2006 geog090 week05 lecture01 discrete distribution

7/31/2019 2006 Geog090 Week05 Lecture01 Discrete Distribution

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Office

310

320. . .

. . .


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Probability-Related Concepts

How to Assign Probabilities to Experimental

Outcomes

Probability Rules

Discrete Random Variables

Continuous Random Variables

Probability Distribution & Functions

Topics Covered


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Concepts

An event Any phenomenon you can observe thatcan have more than one outcome (e.g., flipping a

coin)

An outcome Any unique condition that can be

the result of an event (e.g., flipping a coin: heads or

tails), a.k.a simple event or sample points

Sample space The set of all possible outcomes

associated with an event

Probability is a measure of the likelihood of each

possible outcome


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Probability Distributions

The usual application of probability distributions

is to find a theoretical distribution

Reflects a process that explains what we see in

some observed sample of a geographic

phenomenon

Compare the form of the sampled information and

theoretical distribution through a test of

significance

Geography: discrete random events in space and

time (e.g. how often will a tornado occur?)


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Discrete Probability Distributions

Discrete probability distributions

The Uniform Distribution

The Binomial Distribution

The Poisson Distribution

Each is appropriately applied in certainsituations and to particular phenomena


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Source: http://davidmlane.com/hyperstat/A12237.html


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Describes the situation where the probability of all

outcomes is the same

noutcomesP(xi) = 1/n

e.g. flipping a coin:

P(xheads) = 1/2 = P(xtails)0

0.25

0.50

P(xi)

heads

xi

tails

A uniform probability

mass function


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0

/1)1/(1

)(

nab

xf

1

)1/()1(

0

)()( abaxxXPxF

a


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A little simplistic and perhaps useless

But actually well applied in two situations

1. The probability of each outcome is trulyequal (e.g. the coin toss)

2.No prior knowledge of how a variable is

distributed (i.e. complete uncertainty), the firstdistribution we should use is uniform (no

assumptions about the distribution)


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However, truly uniformly distributed geographic

phenomena are somewhat rare

We often encounter the situation of not knowing

how something is distributed until we sample it

When we are resisting making assumptions

we usually apply the uniform distribution as asort of null hypothesis of distribution


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Example Predict the direction of the prevailing

wind with no prior knowledge of the weathersystems tendencies in the area

We would have to begin with the idea that

P(xNorth) = 1/4P(xEast) = 1/4

P(xSouth) = 1/4

P(xWest

) = 1/4

Until we had an opportunity to sample and find

out some tendency in the wind pattern based on

those observations

0

0.25

P(xi)

N E S W

0.125


12/42(Lillesand et al. 2004)

(Binford 2005)

Remote Sensing

Supervised classification


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Provides information about the probability of therepetition of events when there are only two

possible outcomes,

e.g. heads or tails, left or right, success or failure, rain

or no rain

Events with multiple outcomes may be simplified as

events with two outcomes (e.g., forest or non-forest)

Characterizing the probability of a proportion of

the events having a certain outcome over a

specified number of events


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A binomial distribution is produced by a set ofBernoulli trials (Jacques Bernoulli)

The law of large numbers for independent trials

at the heart of probability theory Given enough observed events, the observed

probability should approach the theoretical values

drawn from probability distributions e.g. enough coin tosses should approach the P = 0.5

value for each outcome (heads or tails)


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How to test the law?

A set of Bernoulli trials is the way to operationally

test the law of large numbers using an event with

two possible outcomes:

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

event like a coin toss) are performed

(2) Every trial must have the same set of possibleoutcomes (e.g., heads and tails)


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Bernoulli Trials

(3) The probability of each outcome must be the

same for all trials, i.e. P(xi) must be the same

each time for both xivalues

(4) The resulting random variable is determined

by the number of successes in the trials

(successes one of the two outcomes)

p = the probability of success in a trial q = (p1) as the probability of failure in a trial

p + q = 1


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Bernoulli Trials

Suppose on a series of successive days, we will

record whether or not it rains in Chapel Hill We will denote the 2 outcomes using R when it

rains and N when it does not rain

n 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


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Bernoulli Trials

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

above equations to get the probability of each outcomefrom a series of successive samples (e.g. p=0.2)

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


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Bernoulli Trials

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

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

3 events, S = (p + q)3

= p3 + 3p2q + 3pq2 + q3

4 events, S = (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 the

binomial expansion of (p + q)n, where n = # of events


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A general formula for calculating the probability

of x successes (n trials & a probability p of

success:

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

combinations of x successes and (nx) failures:

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

C(n,x) =n!

x! * (n x)!


