[SOLVED] IT代写 ISE 562, Dr. Bayes Methods Decision Theory

ISE 562, Dr. Bayes Methods Decision Theory
ISE 562, Dr. differences between discrete/continuous
• Probability is not height of function but area under curve for an interval.
p(x) discrete P(X=a)

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• Total area under the function =1
9/4/2022 2
f(x) continuous
P(aXb) a b • Require f(X)0 for all X (but X can be < 0) ISE 562, Dr. vs. Cumulative ProbabilityPDF’s and CDF’sPDF : P(a  X  b) CDF:P(Xx)x f()d f (X )  d dX•P(aXb) = P(Xb) – P(Xa) •P(Xa) = 1-P(Xa)•P(Xa) = 1-P(Xa) ISE 562, Dr. histogram f(x) Empirical pdfs9 / 4 / 2 0 2 20202 f ( X )  kX 0  X  22 X2 kX2 2 0kXdXk22 2k0 2k  1 so k  12 f ( X )  12 X 0  X  2 4 ISE 562, Dr. pdfs2 1 X32 4 E[X]0X2XdX6 32 E[X2]2 2X21XdX2X42 42 162  8 3 299 0 ISE 562, Dr. pdfs: Cumulative DistributionFunction (CDF) x F(X)x 1d 2 1 X2 024040 X 2 cdf ISE 562, Dr. pdfs: Expected value of a function Y=Cost(X)=aX+bE(Cost(X ))  2 Cost(X ) 12XdX 02(aX b)1XdX 2(aX2 bX)dX 02022aX3 bX2 2 4 6  4  3ab ISE 562, Dr. use properties of expectation: Y=Cost(X)=aX+bE(Cost(X))E(aX b)  E(aX )  E(b) aE ( X )  b 43 a  b ISE 562, Dr.’s with more than one random variablef(X,Y) called the joint probability of X and YMarginal pdfs can be obtained by integrating one variable out of the function:f(X) f(X,Y)dY f(Y) f(X,Y)dX  ISE 562, Dr. we have a reservoir fed by 2 rivers. Let X=stream flow in river X and Y=flow in river Y. The joint pdf of flows is:4000  X f(X,Y)c 4000c  2.5×1070  X  4000 0Y 2000 ISE 562, Dr. Smithf ( X )  2000 c 4000  X dY  c (4000  X ) 0 4000 2f (Y )  4000 c 4000  X dX  2000c 0 4000E[(X Y)] 20004000X Yc 4000 X dXdY 00 4000 ISE 562, Dr. Smithf(X|Y) f(X,Y) f (Y )Conditional Probability Bayes rule is derived from the above conditional probability relationship where Y represents the sample information (observations) and X is the random (decision) variable.9/4/2022 12 ISE 562, Dr. Smithi) Conditional Probabilityii) As in discrete case: iii) Integrating out the Bayes for Continuous Random Variablesf(|y) f(,y) f (y)f(,y) f()f(y|) f (y)   f (, y)d f()f(y|)d  Substituting (ii) and (iii) in (i)f ( ) f ( y |  )f(|y) f()f(y|)d  ISE 562, Dr. Smithf ( | y ) posterior pdf   ( prior pdf )(likelihood )9/4/2022 14f ( ) f ( y |  )  f()f(y|)d( prior pdf )(likelihood )ISE 562, Dr. Smith• We denote an uncertain decision variable as  and assume that sample information involving  can be summarized by a sample statistic y.• If y has all the information from the sample relevant to the uncertainty about , then y is called a sufficient statistic.• Example: for a Bernoulli process, the sample information can be summarized by n and r; the actual number of successes and failures adds no additional information about p since p=r/n.9/4/2022 15Sufficiency ISE 562, Dr. do we care about sufficiency?SufficiencyFor Bayes rule this means that knowledge of n and r is sufficient to determine the likelihoods so the posterior distribution of p given n and r is exactly the same as the posterior distribution of p given the entire sequence of observations.Simply put, everything we need to know about the sample is contained in the likelihood function.9/4/2022 16ISE 562, Dr. Smith• Example: Let =market share of a new product(01); assume it is continuous and has prior pdf: Prior pdff() 2 f ( )  2(1   ) 0    1• We take a sample of 5 consumers and 1 buys new brand while other 4 purchase a different brand• Assume binomial likelihood distribution with Likelihood success = “buys new product” sof ( y |  )  P ( r  1 | n  5,    ) 5! 