Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Student t distributionExponential DistributionChi-Square distributionF distributionRandom VariablesA random variable is a variable whose value is determined by the outcome of a random experiment.Example: In a single die toss experiment, the sample space consists of six elements, 1, …, 6, denoted by i, i=1,…,6. A random variable may be defined for this experiment: let the value of the random variable be equal to the value of the dice, i.e x(i) = i, i=1, 2, …, 6The random variables discussed in our example could take on only a set of discrete numbers. Such random variables are know as discrete random variables (i.e. variables assume countable values). Random variables of another type, know as continuous random variables, may take on values anywhere within a continuous ranges.Binomial DistributionIn many Geographic studies, we often face a situation where we deal with a random variable that only takes two values, zero-one, yes-no, presence-absence, over a given period of time. Since there are only two possible outcomes, knowing the probability of one knows the probability of the other. P(1)=p P(0)=1-p=qIf the random experiment is conducted n times, then the probability for the event to happen x times follow binomial distribution: xnxxnxqpxnxnqpxnxP)!(!!)(Where n! the factorial of n. e.g. 5!=5*4*3*2*1=120.For example, the presence-absence of drought in a year directly influence the profit of agriculture due to irrigation costs added in a dry year. Suppose a geographer is hired to do risk analysis for an Ag. Company whether a piece of land is profitable for agriculture. Past experience shows that irrigation can be afforded only one year in five. According to weather record, 4 out of the last 25 years suffered from drought in the area. Let 1 denote drought presence, and 0 denote drought absence, then P(1)=4/25=0.16, so P(0)=1-0.16=0.84.For 5 years, there are six possibilities of drought occurrence: 0, 1, 2, 3, 4, 5. Binomial Distribution ExampleDrought occurrence probability in 5 years (probability mass function):418.084.016.0)!05(!0!5)0(050P398.084.016.0)!15(!1!5)1(151P152.084.016.0)!25(!2!5)2(252P029.084.016.0)!35(!3!5)3(353P003.084.016.0)!45(!4!5)4(454P000.084.016.0)!55(!5!5)5(555PThe probability of profitable agriculture is summation of probabilities of no drought and one drought in five years, i.e. 0.418+0.398=0.816This the risk is 18.4% in five years.Poisson DistributionA discrete random variable is said to follow Poisson Distribution if its probability mass function is !)(xexPxWhere x = 0, 1, 2, …, and  is the mean.Poisson Distribution Example Suppose a geographer is assessing the risk of summer wheat yields to devastating hailstorms in a particular geographic location. Weather records show that in the past 35 years show that 10 years with no hailstorm, 13 years with one hailstorm, 8 years with 2 hailstorms, 3 years with 3 hailstorms, and 1 year with 1 hailstorm. Assume the occurrence of hailstorm is independent of past or future occurrences and can be considered random. Then the number of hailstorms happening in any given year follows Poisson distribution. In the above example, there are 42 hailstorms in 35 years, thus the mean number of hailstorms in a year is 1.2, then301.0!02.1)0(02.1eP361.0!12.1)1(12.1eP217.0!22.1)2(22.1eP 087.0!32.1)3(32.1eP026.0!42.1)4(42.1ePNormal Distribution A continuous random variable is said to be normally distributed if its pdf is 222)(21)(xexfWhere (, ) are the distribution parametersxf(x)What Does the Mean Tell Us?For a random variable that follows normal distribution (, ), f(x)x1 2 The mean value tells us where the value x is concentrated most.f(x)x The variance tell how the value is spread. The larger the variance, the more even the value spreads over a large range. Is this good or bad?122 > 1What Does the Variance Tell Us?f(x)x xf(x)xxDoes the variance change here? Why?Prob -3 -2 -1 0 1 2 3 68.3%95.5%99.7%Standard Normal DistributionHypothesis TestingOne application for the probability distribution is hypothesis testing, which is a standard statistical analysis for “difference” or “effect”. For example, before a new drug is put into market, FDA requires a detailed statistical analysis report on how effective the new drug is. This often requires a lot of random experiments. What they do is they usually ask a group of volunteers to test the new drug, and in the meantime, they have another group who may not take anything or a traditional drug for the same purpose. Then they test how effective the new drug is. The way they do this analysis is based on two statements: Statement 1: The new drug is not effective Statement 2: The new drug is effectiveThese two statement is mutually exclusive, meaning negation of statement 1 naturally goes to statement 2. These two statements are hypotheses. The experimental results from the volunteers will be used to test which statement is acceptable, or we call it hypothesis testing.Prob -3 -2 -  + +2 +3 68.3%95.5%99.7%Hypothesis TestingNull Hypothesis (H0): no effect or no differenceAlternative Hypothesis(H1): The null hypothesis is given on which a probability distribution will be developed and its probability is then used to test the hypotheses.However, the decision made based on the statistics is not always correct. We make mistakes. Types of error we have.Type I: H0 is true,but is rejectedType II: H0 is false, but is not rejectedStudent t distributionProbability Density Function:2121221)(kkxkkkxfWhere k is degrees of freedomMean: 0Variance: k/(k-2), k>2Exponential DistributionProbability density functionxexf)(Mean: 1/λVariance: 1/ λ2Chi-Square distributionProbability Density Function: 2122221)(xkkexkxfWhere k is degrees of freedom, and x≥0Mean: kVariance: 2kF distributionProbability Density function:2211)(dUdUxf Where U1 and U2 are