Introduction to least square fits for Geant 4 Simulation and ROOT Analysis of a Silicon Beam Telescope Mar , DESY Hamburg Olaf Behnke, DESY

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1 Introduction to least square fits for Geant 4 Simulation and ROOT Analysis of a Silicon Beam Telescope Mar 4-6 2, DESY Hamburg Olaf Behnke, DESY Literature: Roger Barlow: Statistics, A Guide To The Use Of Statistical Methods In The Physical Sciences Wiley & Sons, 994 Jay Orear: Notes on Statistics for Physicists, Revised, 958, http : //

2 y (cm) Example: for possible discovery Z µ + µ need precise muon track fits ZeVis Zeus Run Event 366 date: time: 4::4 5 XY View Introductory Track fit example Track: coordinates y i measured in n detector layers at fixed positions x i perform track fit Typical Assumptions: Measurements with gaussian uncertainties Linear(ized) model, here: y = a +a x+a 2 x 2 (but could also use exact track helix model) Construct χ 2 : χ 2 = [y i (a +a x+a 2 x 2 )] 2 i σi 2 Determine a, a, a 2 by finding χ 2 minimum (normal equations) Check consistency: use χ 2 and χ 2 -fit probability reject outliers Analyse results: parameters, errors and correlations (error ellipses), track trajectory error band calculate momentum (error propagation) x (cm)

3 y (cm) Example: for possible discovery Z µ + µ need precise muon track fits ZeVis Zeus Run Event 366 date: time: 4::4 5 XY View Introductory Track fit example Track: coordinates y i measured in n detector layers at fixed positions x i perform track fit Typical Assumptions: Measurements with gaussian uncertainties Linear(ized) model, here: y = a +a x+a 2 x 2 (but could also use exact track helix model) Construct χ 2 : χ 2 = [y i (a +a x+a 2 x 2 )] 2 i σi 2 Determine a, a, a 2 by finding χ 2 minimum (normal equations) Check consistency: use χ 2 and χ 2 -fit probability reject outliers Analyse results: parameters, errors and correlations (error ellipses), track trajectory error band calculate momentum (error propagation) x (cm)

4 4 Lecture Part Least square χ 2 -fit method introduction Fit of a constant χ 2 min as consistency check

5 Method of least squares fit - Intro Example: Particle trajectory measurement y (cm).4.2 Det Det 2 Det 3 Det 4 n-measurements y i ± σ i at fixed x i.8 Model: y = f(x,a).6 here: y = ax x (cm) how to determine a? Idea: for correct a one expects: y i f(x i,a) < σ i 5

6 Method of least squares fit - Intro Example: Particle trajectory measurement y (cm).4.2 Det Det 2 Det 3 Det 4 n-measurements y i ± σ i at fixed x i.8 Model: y = f(x,a).6 here: y = ax x (cm) how to determine a? Idea: for correct a one expects: y i f(x i,a) < σ i 6

7 7 Method of least squares fit - Intro χ 2 = n i= (y i f(x i,a)) 2 σ 2 i Minimum w.r.t a determine estimator â from dχ2 da = dχ2 da = 2 n i= (y i f(x i,a)) σ 2 i df(x i,a) da = In general not analytically solvable. use iterative (numerical) methods (MINUIT, Mathematica)

8 8 Method of least squares fit - Intro χ 2 = n i= (y i f(x i,a)) 2 σ 2 i Minimum w.r.t a determine estimator â from dχ2 da = dχ2 da a=â = 2 n i= (y i f(x i,a)) σ 2 i df(x i,a) da = In general not analytically solvable. use iterative (numerical) methods (MINUIT, Mathematica)

9 9 Method of least squares fit Most general case y i, y j correlated measurem. with cov. V ij m fitparameters a χ2 = n i,j= (y i f(x i, a))v ij (y j f(x j, a)) = ( y f( a)) t V ( y f( a)) (vector-notation)

10 Method of least squares fit Most general case y i, y j correlated measurem. with cov. V ij m fitparameters a χ2 = n i,j= (y i f(x i, a))v ij (y j f(x j, a)) = ( y f( a)) t V ( y f( a)) (vector-notation)

