Office 310 320.... Probability-Related Concepts How to Assign Probabilities to Experimental Outcomes...

transcript

Office

320. . .

• Probability-Related Concepts

• How to Assign Probabilities to Experimental Outcomes

• Probability Rules

• Discrete Random Variables

• Continuous Random Variables

• Probability Distribution & Functions

Topics Covered

Concepts

• An event – Any phenomenon you can observe that can 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

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?)

Discrete Probability Distributions

• Discrete probability distributions

– The Uniform Distribution

– The Binomial Distribution

– The Poisson Distribution

• Each is appropriately applied in certain

situations and to particular phenomena

The Uniform Distribution

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

• Describes the situation where the probability of all outcomes is the same

• n outcomes P(xi) = 1/n

• e.g. flipping a coin:

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

i)heads

A uniform probability mass function

/1)1/(1)(

)1/()1(

)()( abaxxXPxF

a<=x<=b

otherwise

a<=x<=b

Source: http://en.wikipedia.org/wiki/Uniform_distribution_(discrete)

• A little simplistic and perhaps useless

• But actually well applied in two situations

• 1. The probability of each outcome is truly

equal (e.g. the coin toss)

• 2. No prior knowledge of how a variable is

distributed (i.e. complete uncertainty), the first

distribution we should use is uniform (no

assumptions about the distribution)

• 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 a

sort of null hypothesis of distribution

• Example – Predict the direction of the prevailing wind with no prior knowledge of the weather system’s tendencies in the area

• We would have to begin with the idea that

P(xNorth) = 1/4

P(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

N E S W

(Lillesand et al. 2004)

(Binford 2005)

Remote SensingSupervised classification

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 …

– 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

• A binomial distribution is produced by a set of

Bernoulli 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)

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 possible

outcomes (e.g., heads and tails)

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 xi values

• (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 = (p –1) as the probability of failure in a trial

• p + q = 1

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

1 R 1 p p

N 0 (1 - p) q

2 RR 2 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

• 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. 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

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

• 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 (n –x) failures:

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

C(n,x) = n!

x! * (n – x)!

The Binomial Distribution – Example

• e.g., the probability of 2 successes in 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

• 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

• Naturally, we can plot the probability mass function produced by this binomial distribution:

xi P(xi)

0 0.4096

1 0.4096

2 0.1536

3 0.0256

4 0.00160

1 2 3 4

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 p and n = 100

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

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

Rare Discrete Random Events• Some discrete random events in question happen

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

• The greatest probability is zero occurrences at a certain time or place, with a small chance of one occurrence, an even smaller chance of two occurrences, etc.

• heavily peaked and skewed:

1 2 3 4xi

The Poisson Distribution

• 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 distribution’s 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)

• 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 events per

unit (λ) is found by past experience (observations)

• Poisson formulated his distribution as follows:

P(x) = e- * x

• To calculate a Poisson distribution, you must know λ

where e = 2.71828 (base of the natural logarithm)

λ = the mean or expected value

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

x! = x * (x – 1) * (x – 2) * … * 2 * 1

• Poisson distribution

P(x) = e- * x

x! • The shape of the distribution depends strongly

upon the value of λ, because as λ increases, the distribution becomes less skewed, eventually approaching a normal-shaped distribution as it gets 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)

P(x) = e- * x

x! • The shape of the distribution depends strongly

upon the value of λ, because as λ increases, the distribution becomes less skewed, eventually approaching a normal-shaped distribution as l gets 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)

P(x) = e- * x

x! • The Poisson distribution can be defined as the

limiting case of the binomial distribution:

np constant

• The Poisson distribution is sometimes known as the Law of Small Numbers, because it describes the behavior of events that are rare

• 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:

Murder rates in Fayetteville

• Fitting a Poisson distribution to the 24-hour murder rates in Fayetteville in a 31-day month (to ask the question ‘Do murders randomly occur in time?’)

# of Murders Days (Frequency)

Total 31 days

= mean murders per day = 22 / 31 = 0.71

# of Murders Days # of Murders* # of Days

0 17 0

Total 31 days 22

71.0*)(

31*)(exp xPF

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

0 17 015.2

1 9 910.9

2 3 63.7

3 1 30.9

4 1 40.2

Total 31 2230.9

• We can compare Fobs to Fexp using a X2 test to see if observations do match Poisson Dist.

The Poisson Distribution• 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 (observed frequency) and x * Fobs

• (4) Compute λ = Σ (x * Fobs) / Σ Fobs

• (5) Compute P(x) values using the equation or a table

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

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

Office 310 320.... Probability-Related Concepts How to Assign Probabilities to Experimental Outcomes...

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