1 / 31

Probability function P on subspace of S P(S)=1 For every event A in S, P(A)≥0

Probability function P on subspace of S P(S)=1 For every event A in S, P(A)≥0 If A 1 , A 2 , .. are mutually exclusive, then P(  A i )=∑ P(A i )

kobe
Download Presentation

Probability function P on subspace of S P(S)=1 For every event A in S, P(A)≥0

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Probability function P on subspace of S P(S)=1 For every event A in S, P(A)≥0 If A1, A2, .. are mutually exclusive, then P(Ai)=∑ P(Ai) A random variable X is a function that assigns a value to each outcome s in the sample space S (realizations of the random variable). Example: dart, bulls eye counts 50, i.e. X(s)=50, s = bulls eye location Example: measurement of a mass five times, yielding the true value m + random measurement errors ei (m+e1, m+e2, …)

  2. Probability Density Function (PDF), fX(x) The relative probability of realizations for a random variable P(X≤a)=∫ fX(x)dx ∫ fX(x)dx = 1 PDF for common functions: Uniform random variable on [a,b]: fU(x)=1/(b-a), a≤x≤b; 0, x<a or x>b a - - -

  3. PDF for common functions: 2. Normal (Gaussian) function: fN(x)=1/√2 exp(-0.5(x-)2/2) N(2) normal distribution 3. Exponential function: fexp(x)= exp(-x), x≥0; 0, x<0

  4. 4. Double-sided Exponential function: fdexp(x)=[1/(23/2] exp(-√2|x-) 5.  function: f  x 0.5-1 exp(-x/2) with xx-1e-d n independent variable with standard normal distributions, Z=∑Xi2 is arandom variable with =n degrees of freedom.  0

  5. 6 . Student's t distribution with  degrees of freedom: ft(x)/2) 1/√ (1+x2/)-(+1)/2 with xx-1e-d Approaches a standard normal distribution for large number of degrees of freedom Cumulative distribution functions (CDF): FX(a)=P(X≤a)=∫ fX(x)dx P(a≤X≤b) = ∫f(x)dx a - b a

  6. Characterization of PDF’s: Expected value of random variable x: E[X] = ∫ xfX(x)dx, E[g(X)] = ∫ g(x)fX(x)dx Peak of distribution: XML (maximum likelihood) Variance: Var(X)=X2 = E[(X-x)2] = E[X2]-X2 = ∫ (x-x)2 fX(x)dx, x=√Var(X) Variance measures the width of PDF’s. Wide PDF’s indicate noisy data, narrow indicate relatively noise-free data.  -  -  -

  7. Joint PDF’s: The joint PDF can quantify the probability that a set of random variables will take on a given value. f(X≤a,Y≤b)= ∫ ∫ f(x,y)dydx Expected value for joint PDF: E[g(X,Y)]=∫ ∫ g(x,y)f(x,y)dydx If X and Y are independent, f(x,y)=fX(x)fY(y) a - b -  -  -

  8. Covariance of X and Y with Joint PDF: Cov(X,Y)=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y] X and Y independent, E[XY]=E[X]E[Y] and Cov(X,Y)=0 Covariance of a variable with itself = variance If Cov(X,Y)=0, X and Y are called uncorrelated Correlation of X and Y (X,Y)=Cov(X,Y)/√Var(X) Var(Y) (correlation is a scaled covariance)

  9. Model covariance matrix cov(m): The general model mest=Md + v has covariance matrix M[cov d]MT (B.64) The least-squares solution mest=[GTG]-1GTd has covariance matrix [cov m] = [GTG]-1GT [cov d] ([GTG]-1GT)T If data is uncorrelated and has equal variance d2 then [cov m] = d2 [GTG]-1

  10. More on Gaussian (Normal) distributions: Central limit theorem: Let X1, X2, …, Xn be independent and identically distributed random variables with finite expected value  and variance 2. Let Zn=[X1+X2+..+Xn-n/√n In the limit as n approaches infinity, Zn approaches the standard normal distribution Thus many summed variables in nature are normal distributions, thus LS solutions OK