11 Example for two correlated measurements y (cm) y =.3+-. Measure track in two detector layers with global position uncertainty. y 2 =-.+-. V =.2 + σ 2 corr σ 2 corr -. σ 2 corr. 2 + σ 2 corr σ corr x (cm)

12 Fit of a constant y (cm).4.3 Det Det 2 Det 3 Measure position of horizontally flying particle x (cm) Averaging of n measurements y i ± σ i χ 2 = n (y i a) 2 σ 2 i i 2

13 3 Fit of a constant (one measurement) Idiot example of one measurement y ± σ : y (cm) Det χ 2 = (y a) 2 σ 2 Min.χ 2 dχ 2 : da = -.2 Estimated value: â = y Error propagation: σâ = σ x (cm)

14 True and inverse probability densities for one measurement with gaussian uncertainty: â = y, σâ = σ True probability density to observe â for given true value a o : p = e (â a o) 2 2σ 2 2πσ But what if we don t know a? Estimate inverse probability density for a o from the measurement â ± σâ : p = 2πσâ e (â ao)2 2σ â 2 a o â Note: this is not a real prob. density! from now on we will use a as synonym for a! 4

15 Fit of a constant (one measurement) Inverse probability density for true a: p p e (a â)2 2σ â 2 with χ 2 = (a â)2 σâ 2 p e χ2 /2 σ σ 2 â = 2 d 2 χ 2 da 2 a=â â σ χ 2 (â ± σâ) = â Note: These two relations hold for a large class of one parameter χ 2 -fitproblems! 5

16 6 Fit of a constant - many measurements Probability for true value a to observe measurements y i, with i =,n: p(y,y 2,...,y n a) n = e 2 n i= (y i a) 2 σ 2 i i= = e χ2 2 (y i a)2 2σ e i 2 but we don t know true a, so let s turn the whole thing around to estimate probability density for true a from the measurements

17 Fit of a constant - many measurements Recalling p(y,y 2,...,y n a) = e χ2 /2 Expand χ 2 around its minimum at â: χ 2 = χ 2 (â) + dχ2 da a=â }{{} = (a â) + 2 = χ 2 (â)+h (a â) 2 with H = 2 d 2 χ 2 da 2 a=â d 2 χ 2 da 2 a=â (a â) 2 Hesse matrix (for one par. a number) p(y,y 2,...,y n a) e χ2(â) 2 }{{} Fit consistency e 2 H (â a)2 }{{} gaussian density interpreted as inverse probability density for true a: Gaussian distribution around â with width σ = H /2 7

18 Fit of a constant - many measurements Recalling p(y,y 2,...,y n a) = e χ2 /2 Expand χ 2 around its minimum at â: χ 2 = χ 2 (â) + dχ2 da a=â }{{} = (a â) + 2 = χ 2 (â)+h (a â) 2 with H = 2 d 2 χ 2 da 2 a=â d 2 χ 2 da 2 a=â (a â) 2 Hesse matrix (for one par. a number) p(y,y 2,...,y n a) e χ2(â) 2 }{{} Fit consistency e 2 H (â a)2 }{{} gaussian density interpreted as inverse probability density for true a: Gaussian distribution around â with width σ = H /2 8

19 Generalisation to any one-parameter (linear) fit χ 2 (a) = χ 2 (a â)2 (â) + σâ 2 χ 2 (â ± σâ) = χ 2 (â) + = χ 2 + min χ 2 (a).5 σ χ 2 min+ Read error directly from χ 2 curve.5 χ 2 min ˆa a-σ a+σ a 9

20 Mini-exercise Averaging of two meas. via χ 2 parabolas Two measurements y and y 2 of the observable a are represented in the figure by χ 2 parabolas: χ 2 i = (y i a) 2 /σ 2 i ; i =, 2 Determine (yes by eye!) from the two χ 2 curves the values y, σ and y 2, σ 2 Draw the total χ 2, i.e. the sum of the two parabolas (yes, do it by hand :-)) and determine â and σâ χ 2 2 (use χ 2 min and χ 2 = χ 2 min + ) How much is the error σâ reduced compared to σ and σ 2? Relax your eyes and hands ;-) χ

