R. Kass/Sp07P416/Lecture 71 More on Least Squares Fit (LSQF) In Lec 5, we discussed how we can fit our data points to a linear function (straight line) and get the "best" estimate of the slope and intercept. However, we did not discuss two important issues: I) How to estimate the uncertainties on our slope and intercept obtained from a LSQF? II) How to apply the LSQF when we have a non-linear function? Estimation of Errors on parameters determined from a LSQF Assume we have data points that lie on a straight line: y =  +  x Assume we have n measurements of the x’s and y's. For simplicity, assume that each y measurement has the same error, . Assume that x is known much more accurately than y.  ignore any uncertainty associated with x. Previously we showed that the solution for the intercept  and slope  is: Some Advanced Topics using Propagation of Errors and Least Squares Fitting

R. Kass/Sp07P416/Lecture 72 Since  and  are functions of the measurements (y i 's) we can use the Propagation of Errors technique to estimate   and  . Assume that:a) Each measurement is independent of each other (  xy =0). b) We can neglect any error (  ) associated with x. c) Each y measurement has the same , i.e.  i = . See lecture 4 (general formula for 2 variables) assumption a) assumption b) assumption c) plugging our formula for 

R. Kass/Sp07P416/Lecture 73 We can find the variance in the slope (  ) using exactly the same procedure: variance in the intercept

R. Kass/Sp07P416/Lecture 74 n If we don't know the true value of  we can estimate the variance using the spread between the measurements (y i ’s) and the fitted values of y: n  2= number of degree of freedom = number of data points  number of parameters (  ) extracted from the data n If each y i measurement has a different error  i : The above expressions simplify to the “equal variance” case. Don't forget to keep track of the “n’s” when factoring out equal  ’s. For example: variance in the slope weighted slope and intercept

R. Kass/Sp07P416/Lecture 75 x y  y x Previous fit to: f(x,b)=1+bx, b=1.05 New fit to: f(x,b)=a+bx, a =1.158 ± b=1.011 ± Example: Fit to straight line (same data as Lec 6) Errors on a and b calculated using: a and b calculated using weighted fit (Taylor, P 198) χ 2 =0.65 for 2 degrees of freedom P(χ 2 >0.65 for 2dof) = 72%

R. Kass/Sp07P416/Lecture 76 l LSQF with non-linear functions: u For our purposes, a non-linear function is a function where one or more of the parameters that we are trying to determine (e.g. ,  from the straight line fit) is raised to a power other than 1. n Example: functions that are non-linear in the parameter  : However, these functions are linear in the parameters A. u The problem with most non-linear functions is that we cannot write down a solution for the parameters in a closed form using, for example, the techniques of linear algebra (i.e. matrices). n Usually non-linear problems are solved numerically using a computer. n Sometimes by a change of variable(s) we can turn a non-linear problem into a linear one. Example: take the natural log of both sides of the above exponential equation: lny=lnA-x/  =C+Dx Now a linear problem in the parameters C and D! In fact it’s just a straight line! To measure the lifetime  (Lab 7) we first fit for D and then transform D into . u Example: Decay of a radioactive substance. Fit the following data to find N 0 and  : n N represents the amount of the substance present at time t. n N 0 is the amount of the substance at the beginning of the experiment (t = 0). n  is the lifetime of the substance.  = -1/D

R. Kass/Sp07P416/Lecture 77 i titi NiNi y i = ln N i n The intercept is given by: C = 4.77 = ln A or A = Lifetime Data time, t ln[N(t)]

R. Kass/Sp07P416/Lecture 78 u Example : Find the values A and  taking into account the uncertainties in the data points. n The uncertainty in the number of radioactive decays is governed by Poisson statistics. n The number of counts N i in a bin is assumed to be the average (  ) of a Poisson distribution:  = N i = Variance n The variance of y i (= ln N i ) can be calculated using propagation of errors: n The slope and intercept from a straight line fit that includes uncertainties in the data points: If all the  's are the same then the above expressions are identical to the unweighted case.  =4.725 and  =  = -1/  = -1/ = sec n To calculate the error on the lifetime (  ), we first must calculate the error on  : The experimentally determined lifetime is:  = ± 12.3 sec. Taylor P198 and problem 8.9

R. Kass/Sp07P416/Lecture 79 Least Squares Fitting-Exponential Example We can calculate the  2 to see how “good” the data fits an exponential decay distribution: For this problem: lnA=4.725  A= and  = sec Poisson approximation The chi sq per dof is 1.96 The chi sq prob. is 4.9 % Mathematica Calculation: Do[{csq=csq+(cnt[i]-a*Exp[-x[i]/tau])^2/(a*Exp[-x[i]/tau])},{i,1,10}] Print["The chi sq per dof is ",csq/8] xvt=1-CDF[ChiSquareDistribution[8],csq]; Print["The chi sq prob. is ",100*xvt,"%"] This is not such a good fit since the probability is only ~4.9%.

R. Kass/Sp07P416/Lecture 710 The Error on the Mean Error on the mean (review from Lecture 4) Question: If we have a set of measurements of the same quantity: What's the best way to combine these measurements? How to calculate the variance once we combine the measurements? Assuming Gaussian statistics, the Maximum Likelihood Method says combine the measurements as: If all the variances (   1 =   2 =   3..) are the same: The variance of the weighted average can be calculated using propagation of errors: weighted average unweighted average  x =error in the weighted mean And the derivatives are:

R. Kass/Sp07P416/Lecture 711 If all the variances are the same: The error in the mean (  x ) gets smaller as the number of measurements (n) increases. Don't confuse the error in the mean (  x ) with the standard deviation of the distribution (  )! If we make more measurements: The standard deviation (  ) of the (gaussian) distribution remains the same, BUT the error in the mean (  x ) decreases Lecture 4 Example: Two experiments measure the mass of the proton: Experiment 1 measures m p =950±25 MeV Experiment 2 measures m p =936±5 MeV Using just the average of the two experiments we find: m p =( )/2=943MeV Using the weighted average of the two experiments we find: m p =(950/ /5 2 )/(1/ /5 2 )=936.5MeV and the variance:  2 =1/(1/ /5 2 )=24 MeV 2  =4.9 MeV Since experiment 2 was more precise than experiment 1 we would expect the final result to be closer to experiment 2’s value. m p =(936.5±4.9) MeV