Ex. 3. Mathematica knows about all standard probability distributions. c) The probability is just the pdf integrated over the interval

Transcription

1 Ex. 3 Mathematica knows about all standard probability distributions a) In[336]:= p.4; n 6; r 5; PDF BinomialDistribution n, p r BarChart Table PDF BinomialDistribution n, p, k, k,, n Out[339]= Out[34]= b) In[341]:= Ν 2.4; PDF PoissonDistribution Ν r BarChart Table PDF PoissonDistribution Ν, k, k,, n Out[342]= Out[343]= c) The probability is just the pdf integrated over the interval

2 2 Session3.nb In[344]:= Μ 2.4; p.4; n 6; Σ n p 1 p distr NormalDistribution Μ, Σ ; CDF distr 5.5 CDF distr 4.5 Plot PDF distr x, x, 1.2, 6, AxesOrigin, ; Show, Plot PDF distr x, x, 4.5, 5.5, AxesOrigin,, Filling Axis Out[344]= 1.2 Out[346]= Out[348]= d) In[349]:= Μ 2.4; Σ Ν distr NormalDistribution Μ, Σ ; CDF distr 5.5 CDF distr 4.5 Plot PDF distr x, x, 1.2, 6, AxesOrigin, ; Show, Plot PDF distr x, x, 4.5, 5.5, AxesOrigin,, Filling Axis Out[349]= Out[351]= Out[353]= Ex 4 In[354]:= Needs "MultivariateStatistics`" Clear Μ, Σ

3 Session3.nb 3 Covariance matrix in 2 dimension can be written as a function of Ρ, Σ x, Σ y In[356]:= meanvector Μ x, Μ y ; covmatrix Σ x 2, Ρ Σ x Σ y, Ρ Σ x Σ y, Σ y 2 ; MatrixForm distr MultinormalDistribution meanvector, covmatrix ; Out[358]//MatrixForm= mean and covariance are as expected In[36]:= Mean distr Covariance distr Out[36]= Out[361]= Μ x, Μ y 1,.5,.5, 1 a) For simplicity center around the origin. Matrix is symmetric, and so are the contours In[362]:= meanvector, ; Ρ ; Σ x 1; Σ y 1; covmatrix Σ x 2, Ρ Σ x Σ y, Ρ Σ x Σ y, Σ y 2 ; MatrixForm distr MultinormalDistribution meanvector, covmatrix ; Out[365]//MatrixForm= 1 1 Quick check: the contours are circles In[367]:= ContourPlot PDF distr, x, y, x, 2, 2, y, 2, 2, ColorFunction "Rainbow", Contours Out[367]= Analytical solution: the pdf decreases monotonically in radial direction, so we need to find the radius R such that 68% (95%) is contained inside of the circle.

4 4 Session3.nb Analytical solution: the pdf decreases monotonically in radial direction, so we need to find the radius R such that 68% (95%) is contained inside of the circle. 1) Transform to polar coordinates In[368]:= Clear r, Φ ; var r, Φ ; trafo r Cos Φ, r Sin Φ ; Table D trafo i, var j, i, 2, j, 2 MatrixForm Jacobian matrix jacobian Det FullSimplify distrtransformed PDF distr, x, y jacobian. x trafo 1, y trafo 2 FullSimplify Out[371]//MatrixForm= Out[372]= Cos Φ r Sin Φ Sin Φ r Cos Φ r Out[373]= r2 2 r 2 Π 2) Integrate over the pdf up to a radius R Choose a starting value of 1 to find the solution numerically In[374]:= R68 R. FindRoot Integrate distrtransformed, r,, R, Φ,, 2 Pi.68, R, 1 R95 R. FindRoot Integrate distrtransformed, r,, R, Φ,, 2 Pi.95, R, 1 Out[374]= Out[375]= ) Make sure we didn't commit a simple error: the pdf should be normalized In[376]:= Integrate distrtransformed, r,, Infinity, Φ,, 2 Pi Out[376]= 1 4) Now plot the contours. Red circle is the 68% region, the blue circle the 95% region.

