Es272 ch5a

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Es272 ch5a

  1. 1. Part 5a: LEAST-SQUARES REGRESSION – – – – – Simple Linear Regression Polynomial Regression Multiple Regression Statistical Analysis of L-S Theory Non-Linear Regression
  2. 2. Introduction: Consider the falling object in air problem: t0 m v0 t1 m “Best fit” v1 v t tn m vn  (t) values considered to be error-free.  Every measurement of (v) contain some error.  Assume error in (v) are normally distributed (random error).  Find “best fit” curve to represent v(t)
  3. 3. Simple Linear Regression  Consider a set of n scattered data  Find a line that “best fits” the scattered data y a0 a0= intercept a1= slope a1 x  There are a number of ways to define the “best fit” line. However we want to find one that is unique, i.e., for a particular set of data.  A uniquely defined best-fit line can be found by minimizing the sum of the square of the residuals from each data point: n Sr n ( ymeas i 1 y fit ) 2 ( yi a0 a1 xi ) 2 i 1 Find a0 and a1 that minimizes Sr (least-square) sum of the square of the residuals (or spread)
  4. 4. To minimize Sr (a0 , a1), differentiate and set to zero: n Sr a0 2 ( yi a0 a1 xi ) 0 [( yi a0 a1 xi ) xi ] 0 i 1 n Sr a1 2 i 1 or 0 na0 xi a0 yi a0 xi a1 xi2 a1 a1xi yi 0 yi xi a0 xi a1 xi2 Normal equations for simple linear L-S regression xi yi Need to solve these simultaneous equations for the unknowns a0 and a1
  5. 5. Solution for a1 and a0 gives n a1 xi yi n xi 2 i x xi yi 2 and yi a0 n xi a1 n y a1 x EX: Find linear fit for the set of measurements: x 1 0.5 2 2.5 3 2.0 4 4.0 5 3.5 6 6.0 7 y y 5.5 0.0714 0.839x n 7 xi 28 x 4 xi yi 119.5 a1 a0 yi y 3.4286 xi2 140 7(119 .5) 2(24 ) 7(140 ) (28 ) 2 3.4286 24 0.839 (4) 0.839 0.0714
  6. 6. Quantification of Error: Sum of the square of the residuals for the mean 2 n St sum of the square of the residuals for the linear regression ( yi y) Sr ( yi a0 a1 xi ) i 1 i 1 standard deviation sy 2 n St n 1 standard error of the L-S estimate sy/ x Sr n 2 All these approaches are based on the assumptions: x > error-free y > normal error
  7. 7. “Coefficient of determination” is defined as r St 2 Sr r St Sr =0 (r=1) Sr =St (r=0) “correlation coefficient” Perfect fit No improvement by fitting the line Alternative formulation for the correlation coefficient n r n 2 i x xi yi xi xi 2 n yi 2 i y yi Note: r 1 does not always necessarily mean that the fit is “good”. You should always plot the data along with the regression curve to see the goodness of the fit. 2 Four set of data with same r=0.816
  8. 8. Linearization of non-linear relationships:  Many engineering applications involve non-linear relationships, e.g., exponential, power law, or saturated growth rate. exponential power-law saturated growth-rate y a1e b1x y a2 x b2 y a3 x b3 x  These relationships can be linearized by some mathematical operations: ln y ln a1 b1 x log y b2 log x log a2 1 y b3 1 a3 x  Linear L-S fit can be applied to find the coefficients. 1 a3
  9. 9. EX: Fit a power law relationship to the following dataset: x y 1 0.5 2 1.7 3 3.4 4 5.7 5 8.4 Power law model y log y a2 x b2 b2 log x log a2 (find a2 and b2) Calculate logarithm of both data: log x log y 0 -0.301 0.301 0.226 0.477 0.534 0.602 0.753 0.699 0.922 Applying simple linear regression gives; slope=1.75 and intercept=-0.300 b2 1.75 log a2 y 0.5 x1.75 0.300 a2 0.5
  10. 10. Polynomial Regression  In some cases, we may want to fit our data to a curve rather than a line. We can then apply polynomial regression (In fact, linear regression is nothing but an n=1 polynomial regression). Data to fit to a second order polynomial: y a0 a1 x a2 x 2 Sum of the square of the residuals (spread) 2 n Sr ( yi ,obs yi , fit ) i 1 2 n ( yi a0 a1 xi a2 xi2 ) i 1 To minimize Sr(a0, a1, a2), take derivatives and equate to zero: Sr a0 n 2 ( yi i 1 a0 a1 xi a2 xi2 ) 0
  11. 11. Sr a1 Sr a2 n xi ( yi a0 a1 xi a2 xi2 ) 0 xi2 ( yi 2 a0 a1 xi a2 xi2 ) 0 i 1 n 2 i 1 Three linear equations with three unknowns a0, a1, a2 : (n)a0 xi a1 xi2 a2 yi xi a0 xi2 a1 xi3 a2 xi yi “normal equations” xi2 a0 xi3 a1 xi4 a2 xi2 yi all summations are i=1..n  This set of equations can be solved by any linear solution techniques (e.g., Gauss elimination, LU Dec., Cholesky Dec., etc.)