21 2 Averaging several measurements n measurements y i ± σ i (Quiz question: Why is n Σy i not the best average? (Note: σ σ 2, etc.) y (cm).4.3 Det DUT Det 3 Measure position of horizontally flying particle x (cm)

22 Averaging several measurements n measurements y i ± σ i : dχ 2 da = = n i= χ 2 = n i= 2(y i a) σ 2 i â = n σ 2 â = 2 i= (y i a) 2 σ 2 i = 2 [ yi σ 2 i n i= ] / n i= d 2 χ 2 da 2 = n i= [ σ 2 i y i σ 2 i σ 2 i ] + 2a n i= σ 2 i 22

23 Averaging several measurements n measurements y i ± σ i : dχ 2 da = = n i= χ 2 = n i= 2(y i a) σ 2 i â = n σ 2 â = 2 i= (y i a) 2 σ 2 i = 2 [ yi σ 2 i n i= ] / n i= d 2 χ 2 da 2 = n i= [ σ 2 i y i σ 2 i σ 2 i ] + 2a n i= σ 2 i 23

24 Averaging - just reformulated Single measurements contribute with weight G i = σ i 2 Define G s := n G i ; Hesse matrix H = d 2 χ 2 2 = G da 2 s i= â = n i= G i n i= G i y i = G s σâ from simple error propagation: σâ 2 = n ( ) 2 dâ dy i σ 2 i = n = G 2 s i= n i= i= G i = G s = n i= /σ 2 i n i= ( Gi G s ) 2 σ 2 i G i y i ; Corollar: least square fitting is nothing else than a clever mapping of measurements to the fitparameters and obtaining fitparameter uncertainties using error propagation 24

25 The role of the Hesse matrix illustrated for weighted average (just a number) H = 2 d 2 χ 2 da 2 = n i= σ 2 i + σ 2 σ 2 2 H grows with each measurement σ 2 σ2 2 σ3 2 σ 4 2 σ 5 2 H is counting the information from the measurements Finally V = H Note: all this holds also for fits with many parameters 25

26 Mini exercise weighted average y (cm) Track trajectory y =.+-.2 Weighted average of two measurements: ŷ = σŷ = ( ).2 2 = = y 2 = x (cm) 26

27 Mini exercise weighted average y (cm) Track trajectory y =.+-.2 Weighted average of two measurements: ŷ = σŷ = ( ).2 2 = = y 2 = x (cm) 27

28 28 Mini summary of what we have learnt One parameter fits: Least square expression for independent measurements: χ 2 = n (y i f(x i,a)) 2 i= σ i 2 get estimator â from minimum χ 2 dχ 2 /da a=â = True physics parameters have a definite value, so true probability densities exist only for the measurements, fitting means estimating (inverse) probability densities for the true parameters = d 2 χ 2 (general relation) σâ 2 2 da 2 a=â χ 2 (â ± σâ) = (general relation) Averaging several measurements can be easily done graphically by adding individual χ 2 parabolas Least square fitting is nothing else than clever mapping of measurements to fitparameters; errors of fitparameters can be obtained from simple errorpropagation The Hesse matrix H = 2 measurements d 2 χ 2 da 2 a=â counts the information from the

29 29 Consistency of measurements Recall inverse probability density for averaging n measurements: p(y,y 2,...,y n a) e χ2 (â) 2 }{{} Fit consistency Now lets have a closer look at the first term e 2 H (â a)2 }{{} gaussian density

30 Consistency of measurements Example: Two measurements y ± σ and y 2 ± σ 2 ; the true value a be known, are the measurements consistent with a?: χ 2 = 2 (y a) 2 + (y 2 a) 2 = 2 χ 2 = 8 σ 2 σ2 2 χ 2 is a measure of consistency But how should χ 2 be distributed? 3

31 Consistency of measurements Example: Two measurements y ± σ and y 2 ± σ 2 ; the true value a be known, are the measurements consistent with a?: χ 2 = 2 (y a) 2 + (y 2 a) 2 = 2 χ 2 = 8 σ 2 σ2 2 χ 2 is a measure of consistency But how should χ 2 be distributed? 3