5 Session3.nb 5 In[377]:= ContourPlot distr, r,, R95 1, Φ,, 2 Pi, ColorFunction "Rainbow" ; ContourPlot PDF distr, x, y, x, 3, 3, y, 3, 3, ColorFunction "Rainbow", Contours distrtransformed. r R68, distrtransformed. r R Out[378]= b) Positive correlation, ellipsoids tilted. We omit the exact results from now on, just show the form of the contours

6 6 Session3.nb In[379]:= meanvector, ; Ρ.5; Σ x 1; Σ y 1; covmatrix Σ x 2, Ρ Σ x Σ y, Ρ Σ x Σ y, Σ y 2 ; MatrixForm distr MultinormalDistribution meanvector, covmatrix ; ContourPlot PDF distr, x, y, x, 2, 2, y, 2, 2, ColorFunction "Rainbow", Contours 7 Out[382]//MatrixForm= Out[384]= c) negative correlations, ellipsoids point towards negative values. Variance larger in y direction, thus ellipsoids stretched

7 Session3.nb 7 In[385]:= meanvector, ; Ρ.5; Σ x 1; Σ y 3; covmatrix Σ x 2, Ρ Σ x Σ y, Ρ Σ x Σ y, Σ y 2 ; MatrixForm distr MultinormalDistribution meanvector, covmatrix ; ContourPlot PDF distr, x, y, x, 2, 2, y, 2, 2, ColorFunction "Rainbow", Contours 7 Out[388]//MatrixForm= Out[39]= Ex 1 fix the Random Number generator to get reproducible results In[391]:= SeedRandom 1232 ; the histogram range and bin width In[392]:= xmin ; xmax 1; width.3; the numbers from the instructions In[393]:= Nsamples 1; Nexperiments 1, 1, 1 ; 1: Not quite Gaussian yet

8 8 Session3.nb In[395]:= Table RandomReal, 1, Nsamples Mean, k, 1, Nexperiments 1 ; Histogram, xmin, xmax, width Out[396]= : pretty Gaussian In[397]:= Table RandomReal, 1, Nsamples Mean, k, 1, Nexperiments 2 ; Histogram, xmin, xmax, width 2 15 Out[398]= 1 5 1: Now it looks good In[399]:= Table RandomReal, 1, Nsamples Mean, k, 1, Nexperiments 3 ; Histogram, xmin, xmax, width Out[4]= : not much better than with 1

9 Session3.nb In[41]:= Table RandomReal, 1, Nsamples Mean, k, 1, 1 ; Histogram, xmin, xmax, width Out[42]= Ex 2 a) create all random numbers as a matrix where each column is one experiment with 1 samples In[43]:= Nexperiments 1; In[44]:= randomnumbers Table RandomReal i 1, i, i, 1, Nsamples, k, 1, Nexperiments ; and then compute averages over each column. Mathematica indexes matrices by rows, so we need to transpose In[45]:= averages Table Transpose randomnumbers k Mean, k, 1, Nexperiments ; Distribution sharply peaked at 5 In[46]:= Histogram averages,, 1, Out[46]= 2 1 b) Same as before, just rescale differently In[47]:= randomnumbers Table RandomReal, 1 2 i 1, i, 1, Nsamples, k, 1, Nexperiments ; averages Table Transpose randomnumbers k Mean, k, 1, Nexperiments ;

10 1 Session3.nb In[49]:= Histogram averages,, 2, Out[49]= 1 5 In[41]:= Now what happens if only one number is large, all others between and 1? Now the distribution is very much uniform, the big number with power 2 9 dominates the other distributions. Central Limit Theorem evidently doesn't apply In[411]:= randomnumbers Table RandomReal, 1, i, 1, Nsamples, k, 1, Nexperiments ; randomnumbers ; averages Table Transpose randomnumbers k Mean, k, 1, Nexperiments ; In[414]:= Histogram averages,, 2, Out[414]= 1 5 Ex 5 We generate RN uniform in a square of length 2, closing the unit circle, so the ratio of the two areas is the probability of a success p Πr2 2 r 2 Π 4 we do not take too many events to start with. More events will simply make your posterior sharper, and your analysis more precise In[415]:= nevents 2; SeedRandom 234 ;