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The Binomial Distribution Example

e.g., the probability of 2 successesin 4 trials,given p=0.2 is:

P(x) =

4!

2! * (4 2)! * (0.2)2 *(1 0.2)4 - 2

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

number of rain days out of four days (p = 0.2):

The chance of having one or more days of rain

out of four: P(1) + P(2) + P(3) + P(4) = 0.5904

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 = 0.4096

2 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


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

xiP(x

i)

0 0.4096

1 0.4096

2 0.1536

3 0.0256

4 0.00160

0.25

0.50

P(xi)

1 2 3 4

xi

0


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Source: http://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm

The following is the plot of the binomial probability density

function for four values of pand n= 100


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Source: http://home.xnet.com/~fidler/triton/math/review/mat170/probty/p-dist/discrete/Binom/binom1.htm


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Source: http://www.mpimet.mpg.de/~vonstorch.jinsong/stat_vls/s3.pdf


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Rare Discrete Random Events

Some discrete random events in question happen

rarely (if at all), and the time and place of theseevents are independent and random (e.g.,

tornados)

The greatest probability is zero occurrences at acertain time or place, with a small chance of one

occurrence, an even smaller chance of two

occurrences, etc.

heavily peaked andskewed:

0

0.25

0.5

P(xi)

1 2 3 4

xi

0


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In the 1830s, S.D. Poisson described a distribution

with these characteristics

Describing the number of events that will occur

within a certain area or duration (e.g. # of

meteorite impacts per state, # of tornados per year,

# of hurricanes in NC)

Poisson distributions characteristics:

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)


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Poisson formulated his distribution as follows:

P(x) =e-l * lx

x!

To calculate a Poisson distribution, you mustknow

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) * (x2) * * 2 * 1


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Poisson distribution

P(x) =e-l * lx

x!

The shape of the distribution depends stronglyupon the value of , because as increases, the

distribution becomes less skewed, eventually

approaching a normal-shaped distribution as itgets quite large

We can evaluate P(x) for any value of x, but large

values of x will have very small values of P(x)
http://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.html


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P(x) =e-l * lx

x!

The shape of the distribution depends stronglyupon the value of , because as increases, the

distribution becomes less skewed, eventually

approaching a normal-shaped distribution as lgets quite large

We can evaluate P(x) for any value of x, but large

values of x will have very small values of P(x)
http://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.htmlhttp://www.capdm.com/demos/software/html/capdm/qm/poissondist/usage.html


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n

P(x) =e-l * lx

x! The Poisson distribution can be defined as thelimiting case of the binomial distribution:

lnp constant


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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

We can observe the frequency of some rarephenomenon, 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:

M d i F ill


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Murder rates in Fayetteville

Fitting a Poisson distribution to the 24-hourmurder rates in Fayetteville in a 31-day month (to

ask the question Do murders randomly occur in

time?)

# of Murders Days (Frequency)

0 17

1 9

2 3

3 1

4 1

Total 31 days

M d i F ill


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= mean murders per day = 22 / 31 = 0.71

!

*

)( x

e

xP

xl

l

# of Murders Days # of Murders* # of Days

0 17 0

1 9 9

2 3 6

3 1 34 1 4

Total 31 days 22


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!*)(x

exPx

l

l

!

71.0*)(

71.0

x

exP

x

31*)(exp xPF


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x (# of Murders) Obs. Frequency (Fobs) x*Fobs Fexp

0 17 0 15.2

1 9 9 10.9

2 3 6 3.7

3 1 3 0.9

4 1 4 0.2Total 31 22 30.9


We can compare Fobs to Fexp using a X2 test to see

if observations do match Poisson Dist.


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Th P i Di t ib ti


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Procedure for finding Poisson probabilities and

expected 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 (observedfrequency) and x * Fobs

(4)Compute

= (x * Fobs) / Fobs (5) Compute P(x) values using the equation or a

table

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


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2006 geog090 week05 lecture01 discrete distribution

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