32 32 χ 2 for two measurements and known true value Expected density for (y,y 2 ) (simple case a = ;σ = σ 2 = ): f(y,y 2 ) = 2π e y2 /2 e y2 2 /2 = 2π e r2 /2 with r = y 2 + y2 2 = χ 2 Probability to find value between r and r + dr: f(r)dr = 2πr 2π e r2 /2 dr = re r2 /2 dr z = r 2 : f(z)dz = f(r) dr dz dz = 2 e z/2 dz introduces χ 2 -distribution for z = χ 2 and two imensions (ndf=2):

33 33 χ 2 for two measurements and known true value Expected density for (y,y 2 ) (simple case a = ;σ = σ 2 = ): f(y,y 2 ) = 2π e y2 /2 e y2 2 /2 = 2π e r2 /2 with r = y 2 + y2 2 = χ 2 Probability to find value between r and r + dr: f(r)dr = 2πr 2π e r2 /2 dr = re r2 /2 dr z = r 2 : f(z)dz = f(r) dr dz dz = 2 e z/2 dz introduces χ 2 -distribution for z = χ 2 and two imensions (ndf=2):

34 34 χ 2 for two measurements and known true value Expected density for (y,y 2 ) (simple case a = ;σ = σ 2 = ): f(y,y 2 ) = 2π e y2 /2 e y2 2 /2 = 2π e r2 /2 with r = y 2 + y2 2 = χ 2 Probability to find value between r and r + dr: f(r)dr = 2πr 2π e r2 /2 dr = re r2 /2 dr z = r 2 : f(z)dz = f(r) dr dz dz = 2 e z/2 dz introduces χ 2 -distribution for z = χ 2 and two dimensions (ndf=2): f(z, 2) = 2 e z/2

35 χ 2 -function for n degrees of freedom maps the χ 2 in n dimensions into probability density for χ 2. χ 2 distribution f(χ 2,n) = Γ(n/2)2 n/2 (χ2 ) n/2 e χ2 /2 with Γ(n/2) = n dof = n dof =2 n dof =3 n dof =4 n dof =5 dte t t n/2 Properties: f(χ 2, n)dχ 2 = χ 2 = n V (χ 2 ) = 2n; σ(χ 2 ) = 2n χ 2 /n = V (χ 2 /n) = 2; σ(χ 2 /n) = 2/n χ 2 35

36 36 χ 2 distributions for various n χ 2 distribution n dof = n dof =2 n dof =3 n dof =4 n dof = χ 2 distribution n dof = n dof =2 n dof =3 n dof =4 n dof =5 χ 2 distr χ χ χ 2 /n dof distribution n dof = n dof =2 n dof =3 n dof =4 n dof = χ 2 /n dof distribution n dof = n dof =2 n dof =3 n dof =4 n dof =5 χ 2 /n distr χ 2 /n dof χ 2 /n dof

37 37 f(χ 2, 2) function and prob(χ 2, 2) f(χ 2, 2) = 2 e χ2 /2 prob(χ 2, 2) = χ 2 f(χ 2, 2)dχ 2 Probability to observe for repeated experiments a χ 2 greater or equal than the current one is a common measure for consistency Here prob(χ 2 = 3, 2) = TMath :: Prob(3, 2) = 22% }{{} ROOT F unction

38 prob(χ 2, n)-function for n degrees of freedom prob(χ 2,n) = χ 2 f(χ 2,n)dχ 2 = Γ(n/2) χ 2 /2 dte t t n/2 prob(χ 2,n) n dof = n dof =2 n dof =3 n dof =4 n dof = χ 2 Note: for repeated experiments expect the observed values of prob(χ 2,n) to be flatly distributed over interval [, ] 38

39 χ 2 for averaging measurements y (cm) Weighted Average x (cm) The figure shows the result of a fit of a constant using n measurements. When repeating the fit many times the resulting χ 2 min distribution should follow a χ 2 distribution with n degrees of freedom. One degree of freedom is sacrificed to determine the weighted average. A prove for this (for n = 2) is given in the appendix. 39