  12. 12.  The approach can be generalized to order (m) polynomial following the same way. Now, the fit function becomes y a1 x a2 x 2 .. am x m a0  This will require the solution of an order (m+1) system of linear equations. The standard error becomes Because (m+1) degrees of freedom was lost from data of (n) due to extraction of (m+1) coefficients . Sr n (m 1) sy / x EX 17.5: Fit an 2nd order polynomial to the following data x i yi 152.6 m 2 2 i x 55 xi yi xi2 yi 3 i x 225 2488.8 4 i x 979 2.1 7.7 2 13.6 3 585.6 0 1 xi 15 n 6 yi 27.2 4 40.9 5 61.1
  13. 13. System of linear equations: 6 15 55 a0 152.6 15 55 225 a1 585.6 55 225 979 a2 2488.8 We get a0 2.47857 a1 2.35929 a2 1.86071 Then, the fit function: y 2.47857 2.35929 x 1.86071 x 2 Standard error: Sr n (m 1) sy / x where 3.74657 1.12 6 (3) 2 6 Sr ( yi i 1 2.47857 2.35929xi 1.86071xi2 ) 3.74657
  14. 14. Multiple Linear Regression  In some cases, data may have two or more independent variables. In this example, for a function of two x 2 variables, the linear regression gives a planar fit function. y ( x1 , x2 ) x1 Function to fit y a0 a1 x1 a2 x2 Sum of the square of the residuals (spread) 2 n Sr ( yi ,obs i 1 2 n yi , fit ) ( yi a0 a1 x1i a2 x2i ) i 1
  15. 15. Minimizing the spread function gives: n Sr a0 2 ( yi a0 a1 x1i a 2 x2 i ) 0 i 1 n Sr a1 2 x1i ( yi a0 a1 x1i a 2 x2 i ) 0 x2 i ( yi a0 a1 x1i a 2 x2 i ) 0 i 1 n Sr a2 2 i 1 The system of equations to be solved: n x1i 2 1i x1i x x2 i x1i x2i x2 i a0 yi x1i x2i a1 x1i y1i 2 x2 i a2 x 2 i yi Normal equations for multiple linear regression
  16. 16. EX 17.7: Fit a planar surface to the following data x1 x2 y 0 0 5 2 1 10 2.5 2 9 1 3 0 4 6 3 7 2 27 We first do the following calculations: y x1 x2 x1x1 x2x2 x1x2 x1y x2y 5 0 5 0 0 0 0 0 10 1 10 4 1 2 20 10 9 2 9 6.25 4 5 22.5 18 0 3 0 1 9 3 0 0 3 6 3 16 36 24 12 18 27 2 27 49 4 14 189 54 54 16.5 14 76.25 54 48 243.5 100
  17. 17. The system of equations to calculate the fit coefficients: 6 16.5 14 a0 16.5 76.25 48 a1 14 48 54 a2 54 243.5 100 returns a0 a1 5 The fit function y 4 a2 3 5 4 x1 3x2  For the general case of a function of m-variables, the same strategy can applied. The fit function in this case: y a0 a1 x1 a2 x2 .. am xm Standard error: sy / x Sr n (m 1)
  18. 18.  A useful application of multiple regression is for fitting a power law equation of multiple variables of the form: y a a a0 x1a1 x2 2 .. xmm Linearization of this equation gives log y log a0 a1 log x1 ... am log xm  The coefficients in the last equation can be calculated using multiple linear regression, and can be substituted to the original power law equation.