1(1)4 5(1)4 9/4/2022 4!1! ISE 562, Dr. Smith• Applying Bayes rule we substitute the prior and likelhood functions:f ( | y )  f ( ) f ( y |  )  2 (1   ) 5 (1   ) 4  f()f(y|)d 12(1)5(1)4d  0 10(1)5  (1)5 101(1)5d 1(1)5dPosterior pdfSince  x (1 x) dx  where (n)  (n 1)! f(|y)1 m1 n1 (m)(n)  0 (mn)  2  (1)5  42(1)5 Notice this is a function of . In the discrete Bayes case we had specific values of p for the states—here we can have any value in [0,1] for the state variable. ISE 562, Dr. Smith• As shown the process appears somewhat difficult and computationally challenging• May not find the integral is computable in closed form; may need numerical quadrature techniques• There is another way!!!•  conjugate families of distributions that simplify the process of combining priors and likelihoods9/4/2022 19 ISE 562, Dr. of conjugate families of pdfs1. Tractability:easytospecifyposterior given the prior and likelihood function2. Richness:thepriorshouldreflectthe prior information (this is done with parameters to fit the distribution to the information)3. Easeofinterpretation:priorshouldbe interpretable in terms of previous sample results9/4/2022 20ISE 562, Dr. Smith2 conjugate families studied here:1. SamplingfromaBernoulliprocess whose conjugate is the family of beta distributions2. Sampling from a normal distributed process with known variance whose conjugate distribution is the family of normal distributions.LikelihoodPrior Likelihood ISE 562, Dr./binomial“Conjugate” family refers to the relationship between the prior and the likelihood function.1. Sampling from a Bernoulli process (the likelihood function) has as its conjugate, the family of beta distributionsf(p) (n1)! pr1(1 p)nr1 0 p1 (r1)!(nr1)!(Note that the random variable p, varies from zero to one for the beta pdf)9/4/2022 Beta pdf 22ISE 562, Dr./binomial “If n, r, not integers we must use gamma functionsf(p) (n) (r)(n  r)(t)   xt1exdxnote : if t an integer(t)  (t 1)! 9/4/2022pr1(1 p)nr1 t  0ISE 562, Dr./binomial Mean and variance of the beta distribution:  E ( pˆ | r , n )  nr2 V(pˆ|r,n) r(nr)To calculate probabilities we use fractiles:The f fractile of the pdf of a continuous randomvariable is the value xf where P(X xf)=f9/4/2022 24 ISE 562, Dr./binomialSuppose we have a beta pdf of p with r=5 and n=8 and we want P(p0.562|r=5,n=8) and P(.265p 0.562|r=5,n=8)9/4/2022 252.5000 2.0000 1.5000 1.0000 0.5000 0.0000ISE 562, Dr./binomialSuppose we have a beta pdf of p with r=5 and n=8• Can use rhs of beta calculator (back of book)P(p0.562|r=5,n=8)=0.3402• For P(.265p 0.562|r=5,n=8) use:• P(a x b)=P(xb) – P(xa)• = P(p.562) – P(p.265)• =0.3402 – 0.0167• Can use fractiles (on lhs) to enter probabilityand then look up p.9/4/2022 Calculator 262.5000 2.0000 1.5000 1.0000 0.5000 0.0000 ISE 562, Dr. Alert!• Prior pdfs and likelihoods denoted with single primes; e.g., p’, f’(), E’[], ’,…• Posterior pdfs and parameters of posterior pdfs denoted with double primes, (p”), f”(), E”[], ”,…9/4/2022 27ISE 562, Dr. SmithNow the big advantage of conjugate pdfs… ISE 562, Dr. a binomial process– Stationarity and Independency– Beta prior pdf of the form:f'(p) (n’1)! pr’1(1 p)n’r’1 (r’1)!(n’r’1)!Beta/binomial – We draw a sample for binomial likelihood function with r successes in n trials. Then the posterior pdf becomes:f”(p) (n”1)! pr”1(1p)n”r”1 0 p1 (r”1)!(n”r”1)!with n” = n’ + n and r” = r’ + r ISE 562, Dr./binomial• Example: Let =market share of a new product(01); assume it is continuous and has prior pdf: Prior pdff() 2 f ( )  2(1   ) 0    1• We take a sample of 5 consumers and 1 buys new brand while other 4 purchase a different brand• Assume binomial likelihood distribution with Likelihood success = “buys new product” sof ( y |  )  P ( r  1 | n  5,    ) 5! 