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David Tenenbaum GEOG 090 UNC-CH Spring 2005

Probability Some Definitions

Probability Refers to the likelihood that something

(an event) will have a certain outcome

An Event Any phenomenon you can observe that canhave more than one outcome (e.g. flipping a coin)

An Outcome Any unique condition that can be the

result of an event (e.g. the available outcomes when

flipping a coin are heads and tails), a.k.a. simple events

or sample points Sample Space The set of all possible outcomes

associated with an event (e.g. the sample space for

flipping a coin includes heads and tails)

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For example, suppose we have a data set where in six

cities, we count the number of malls located in that city

present:City # of Malls

1 1

2 4

3 4

4 45 2

6 3

We might wonder if we randomly pick one of these sixcities, what is the chance that it will have n malls?

Probability An Example

Outcome #1

Outcome #2

Outcome #3

Outcome #4

Sample

Space

Each

count of

the # of

malls in

a city is

an event

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What we have here is a random variable

defined as variable X whose range of values xiare

sampled randomly from a population

To put this another way, a random variable is a

function defined on the sample space thismeans that we are interested in all the possible

outcomes The question was: If we randomly pick one of

the six cities, what is the chance that it will have

n malls?

Random Variables and

Probability Functions

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4/35David Tenenbaum GEOG 090 UNC-CH Spring 2005

To answer this question, we need to form a

probability function (a.k.a. probability

distribution) from the sample space that gives allvalues of a random variable and their probabilities

A probability distribution expresses the relativenumber of times we expect a random variable to

assume each and every possible value

We either base a probability function on either a

very large empirically-gathered set of outcomes,

or else we determine the shape of a probabilityfunction mathematically

Random Variables and

Probability Functions

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Here, the values of xiare drawn from the four outcomes,

and their probabilities are the number of events with each

outcome divided by the total number of events:City # of Malls

1 1

2 4

3 4

4 45 2

6 3

Probability An Example, Part II

Outcome #1

Outcome #2

Outcome #3

Outcome #4

xi P(x

i)

1 1/6 = 0.167

2 1/6 = 0.167

3 1/6 = 0.1674 3/6 = 0.5

The probability of an outcome P(xi) =

# of times an outcome occurred

Total number of events

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We can plot this probability distribution as a probability

mass function:

Probability An Example, Part III

xi P(x

i)

1 1/6 = 0.167

2 1/6 = 0.1673 1/6 = 0.167

4 3/6 = 0.50

0.25

0.50

P(x

i)

1 2 3 4

xi

This plot uses thin lines to denote that the probabilities aremassed at discrete values of this random variable

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The random variable from our example is a

discrete random variable, because it has a finite

number of values (i.e. a city can have 1 or 2malls, but it cannot have 1.5 malls)

Any variable that is generated by counting awhole number of things is likely to be a discrete

variable (e.g. # of coin tosses in a row with heads,

questionnaire responses where one of a set ofordinal categories must be chosen, etc.)

A discrete random variable can be described by aprobability mass function

Discrete Random Variables

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Probability mass functions have the following

rules that dictate their possible values:

1. The probability of any outcome must be greaterthan or equal to zero and must also be less than or

equal to one, i.e.0 P(x

i) 1 for i = {1, 2, 3, , k-1, k}

2. The sum of all probabilities in the sample spacemust total one, i.e.

Probability Mass Functions

P(xi) = 1i=1

i=k

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The remainder of todays lecture will introduce

three kinds of discrete probability distributions

that are useful for us to examine when learningabout statistics:

1. The Uniform Distribution2. The Binomial Distribution

3. The Poisson Distribution Each of these probability distributions is

appropriately applied in certain situations and to

particular phenomena

Discrete Probability Distributions

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The uniform distribution describes the situation

where the probability of all outcomes is the same

The Uniform Distribution

Expressed as an equation,given n possible outcomes

for an event, the probabilityof any outcome is P(xi) =

1/n, e.g. if we are flipping a

coin and we find that headsand tails are equally likely

outcomes, then:P(xheads

) = 1/2 = P(xtails

)

0

0.25

0.50

P(xi)

headsxi

tails

A uniform probability

mass function

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While the idea of a uniform distribution might

seem a little simplistic and perhaps useless, it

actually is well applied in two situations:1. When the probabilities are of each and every

possible outcome are truly equal (e.g. the cointoss)

2. When we have no prior knowledge of how a

variable is distributed (i.e. when we are dealing

with complete uncertainty), we first distribution

we should use is uniform, because it makes noassumptions about the distribution


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While truly uniformly distributed geographic

phenomena are somewhat rare (remember

Toblers law, which implies variation from theuniform if it is true), we often encounter the

situation of not knowing how something is

distributed until we sample it

It is in the latter case, when we are resisting

making assumptions about the distribution ofsome geographic phenomenon that we usually

apply the uniform distribution as a sort of nullhypothesis of distribution


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xi

For example, supposing we wanted to wanted to predictthe direction of the prevailing wind at some location