  19. 19. Generalization of L-S Regression:  In the most general form, L-S regression can be stated as y a0 z0 a1 z1 ... am zm In general, this form is called “linear regression” as the fitting coefficients are linearly dependant on the fit function. functions z0 x 0 , z1 z0 1 , z1 x1 , ..., z m x1 , ..., zm xm xm Polynomial regression Multiple regression  Other functions can be defined for fitting as well, e.g., y a0 a1 cos t a2 sin t
  20. 20. For a particular data point y a0 z0 a1 z1 ... am zm e data For n data (in matrix form): y z10 Z Z a e y1 y2 ... yn coefficients a0 a1 ... am residuals z11 ... z1m ... ... zn 0 y a e Calculated based on the measured independant variables zn1 znm m: order of the fit function n: number of data points Z is generally not a square matrix. n m 1 e1 e2 ... en
  21. 21. Sum of the square of the residuals: n 2 m Sr ( yi i 1 a j z ji ) j 0 To determine the fit coefficients, minimize S r (a0 , a1 ,.., am ) This is equivalent to the following: Z T Z a Z T y Normal equations for the general L-S regression  This is the general representation of the normal equations for L-S regression including simple linear, polynomial, and multiple linear regression methods.
  22. 22. Solution approaches: Z T Z a Z T y A symmetric and square matrix of size [m+1 , m+1]  Elimination methods are best suited for the solution of the above linear system: LU Decomposition / Gauss Elimination Cholesky Decomposition  Especially, Cholesky decomposition is fast and requires less storage. Furthermore,  Cholesky decomposition is very appropriate when the order of the polynomial fit model (m) is not known beforehand. Successive higher order models can be efficiently developed.  Similarly, increasing the number of variables in multiple regression is very efficient using Cholesky decomposition.
  23. 23. Statistical Analysis of L-S Theory Some definitions:  If a histogram of the data shows a bell shape curve, normally distributed data.  This has a well-defined statistics n yi y sy 2 sy mean i 1 n yi n 1 St n 1 y 2 Standard deviation variance  For a perfectly normal distribution: mean±std fall about 68% of the total data. mean±2std fall about 95% of the total data. : true mean : true std
  24. 24. Confidence intervals:  Confidence interval estimates intervals within which the parameter is expected to fall, with a certain degree of confidence.  Find L and U values such that PL U 1 true mean significance level For 95% confidence interval =0.05 L U y y sy n sy n t t / 2,n 1 / 2,n 2 t-distribution (tabulated in books); in EXCEL tinv ( ,n) e.g., for =0.05 and n=20 t /2, n-1=2.086  T-distribution is used to compramize between a perfect and an imperfect estimate. For example, if data is few (small n), t-value becomes larger, hence giving a more conservative interval of confidence.
  25. 25. EX: Some measurements of coefficient of thermal expansion of steel (x10-6 1/°F): 6.495 6.665 6.755 6.565 6.595 6.505 6.625 6.515 6.615 6.435 6.715 6.555 6.635 6.625 6.575 6.395 6.485 6.715 6.655 6.775 6.555 6.655 6.605 6.685 n=8 n=16 n=24 Find the mean and corresponding 95% confidence intervals for the a) first 8 measurements b) first 16 measurements c) all 24 measurements. For n=8 L y U y y sy n sy n 6.59 t t sy 0.089921 t / 2,n 1 t0.05 / 2,8 / 2,n 1 6.59 0.089921 2.364623 8 / 2,n 2 0.089921 2.364623 8 6.6652 2.364623 6.5148 6.59 1 6.5148 6.6652 For eight measurements, there is a 95% probability that true mean falls between these values.
  26. 26. The cases of n=16 and n=24 can be performed in a similar fashion. Hence we obtain: n mean(y) 8 6.5900 16 24 sy t L U 0.089921 2.364623 6.5148 6.6652 6.5794 0.095845 2.131451 6.5283 6.6304 6.6000 0.097133 2.068655 6.5590 6.6410 /2,n-1 Results shows that confidence interval narrows down as the number of measurements increases (even though sy increases by increasing n!). For n=24 we have 95% confidence that true mean is between 6.5590 and 6.6410.