40 Mini exercise χ 2 and probability y (cm) The figure shows the result of a fit of a constant. Determine the total χ 2 (from reading the figure) and the χ 2 - probability x (cm) 4

41 Fits with problems: outliers Toy simulations of constant fits through data points No outliers % random outliers (σ) Exemplary fit track.3.2. a Entries Mean 4.95 RMS χ / ndf 2.7 / 9 p.3733 ±.362 track a Entries Mean RMS χ / ndf / 9 p.373 ± χ 2 min distribution for 2 experiments chi2 distr chi2d Entries 2 Mean RMS chi2 distr chi2d Entries 2 Mean 6.5 RMS prob(χ 2 min, 9) distribution for 2 experiments prob(chi2) distr pchi2d Entries 2 Mean.596 RMS.292 prob(chi2) distr pchi2d Entries 2 Mean.2372 RMS χ 2 min and prob(χ2 min,n dof) highly sensitive to wrong measurements 4

42 CDF I D I CDF II World average of W boson mass or how to arrive at a good χ 2 LEP II NuTeV χ 2 min =.8, n dof = 4, probability = M W [GeV] CDF I D I CDF II LEP II NuTeV Taking out NuTeV result: χ 2 min =.7, n dof = 3, probability =.64 Outlier rejection, is this allowed? M W [GeV] CDF I D I CDF II LEP II NuTeV M W [GeV] Scaling all errors by S = χ 2 min /n dof =.64 χ 2 min = 4., n dof = 4, probability =.4 Standard procedure by Particle Data group destroying the hard work of many experimentalists 42

43 Mini summary of what we have learnt The χ 2 min of a fit is a consistency check Expect χ 2 min/n dof for good fits 43 if χ 2 min/n dof significantly larger than one then suspect data could contain outliers or errors are (generally) underestimated the fitfunction might not be the correct model for the data for repeated experiments (e.g. many track fits) expect for good fits mean value of χ 2 min/n dof distribution and flat prob(χ 2 min, n dof ) distribution in interval [,]

44 44 Lecture part 2 General solution for linear least square fits (normal equations) Straight line fits

45 Our study object y (cm) Det Det 2 Det 3 The main trajectories we will study in this workshop are straight lines: y i = a + a x i This is a classical linear least square fit problem x (cm) 45

46 y vector of n measurements Linear least square fits y (x ). y n (x n ) with cov-matrix V Linear model y : = A a, a vector of m fitparameters Example: y = a ; a = (a ); A =.. In general: A = A( x), but no dependence on a Master formula : χ 2 = ( y A a) t V ( y A a) a. a m to be minimised w.r.t a obtain estimators ˆ a and covariance matrix Vˆ a 46

47 y vector of n measurements Linear least square fits y (x ). y n (x n ) with cov-matrix V Linear model y : = A a, a vector of m fitparameters Example: y = a ; a = (a ); A =.. In general: A = A( x), but no dependence on a Master formula : χ 2 = ( y A a) t V ( y A a) a. a m to be minimised w.r.t a obtain estimators ˆ a and covariance matrix Vˆ a 47

48 Examples for linear least square fits Linear means that y depends linearly on the fitparameters a i. a = ( a a ) ; A = x.. x n a = (a); A = e x.. e x n Watch out: function can be highly non-linear in x 48

49 49 General solution via normal equations χ 2 = ( y A a) t V ( y A a) = y t V y 2 a t AV y + a t A t V A a Min. χ 2 dχ2 d a t = 2A t V y + 2A t V A a = Solution: ˆ a = (A t V A) A t V y = H A t V y = UA t V y with U = H = Cov(ˆ a) Normal equations: Powerful & simple linear algebra to solve fit!

50 5 General solution via normal equations χ 2 = ( y A a) t V ( y A a) = y t V y 2 a t AV y + a t A t V A a Min. χ 2 dχ2 d a t = 2A t V y + 2A t V A a = Solution: ˆ a = (A t V A) A t V y = H A t V y = UA t V y with U = H = Cov(ˆ a) Normal equations: Powerful & simple linear algebra to solve fit!