1(1)4 5(1)4 9/4/2022 4!1!0.001 0.150.001 0.15 ISE 562, Dr./binomial • Example: Let =market share of a new product(01) is a beta prior: f()2(1) 0 1E [  ]   1  2 ( 1   ) d   13 0 • The equivalent mean for the beta is E[p]=r/n=1/3 so r=1 and n=3:9/4/2022 31 2.5000 2.0000 1.5000 1.0000 0.5000 0.0000ISE 562, Dr./binomial• The sample results were r=1 and n=5 so the posterior pdf has parameters• r”=r’+r= 1+1=2• n”=n’+n=3+5=8• So mean of prior went from 1/3 (0.33) to posterior of 2/8=1/4 (0.25) – it shifted to the left.9/4/2022 Prior Posterior 322.5000 2.0000 1.5000 1.0000 0.5000 0.00003.0000 2.5000 2.0000 1.5000 1.0000 0.5000 0.0000ISE 562, Dr./binomial• Notice also the shift in variance:• Prior variance =r(n-r)/(n2(n+1))=1(3-1)/(9(4))=1/18 = 0.055• Posterior variance=2(8-2)/(64(9))=12/5762.5000 2.0000 1.5000 1.0000 0.5000 0.0000=1/48 = 0.0213.0000 2.5000 2.0000 1.5000 1.0000 0.5000 0.0000 ISE 562, Dr./binomial• Posterior mean always lies between prior mean and the sample mean for the Bernoulli/Beta family• Posterior variance generally smaller than prior variance due to addition of sample information to prior knowledge.9/4/2022 Prior Posterior 342.5000 2.0000 1.5000 1.0000 0.5000 0.00003.0000 2.5000 2.0000 1.5000 1.0000 0.5000 0.0000ISE 562, Dr. Conjugate• “Conjugate” family refers to the relationship between the prior and the likelihood function.• Sampling from a normal process (the likelihood function) has as its conjugate, the family of normal distributions  ( x )2 f(x) 1 e 22  zx f(x|,2) f(z|0,1) f(z) 1 ez2 2 ISE 562, Dr. normal distributionP(Xa)a f(X)dX 0P(z  (a  ))  STDNORMALTABLE((a  )) 0.001 0.150.001 0.150.001 0.150.001 0.150.001 0.150.001 0.150.001 0.15 ISE 562, Dr. SmithTable used forThis area =u=0 Z=(a-u)/ ISE 562, Dr. on form of table, can use symmetry properties of normal to calculate various probabilities:P(aXb) = P(Xb) – P(Xa) P(Xa) = 1-P(Xa)P(Xa) = 1-P(Xa)9/4/2022 38 ISE 562, Dr. Normal Table ISE 562, Dr. types of normal pdf problems:1. Given X, u, 2 , find P(X)a) Transform question to probability statementb) Translate X to Z and P(Z) using Z=(X-u)/c) Look up P(Z) in standard normal tableLookup P(Z) in body of table to find Z Solve Z=(X-u)/ for unknown quantityGiven P(X), find X, u, or 2 ISE 562, Dr. carpet warehouse keeps 6000 yards of carpet in stock during a month. The average demand is normally distributed with mean 4500 yards and standard deviation 900 yards. What is the probability a customer order won’t be met? We want P(d6000).P(d6000)=P(z (6000-4500)/900)=P(z  1.67) = 1-P(z1.67) = 1-(0.9525) = 0.0475Std. Normal9/4/2022 Table 41Type 1, Find Probability ISE 562, Dr.mount of coffee a filling machine puts into 4 oz. jars is normally distributed with std. dev, =0.04 oz. If only 2% of the jars are to contain less than 4 oz., what should be the average fill amount?• We want P(X4)=0.02.• Find Z such that P(Zz)=0.02• From table, Z=-2.05• Solve -2.05=(4-u)/.04 for u.• u=4.082 oz.Std. NormalType 2, Find unknown ISE 562, Dr. to Bayes…Normal Conjugate•  and  are summary measures of the normal pdf• For the binomial, r and n are summary measures of the sample info• For the normal, the sample mean and sample variance are summary measures for a sample froma normal populationISE 562, Dr. Smith• If x1, …, xn random variables represent a random sample from a normal population with mean  and , then the sample mean is normally distributed with E[m| ,2]=  and V[m| ,2]= 2/n• Even if population is not normally distributed the Central Limit Theorem shows that as n, the distribution of (m-)/(/n) converges to the normal pdf. This is true for any population with finite mean and variance.9/4/2022 CLT demo 44Normal ConjugateISE 562, Dr. Smith•Be careful—check your data for normal distribution9/4/2022 45 ISE 562, Dr.-squared,2 Goodness-of-fit test• Probability distribution unknown• Sample from the unknown pdf n values• We sort the sample into k class intervals (histogram) and let foi be the observed frequency (count) for interval i.