  27. 27. Confidence Interval for L-S regression:  Using matrix inverse for the solution of (a) is inefficient: a Z T Z 1 Z T y  However, inverse matrix carries useful statistical information about the goodness of the fit. Z T Z 1 Inverse matrix Diagonal terms coefficients variances (var) of the fit Off -diagonal terms the fit coefficients covariances (cov) of 2 var(ai 1 ) uii s y / x cov(ai 1, a j ) ui 2 sy / x 1, j uij: Elements of the inverse matrix  These statistics allow calculation of confidence intervals for the fit coefficients.
  28. 28.  Calculating confidence intervals for simple linear regression: y a0 a1 x For the intercept (a0) L a0 t / 2,n 2 s ( a0 ) U a0 t / 2,n 2 s ( a0 ) For the slope (a1) L a1 t U a1 t / 2,n 2 s (a1 ) / 2,n 2 s (a1 ) Standard error for the coefficient (extracted from the inverse matrix) s(ai ) var(ai )
  29. 29. EX 17.8: Compare results of measured versus model data shown below. a) Plot the measured versus model values. b) Apply simple linear regression formula to see the adequacy of the measured versus model data. c) Recompute regression using matrix approach, estimate standard error of the estimation and for the fit parameters, and develop confidence intervals. a) 60 Model value 8.953 16.405 22.607 27.769 32.065 35.641 38.617 41.095 43.156 44.872 46.301 47.49 48.479 49.303 49.988 50 40 model Measured Value 10 16.3 23 27.5 31 35.6 39 41.5 42.9 45 46 45.5 46 49 50 30 20 10 0 0 20 40 60 measured b) Applying simple linear regression formula gives y 0.859 1.032x x: measured y: model
  30. 30. c) For the statistical analysis, first form the following [Z] matrix and (y) vector 1 Z Then, 10 8.953 1 16.3 .. .. .. 1 16.405 .. y .. 50 Z .. 49.988 T T Z a Z 548.3 a0 552.741 548.3 22191.21 a1 22421.43 15 y Solution using the matrix inversion a a0 a1 0.688414 Z T Z 1 Z 0.01701 T y 552.741 0.85872 0.01701 0.000465 22421.43 1.031592
  31. 31. Standard error for the fit function: Sr n 2 sy / x 0.863403 Standard error for the coefficients: s(a0 ) 2 u11s y / x 0.688414(0.863403) 2 0.716372 s(a1 ) 2 u22 s y / x 0.000465(0.863403) 2 0.018625 For a 95% confidence interval ( =0.05, n=13, Excel returns inv(0.05,13)=2.160368) a0 a0 t / 2, n 2 s(a0 ) 0.85872 2.160368(0.716372) 0.85872 1.547627 a1 a1 t / 2, n 2 s(a1 ) 1.031592 2.160368(0.018625) 1.031592 0.040237 Desired values of slope=1 and intercept=0 falls in the intervals (hence we can conclude that a good fit exist between measured and model values).
  32. 32. Non-linear Regression  In some cases we must fit a non-linear model to the data, e.g., y a0 (1 e a1 x ) parameters a0 and a1 are not linearly dependant on y  Generalized L-S formulation cannot be used for such models.  Same approach of using sum of square of the residuals are applied, but the solution is sought iteratively. Gauss-Newton method:  A Taylor series expansion is used to (approximately) linearize the model. Then standard L-S theory can be applied to estimate the improved estimates of the fit parameters. In most general form y f ( x; a0 , a1 ,..am )
  33. 33. Taylor series around the fit parameters f ( xi ) j f ( xi ) j f ( xi ) 1 a0 f ( xi ) j a0 a1 i: i-th data point j: iteration number a1 Then ymeas f ( xi ) j y fit a0 a0 f ( xi ) j a1 a1 In matrix form: d Zj iteration number d a y1 f ( x1 ) y2 f ( x2 ) ... yn f ( xn ) Zj f1 a0 f2 a0 ... fn a0 f1 a0 f2 a0 ... fn a0 a a0 a1
  34. 34. Applying the generalized L-S formula Zj T Zj a Zj T d  We solve the above system for ( A) for improved values of parameters: a0 , j 1 a0 , j a0 a1, j 1 a1, j a1  The procedure is iterated until an acceptable error: a0 , j a 0 1 a0 , j a0 , j a1, j a 1 1 1 a1, j a1, j 1

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