51 5 Straight line fit through n detector layers y (cm).4.3 Det Det 2 Det 3 χ 2 = n i= y = A a; a = (y i a a x i ) 2 ( ) a a σ 2 x ; A=.. x n ; A t = (.. x.. x n Apply normal equations: ˆ a = (A t V A) A t V y = σ 2 (A t A) i = = ( U =.. i x i x x x 2 ( i x i i x2 i ) ( y xy ) ( i y i i x iy i = σ 2 â cov(â, â ) cov(â, â ) σ 2 â x 2 x 2 ) ) ; V = σ 2. σ 2 y = (A t A) A t y σ 2At ) ( ) ( N Nx = Nx Nx 2 ( x 2 x x ) ( y xy = (A t V A) = σ2 NV [x] Ny Nxy ) ) ( = V [x] x 2 y xxy xy + xy ( ) x 2 x x x (cm) )

52 52 Straight line fit through n detector layers y (cm).4.3 Det Det 2 Det 3 χ 2 = n i= y = A a; a = (y i a a x i ) 2 ( ) a a σ 2 x ; A=.. x n ; A t = (.. x.. x n Apply normal equations: ˆ a = (A t V A) A t V y = σ 2 (A t A) i = = ( U =.. i x i x x x 2 ( i x i i x2 i ) ( y xy ) ( i y i i x iy i = σ 2 â cov(â, â ) cov(â, â ) σ 2 â x 2 x 2 ) ) ; V = σ 2. σ 2 y = (A t A) A t y σ 2At ) ( ) ( N Nx = Nx Nx 2 ( x 2 x x ) ( y xy = (A t V A) = σ2 NV [x] Ny Nxy ) ) ( = V [x] x 2 y xxy xy + xy ( ) x 2 x x x (cm) )

53 53 Straight line fit through n detector layers y (cm).4.3 Det Det 2 Det 3 χ 2 = n i= y = A a; a = (y i a a x i ) 2 ( ) a a σ 2 x ; A=.. x n ; A t = (.. x.. x n Apply normal equations: ˆ a = (A t V A) A t V y = σ 2 (A t A) i = = ( U =.. i x i x x x 2 ( i x i i x2 i ) ( y xy ) ( i y i i x iy i = σ 2 â cov(â, â ) cov(â, â ) σ 2 â x 2 x 2 ) ) ; V = σ 2. σ 2 y = (A t A) A t y σ 2At ) ( ) ( N Nx = Nx Nx 2 ( x 2 x x ) ( y xy ) = (A t V A) = σ2 NV [x] Ny Nxy = V [x] ( ( x 2 x x ) x 2 y xxy xy + xy ) x (cm) )

54 54 Straight line fit through n detector layers y (cm).4.3 Det Det 2 Det 3 χ 2 = n i= y = A a; a = (y i a a x i ) 2 ( ) a a σ 2 x ; A=.. x n ; A t = (.. x.. x n Apply normal equations: ˆ a = (A t V A) A t V y = σ 2 (A t A) i = = ( U =.. i x i x x x 2 ( i x i i x2 i ) ( y xy ) ( i y i i x iy i = σ 2 â cov(â, â ) cov(â, â ) σ 2 â x 2 x 2 ) ) ; V = σ 2. σ 2 y = (A t A) A t y σ 2At ) ( ) ( N Nx = Nx Nx 2 ( x 2 x x ) ( y xy = (A t V A) = σ2 NV [x] Ny Nxy ) ) ( = V [x] x 2 y xxy xy + xy ( ) x 2 x x x (cm) )

55 55 Mini exercise: straight line track-fit The covariance formula ( σ 2 â cov(â, â ) cov(â, â ) σ 2 â ) = σ2 NV [x] ( x 2 x x ) is valid for e.g. a straight line track fit in N detectors of resolution σ: Determine the improvements on the slope error σ a by: a) Doubling the number of detector layers N within the same interval in x b) Distributing the detector layers over an interval in x twice as large c) Buying detectors with measurement uncertainties σ reduced by a factor two

