Discrete Probability DistributionsThe Uniform DistributionThe Uniform DistributionThe Uniform DistributionThe Uniform DistributionThe Binomial DistributionBernoulli TrialsBernoulli TrialsBernoulli Trials – An ExampleBernoulli Trials – An ExampleBernoulli TrialsBernoulli Trials - ExampleBernoulli Trials - ExampleThe Poisson DistributionThe Poisson DistributionThe Poisson DistributionThe Poisson DistributionThe Poisson DistributionThe Poisson DistributionThe Poisson Distribution - ExampleThe Poisson DistributionThe Poisson DistributionContinuous Random VariablesProbability Density FunctionsThe Normal DistributionThe Normal DistributionLST DistributionThe Normal DistributionThe Normal DistributionThe Normal DistributionLST Z-ScoresDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005• The first part of today’s lecture will introduce three kinds of discrete probability distributions that are useful for us to examine:1. The Uniform Distribution2. The Binomial Distribution3. The Poisson Distribution• Each of these probability distributions is appropriately applied in certain situations to help us understand particular geographic phenomenaDiscrete Probability DistributionsDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005• The uniform distribution describes the situation where the probability of all outcomes is the sameThe Uniform Distribution• Expressed as an equation, given n possible outcomes for an event, the probability of any outcome is P(xi) = 1/n, e.g. if we are flipping a coin and we find that heads and tails are equally likely outcomes, then:P(xheads) = 1/2 = P(xtails)00.250.50P(xi)headsxitailsA uniform probability mass functionDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005• 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 coin toss)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 no assumptions about the distributionThe Uniform DistributionDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005• While truly uniformly distributed geographic phenomena are somewhat rare (remember Tobler’s law, which implies variation from the uniform 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 of some geographic phenomenon that we usually apply the uniform distribution as a sort of null hypothesis of distributionThe Uniform DistributionDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005The Uniform Distributionxi• For example, supposing we wanted to wanted to predict the 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 system’s tendencies in the area, we would have to begin with the idea thatP(xNorth) = ¼P(xEast) = ¼P(xSouth) = ¼P(xWest) = ¼until we had an opportunity to sample and establish some tendency in the wind pattern based on those observations00.25P(xi)NESW0.125David Tenenbaum – GEOG 070 – UNC-CH Spring 2005• The binomial distribution provides information about the probability of the repetition of eventswhen 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 a specified number of eventsThe Binomial DistributionDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005•A set of Bernoulli trials is the way to operationally 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 next2. Every trial must have the same set of possible outcomes (heads and tails must be the only available results of coin tosses … using other sorts of experiments, this is a less trivial issue)Bernoulli TrialsDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005Bernoulli trials cont.3. The probability of each outcome must be the same for all trials, i.e. P(xi) must be the sameeach time for both xivalues4. 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 probability of failure in a trial; p + q = 1Bernoulli TrialsDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005• 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)1R 1 p pN0(1 -p)q2RR 2 p2p2RN NR 1 2[p*(1 – p)] 2pqNN 0 (1 – p)2q23 RRR 3 p3p3RRN RNR NRR 2 3[p2 *(1 – p)] 3p2qNNR NRN RNN 1 3[p*(1 – p)2] 3pq2NNN 0 (1 – p)3q3Bernoulli Trials – An Example>1 successive events uses the multiplicative rule (intersection) to calculate the probabilityDavid Tenenbaum – GEOG 070 – UNC-CH Spring 2005n Possible Outcomes # of Rain Days P(# of Rain Days)1R 1 p = 0.2N0q=0.82RR 2 p2=0.04RN NR 1 2pq = 0.32NN 0 q2=0.643 RRR 3 p3= 0.008RRN RNR NRR 2 3p2q = 0.096NNR NRN RNN 1 3pq2= 0.384NNN 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 ExampleΣ = 1Σ = 1Σ = 1David Tenenbaum – GEOG 070 – UNC-CH Spring 2005• We can provide a general formula for calculating the probability of x successes, given n trials and a probability p of success:Bernoulli TrialsP(x) = C(n,x) * px* (1 - p)n - xwhere 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 *