(expressing it in terms of a cardinal direction), and we

had no prior knowledge of the weather systemstendencies in the area, we would have to begin with the

idea that

P(xNorth

) =

P(xEast

) =

P(xSouth) =

P(xWest

) =

until we had an opportunity to sample and establish sometendency in the wind pattern based on those observations

0

0.25

P(xi)

N E S W

0.125

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The binomial distribution provides information

about the probability of the repetition of events

when there are only two possible outcomes, e.g.heads or tails, left or right, success or failure, rain

or no rain any nominal data situation where

there are only two categories / outcomes possible

The binomial distribution is useful for describing

when the same event is repeated over and over,characterizing the probability of a proportion of

the events having a certain outcome over aspecified number of events

The Binomial Distribution

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A binomial distribution is produced by a set of

Bernoulli trials, named after Jacques Bernoulli, a

17th

Century Swiss mathematician who formulateda version of the law of large numbers for

independent trials

The law of large numbers is at the heart of

probability theory, and it states that given enough

observed events, the observed probability shouldapproach the theoretical values drawn from

probability distributions (e.g. enough coin tossesshould approach the P=0.5 value for the outcomes)

The Binomial Distribution

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A set of Bernoulli trials is the way tooperationally test the law of large numbers using

an event that has two possible outcomes:

1. N independent trials of an experiment (i.e. an

event like a coin toss) are performed; using the

word independent here stipulates that the results

of one trial do not influence the result of the next

2. Every trial must have the same set of possibleoutcomes (heads and tails must be the only

available results of coin tosses using other sortsof experiments, this is a less trivial issue)

Bernoulli Trials

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Bernoulli trials cont.3. The probability of each outcome must be the

same for all trials, i.e. P(x

i) must be the sameeach time for both x

ivalues

4. The resulting random variable is determined by

the number of successes in the trials (where we

define successes to be one of the two available

outcomes)

We will use the notation p = the probability of

success in a trial and q = (p 1) as the probabilityof failure in a trial; p + q = 1

Bernoulli Trials

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Suppose on a series of successive days, we will recordwhether or not it rains in Chapel Hill

We will denote the 2 outcomes using R when it rains and

N when it does not rainn Possible Outcomes # of Rain Days P(# of Rain Days)

1 R 1 p p

N 0 (1 - p) q

2 RR 2 p2 p2

RN NR 1 2[p*(1 p)] 2pq

NN 0 (1 p)2 q2

3 RRR 3 p3 p3

RRN RNR NRR 2 3[p2 *(1 p)] 3p2q

NNR NRN RNN 1 3[p*(1 p)2] 3pq2

NNN 0 (1 p)3 q3

Bernoulli Trials An Example

>1 successive events

uses the multiplicative

rule (intersection) tocalculate the probability

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n Possible Outcomes # of Rain Days P(# of Rain Days)

1 R 1 p = 0.2

N 0 q = 0.8

2 RR 2 p2 = 0.04

RN NR 1 2pq = 0.32

NN 0 q2 = 0.64

3 RRR 3 p3 = 0.008

RRN RNR NRR 2 3p2q = 0.096

NNR NRN RNN 1 3pq2 = 0.384

NNN 0 q3 = 0.512

If we have a value for P(R) = p, we can substitute it into

the above equations to get the probability of each

outcome from a series of successive samples, e.g.suppose p=0.2 (and therefore q=0.8, since p + q = 1)

Bernoulli Trials An Example

= 1

= 1

= 1

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Bernoulli Trials

1 event, = (p + q)1 = p + q

2 events, = (p + q)2 = p2 + 2pq + q2

3 events, = (p + q)3

= p3 + 3p2q + 3pq2 + q3

4 events, = (p + q)4= p4 + 4p3q + 6p2q2 + 4pq3 + q4

Source: Earickson, RJ, and Harlin, JM. 1994. Geographic Measurement and Quantitative

Analysis. USA: Macmillan College Publishing Co., p. 132.

A graphical representation: probability# of successes

The sum of the probabilities can be expressed using thebinomial expansion of (p + q)n, where n = # of events

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We can provide a general formula for calculatingthe probability of x successes, given n trials and a

probability p of success:

Bernoulli Trials

P(x) = C(n,x) * px * (1 - p)n - x

where C(n,x) is the number of possible

combinations of x successes and (n x) failures:

C(n,x) = n!x! * (n x)!

where n! = n * (n 1) * (n 2) * * 3 * 2 * 1

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For example, the probability of 2 successes in 4trials, given p=0.2 is:

Bernoulli Trials - Example

P(x) = n!x! * (n x)!