• Then compute the theoretical frequency using normal (or any) distribution for each interval, fti• If any of intervals contain < 5 theoretical observations, combine with next interval• Calculate total deviation of observed values from theoretical values and test for significance9/4/2022 46Testing for Normality ISE 562, Dr. Chi-square statistic is:Testing for Normalityk= no. intervalsp= no. estimated parameters (2 for normal)2 oi ti i1 fti The hypothesis test:Ho: X is normally distributedHa: X is not normally distributedReject Ho if data not normal (if differences toolarge); ie: reject normal if  2   2 ,kp1 ISE 562, Dr.: cell phone company has frequency data for length of calls outside a roaming area. The mean and std. dev are 14.3 and 3.7 minutes respectively. Are the data normally distributed? Use =0.05.9/4/2022 48Length (min)Testing for Normality ISE 562, Dr.(0X5)=P(z(5-14.3)/3.7) – P(z(0-14.3)/3.7)=P(z-2.51)-P(z-3.86)=(1-.9940)-(1-.9999)=0.0069/4/2022 49Testing for NormalityLength (min)Freq (obs) Theor. probTheor freq (np)(fo-ft)2 /ftISE 562, Dr. Smithreject normal if 2  2 ,kp1 2   2   2  3.841,kp1 0.05,421 0.05,1 112.2  3.841so reject normalTesting for Normality ISE 562, Dr. ConjugateReturning to conjugate discussion…Suppose the prior distribution on  is normal with mean m’ and variance ’2:  (m’)2 f ‘()  1 e 2 ‘2  We draw a sample of size n and observe a sample mean of m;the posterior density is then normal:Where y represents the sample results and the posteriorparameters are computed from:f “( | y)  1 e 2 “2 1 m’ n m 11n and m”’2 2(m”)2   ”2 ’2 2Back to 1n p.56 ISE 562, Dr. ConjugateDiscrete prior example: retailer interested in weekly sales, x, at one of their stores• x distributed normally with unknown mean  and variance known, 2=90000• Only 5 potential values to be considered• =1100, 1150, 1200, 1250, 1300• The prior distribution is estimated to be:9/4/2022 52P(=1100)=.15 P(=1150)=.20 P(=1200)=.30 P(=1250)=.20 P(=1300)=.15ISE 562, Dr. ConjugateRetailer wants more information about the store so takes a sample from past sales records assuming weekly sales independent.Takes sample of n=60 weeks and calculates sample mean, m=1240; now calculate the likelihoods*f(1240|1100,f(1240|1150,f(1240|1200,f(1240|1250,f(1240|1300, 9/4/2022 12401100 n 38.73) f 38.73 12401150 n 38.73) f 38.73 12401200 n 38.73) f 38.73 12401250 n 38.73) f 38.73 12401300 n 38.73) f 38.73| 0,1 / 38.73  .0006 / 38.73| 0,1 / 38.73  .0270 / 38.73| 0,1 / 38.73  .2347 / 38.73| 0,1 / 38.73  .3857 / 38.73| 0,1 / 38.73  .1200 / 38.73 * 300/sqrt(60) = 38.73 ISE 562, Dr. ConjugateNote that denominator same for all calculations (see page 35) so we can ignore in table—its optional since it cancels out later9/4/2022 54Prior prob.LikelihoodPrior prob * likelihoodPosterior prob. ISE 562, Dr. ConjugateContinuous prior example: Suppose mgr decides prior is normally distributed with mean, m’=1200 and ’=50.Note that 2=90000 is the variance of weekly sales Note that ’2=2500 is the variance of prior pdf of ,average weekly sales.Using the previous sample information for n=60 weeks with mean m=1240 we calculate the posterior parameters for the posterior pdf:9/4/2022 55ISE 562, Dr. Conjugate1  1  n  1  60  96  “2  ‘2  2 2500 90000 900001m’nm m” ’2 21 1200 60 124090000 1  602500 90000 So the mean and variance of the posterior pdf are 1225 and 90000/96=937.5ISE 562, Dr. Conjugate• The original belief about the mean (prior) was m’=1200 and a population standard deviation of 300• A sample was taken with sample mean of 1240 and standard deviation of 38.73• The posterior mean then moved from the prior value of 1200 to the posterior value of 1225• The posterior standard deviation went from a prior value of 38.73 to sqrt(937.5) = 30.62 (less dispersion due to “learning” from sample information.9/4/2022 57ISE 562, Dr. if no conjugates?• What if the prior and likelihood not conjugate?• All is not lost. In Bayes time until digital computers arrived in the 1950-1960’s, computation of the posterior for an arbitrary prior and likelihood was not feasible.• Today we have software that make these calculations easy (e.g., Mathematica, Matlab, etc.)程序代写 CS代考加微信: assignmentchef QQ: 1823890830 Email: [email protected]