11 Session3.nb Generate all events In[417]:= events Table RandomReal 1, 1, RandomReal 1, 1, i, 1, nevents ; Filter out the successes and failures. x.x is the inner product of the 2 dim. vectors In[418]:= successes DeleteCases events, x_ ; x.x 1 ; failures Cases events, x_ ; x.x 1 ; ListPlot successes, failures, AspectRatio 1.5 Out[42]= In[421]:= Clear f Estimate p=π/4 from the number of successes, r, using Bayes' theorem and a flat prior. The posterior with p limited to [,1] is In[422]:= f p_, r_, n_ : n 1 r n r p r 1 p n r ; p &&p 1 f r nevents, r, nevents N Out[423]= r r 1..5 r r r r r r The maximum of the posterior is at r/n. So our posterior knowledge of the value of Π given the r success observed is completely given by the following plot. The peak becomes narrower as more events are observed

12 12 Session3.nb In[424]:= Out[424]= 148 r Length successes posterior Plot f Π 4, r, nevents, Π, 2.5, 3.7, AxesLabel Π, "Posterior Π r,n ", PlotRange Full ; prior Plot 1 4, x, 1, 4, PlotStyle Red ; Show posterior, prior Posterior Π r,n Out[427]= For the central 68% probability interval around the mode, first find the value of the cumulative distribution function (CDF) at the mode. It is close to.5, since the distribution is almost symmetric around the mode. The CDF is a beta function as discussed in the lecture In[428]:= Clear p ; cdf p_ : If p && p 1, Beta p, 1 r, 1. nevents r else CDFatMode cdf r nevents Plot cdf p, p,, 1 Out[43]= nevents 1. r nevents r,.8.6 Out[431]=.4.2 Now numerically invert the CDF for the lower and upper bounds of the interval

13 Session3.nb In[432]:= level.68; plower1 p. FindRoot CDFatMode cdf p level 2, p,.78 pupper1 p. FindRoot cdf p CDFatMode level 2, p,.78 check results cdf pupper1 cdf plower1 Out[433]=.7952 Out[434]= Out[435]=.68 In[436]:= level.95; plower2 p. FindRoot CDFatMode cdf p level 2, p,.78 pupper2 p. FindRoot cdf p CDFatMode level 2, p,.78 check results cdf pupper2 cdf plower2 Out[437]= Out[438]= Out[439]=.95 Add the limits to the posterior plot. Note the asymmetry from the 95% interval because mode is not exactly equals the median In[44]:= onesigmaband Plot f Π 4, r, nevents, Π, 4 plower1, 4 pupper1, Filling Axis, PlotRange, f r nevents, r, nevents, FillingStyle Directive Opacity.6, Blue ; twosigmaband Plot f Π 4, r, nevents, Π, 4 plower2, 4 pupper2, Filling Axis, PlotRange, f r nevents, r, nevents, FillingStyle Directive Opacity.4, Orange ; Show posterior, onesigmaband, twosigmaband Posterior Π r,n Out[442]= In[443]:= f r nevents.1, r, nevents N Out[443]= Calculate mean, mode, median for Π. Their are very close to one another. 1)Mode: we know it should be at r/n, let's confirm

14 14 Session3.nb In[451]:= FindMaximum f p, r, nevents, p, r nevents N r nevents so for Π 4 Out[451]= , p.74 Out[452]=.74 Out[453]= ) Mean In[448]:= 4 Integrate p f p, r, nevents, p,, 1 N Out[448]= ) Median In[449]:= FindRoot cdf p 1 2, p,.8 ; 4 p. Out[45]=