* px *(1 - p)n - x

P(x) =

24

2 * 2 * (0.2)

2

* (0.8)

2

P(x) = 6 * (0.04)*(0.64) = 0.1536

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Calculating the probabilities of all possible outcomes ofthe number of rain days out of four days, given p = 0.2:

x P(x) C(n,x) px (1 p)n x

0 P(0) 1 (0.2)0 (0.8)4 = 0.4096

1 P(1) 4 (0.2)1

(0.8)3

= 0.40962 P(2) 6 (0.2)2 (0.8)2 = 0.1536

3 P(3) 4 (0.2)3 (0.8)1 = 0.0256

4 P(4) 1 (0.2)4 (0.8)0 = 0.0016


P = product of these terms

We can also calculate the chance of having one or more

days of rain out of four by summing P(1) + P(2) + P(3)

+ P(4) = 0.4096 + 0.1536 + 0.0256 + 0.0016 = 0.5904

i i

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Naturally, we can plot the probability massfunction produced by this binomial distribution:


xi P(xi)

0 0.4096

0.40962 0.1536

3 0.0256

4 0.0016 0

0.25

0.50

P(xi

)

1 2 3 4

xi

0

Th P i Di ib i

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The usual application of probability distributionsis to find a theoretical distribution that reflects a

process such that it explains what we see in someobserved sample of a geographic phenomenon

The theoretical distribution does this by virtue of

the fact that the form of the sampled information

and theoretical distribution can be compared and

be found to similar through a test of significance One theoretical concept that we often study in

geography concerns discrete random events inspace and time (e.g. where will a tornado occur?)

The Poisson Distribution

Th P i Di t ib ti

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The discrete random events in question happenrarely (if at all), and the time and place of these

events are independent and random The greatestprobability is zero occurrences at a

certain time or place, with a small chance of one

occurrence, an even smaller chance of two

occurrences, etc.


A distribution withthese characteristics

will be heavilypeaked and skewed:0

0.25

0.5

P(xi

)

1 2 3 4xi

0

Th P i Di t ib ti

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In the 1830s, French mathematician S.D. Poissondescribed a distribution with these characteristics,

used to describe thenumber of events

that willoccur within a certain area or duration (e.g. #

of meteorite impacts per state, # of tornados per

year in Tornado Alley)

The following characteristics describe the

Poisson distribution:1. It is used to count the number of occurrences

of an event within a given unit of time, area,volume, etc. therefore a discrete distribution


Th P i Di t ib ti

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Poisson distribution cont.:

2. The probability that an event will occur within

a given unit must be the same for all units (i.e.the underlying process governing the

phenomenon must be invariant)

3. The number of events occurring per unit must

be independent of the number of events

occurring in other units (no interactions)

4. The mean or expected number of event per unit

is denoted by and is found by past experience(observations)


Th P i Di t ib ti

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Poisson formulated his distribution as follows:


P(x) =

e- * x

x!where e = 2.71828 (base of the natural logarithm)

= the mean or expected value

x = 1, 2, , n 1, n # of occurrences

x! = x * (x 1) * (x 2) * * 2 * 1 To calculate a Poisson distribution, you must

know and then plug in the values of x whereobservations are likely to occur

Th P i Di t ib ti

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Poisson formulated his distribution as follows:


P(x) =

e- * x

x! The shape of the distribution depends quite

strongly upon the value of , because as

increases, the distribution becomes less skewed,

eventually approaching a normal-shapeddistribution as gets quite large

We can evaluate P(x) for any value of x, but largevalues of x will have very small values of P(x)

The Poisson Distrib tion

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The Poisson distribution is sometimes known asthe Law of Small Numbers, because it describes

the behavior of events that are rare, despite there

being many opportunities for them to occur

We can observe the frequency of some rare

phenomenon, find its mean occurrence, and then

construct a Poisson distribution and compare

our observed values to those from the distribution(effectively expected values) to see the degree to

which our observed phenomenon is obeying the

Law of Small Numbers:


The Poisson Distribution Example

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Fitting a Poisson distribution to the 24-hour murderrates in Fayetteville in a 31-day month (to ask the

question Do murders randomly occur in time?)

x Obs. Frequency x*Fobs P(x) Fexp

0 17 0 0.49 15.2

1 9 9 0.35 10.92 3 6 0.12 3.7

3 1 3 0.03 0.9

4 1 4 0.005 0.2Total values 31 22 1.000 30.9

The Poisson Distribution - Example

days murders

= mean murders per day = 22 / 31 = 0.71

Observed Values Expected Values

P(x) =e- * x

x!

P(x) = e-0.71 * 0.71x

x!

Fexp = P(x) * 31We can compare Fobsto Fexp using a

2 test

to see if observationsdo match Poisson Dist.


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Procedure for finding Poisson probabilities andexpected frequencies:

1. Set up a table with five columns as on the

previous slide

2. Multiply the values of x by their observed

frequencies (x * fobs)3. Sum the columns of fobs (observed frequency)

and x * fobs4. Compute = (x * fobs) / fobs5. Compute P(x) values using the eqn. or a table

6. Compute the values of Fexp = P(x) * fobs



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One characteristic of the Poisson distribution isthat we expect the variance ~ mean (i.e. the two

should have approximately the same values)

When we apply the Poisson distribution to

geographic patterns, we can see how a variance

to mean ratio (

2:x) of about 1 corresponds to arandom pattern that is distributed according to

Poisson probabilities

Suppose we have a point pattern in an (x,y)

coordinate space and we lay down quadrats and

count the number of points per quadrat



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Here, the counts of points per quadrat form thefrequencies we use to check Poisson probabilities:


Regular

Low variance

Mean 12:x is low

Random

Variance

Mean2:x ~ 1

Clustered

Low variance

Mean 02:x is high

2005 geog090 discrete

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