more general look at moments

• Sometimes convenient to calculate similar

integral/expectation for other functions. e.g. the mean of

the random variable Y = g(X)

� E(Y ) = E(g(X)) =∫∞−∞ g(x)f(x)dx

• One important example

• E(X2) =∫∞−∞ x2f(x)dx = V (X) + (E(X))2 = σ2 + µ2

Bo Friis Nielsen – 12/9-2000 20C04141

Reading recommendationsReading recommendations

• for Friday September 8 and Tuesday September 12, read

Chapter 2 lightly, Chapter 3 section: 3.1,3.2,3.3,3.5,

Chapter 4 section: 4.1,4.3,4.4

• For Friday September 15, read 3.8-3.10.

• News on exercises for Friday, not later than Thursday 3pm

on the net.