56 Computer exercise straight line trajectory fit Physics example: A muon track is measured in four layers of streamer tube detectors at x positions of 4., 5., 6. and 7. (in cm), with a measurement precision for y of.5 cm. The goal is to determine its trajectory assuming a straight line. Macro StraightLineFit.C, accessible at http : // õbehnke/stat/gean/straightlinef it.c 56 fits a straight line track trajectory through four measured points. Steering parameters in the macro: xmin, xmax = Interval of the trajectory displayed Output: Tasks: Histogram data (it s of the type TGraphErrors) Plots are drawn of the fitted histogram with error bands error ellipse of the two fitparameters a) Run the macro as it is by.x StraightLineFit.C and fill the fit results for p, p, their errors and correlation into the table below b) Precision of trajectory: Evaluate (by eye) from the shown error bands at which point roughly the trajectory is known best and with which precision (fill the results in the table below) c) Precision of extrapolated trajectory: Evaluate the precision of the extrapolated trajectory at x = (Hint: Change xmax to large value and run the macro again) d) Effect of shift of x coordinate origin: Shift all four xv al points in the macro (simply by overwriting by hand) by a constant value -5.5, set xmin = 4. and xmax = 4. and run the macro again. Fill the fit results in the table. Can you explain why the correlation of p and p has changed? e) Apply a very precise vertex constraint at the origin: Change N to 5 and add a new first point to the measurement points list with xv al =., xerr =., yv al =. and yerr =. (just by hand). Run the macro again and write down the fitted results in the table. How much are the parameter errors reduced by adding this extra point?

57 Task results sheet 57 Task a) Task b) Straight line fit trough four points p = p = corr = x-best precision = y-error = Task c) y-error(x = ) = Task d) Task e) Shifting all x values by -5.5: p = p = corr = Adding vertex constraint at x = : p = p = corr =

58 Mini summary of what we have learnt 58 Linear least square problems: y = A a, y is a linear function of the fitparameters a but can be a linear or nonlinear function of the continuous parameter x. The normal equations are a powerful tool to solve linear least square fit problems ˆ a = (A t V A) A t V y, cov(ˆ a) = (A t V A) Straight line fits are a typical application and there are many others (e.g. parabolas, higher order polynoms, etc.)

59 59 Appendix Content: Proof that χ 2 min for averaging two measurements follows χ 2 -distribution with one degree of freedom Linear least square fits: Covariance matrix of fit parameters

60 χ 2 for two measurements with unknown true value χ 2 min = (y â) 2 σ 2 + (y 2 â) 2 σ 2 2 χ 2 min = G ; â = σ 2 + σ 2 2 ( y σ 2 ( y (G y + G 2 y 2 ) G + G 2 + y ) 2 σ2 2 = G y + G 2 y 2 G + G 2 (with G i := /σ 2 i )) ) 2 ( + G 2 y 2 (G y + G 2 y 2 ) G + G 2 ( ) 2 ( ) 2 (G2 y G 2 y 2 ) (G y 2 G y ) = G + G 2 G + G 2 G + G 2 G G 2 2 = (G + G 2 ) 2(y y 2 ) 2 G 2 G 2 + (G + G 2 ) 2(y y 2 ) 2 = G G 2 (G + G 2 ) (G + G 2 ) 2 (y y 2 ) 2 = G G 2 (y y 2 ) 2 G + G 2 = (y y 2 ) 2 = /G + /G 2 σ 2 + (y y 2 ) 2 σ2 2 ) 2 = y y 2 σ 2+σ2 2 should follow (errorpropagation!) gauss distribution e 2 2 χ 2 = 2 follows -dim χ 2 distr.! One degree of freedom sacrificed for determination of â. General: n-measurements with one unknown a follows χ 2 distribution with n degrees of freedom 6

61 6 Linear least square fits: Covariance Matrix Proof that Covariance matrix U of fit parameters ˆ a is given by U = H Use Normal Equations: ˆ a = B y with B = H A t V Then apply errorpropagation: U = BV B t = H A t V V V AH = H A t V AH = H HH = H