  11. Means and confidence intervals: Given noisy measurements m1, m2, .., mn Estimate the true value m and the uncertainty of the estimate. Assume errors are independent and normally distributed with expected value 0 and some unknown standard deviation  Compute average mave=[m1+m2+..+mn/n s=[∑ (mi-mave)2]1/2/n-1 n i=1

  12. Sampling Theorem: Independent, normally distributed measurements, with expected value m and standard deviated , the random quantity t = (m-mave)/s√n has a Student’s t distribution with n-1 degrees of freedom. If the true standard deviation  is known, we are dealing with a standard normal distribution. The t distribution converges toward the normal distribution for large n

  13. Confidence intervals: Probability that one realization falls within a specified distance of the true mean Let tn-1,0.975 = 97.5 percentile of t distribution tn-1,0.025 = 2.5 percentile of t distribution P(tn-1,0.975 ≤m-mave/(s/√n) ≤tn-1,0.025)=0.95 P(tn-1,0.975 s/ √n ≤m-mave ≤tn-1,0.025 s/√n )=0.95 95% confidence interval Due to symmetry: mave-tn-1,0.975 s/ √n to mave+tn-1,0.975 s/ √n

  14. Confidence intervals related to the  - PDF’s with large  will have large CI, and vice versa Gaussian PDF, the 68% CI is 1 wide and the 95% CI is 2 wide If a particular Gaussian random variable has =1, and if a realization of that variable is 50, there is a 95% chance that the mean of that random variable lies between 48 and 52

  15. Example B.12 illustrates the case where  is estimated If  is known (rarely the case), we have a normal distribution. We can use the t-distribution with an infinite # of observations. E.g., 16 obs, m estimated to be 31.5. is known to be 5. Estimate 80% CI for m. mave-k ≤ m ≤ mave+k k=1.282 x 5/√16 = 1.6 31.5-1.6 ≤ m ≤ 31.5+1.6

  16. Statistical Aspects of LS: PDF for Normal Distribution: fi(di|m)=1/√2 exp(-(di-(Gm)i)2/22) Maximum likelihood function L(m|d) is the product of all individual probability functions: L(m|d)=f1(d1|m)*f2(d2|m)* … * fm(dm|m) Idea: Maximize L(m|d) Maximize log{L(m|d)} Minimize - log{L(m|d)} Minimize -2 log{L(m|d)} min ∑ [di-(Gm)I]2/2 Aside from the 1/2 factor, we have the LS solution

  17. min ∑ [di-(Gm)I]2/2 W=diag(1/1, 1/2, …,1/m) Gw=WG dw=Wd Gwm=dw mL2=[GwTGw)-1 GwTdw ||dw-Gwmw||22= ∑ [di-(Gm)I]2/2 obs2 = ∑ [di-(Gm)I]2/2 obs2 has a2 distribution with m-n degrees of freedom

  18. The probability of finding a 2 value as large or larger than the observed value is p=∫ f2(x)dx p-value test With correct model and independent error, the p-values will be uniformly distributed between 0 and 1. Near-0 or near-1 p-values indicates problems (incorrect model, underestimation of data errors, unlikely realization)  obs2

  19. Multivariate normal distribution Random variables X1, …, Xn have a multivariate normal distribution, then the joint PDF is (B.61) f(x)=(2)-n/2 (det[C])-1/2 exp[-(x-)TC-1(x-)/2] Ci,j=Cov(Xi,Xj)

  20. Eigenvalues and eigenvectors Axl x (A-lIx=0 det (A-lI) = 0 The roots l are the eigenvalues

  21. SVD G=USVT U (m x m) orthogonal spanning data space V (n x n) orthogonal spanning model space S (m x n) eigenvalues along diagonal Let rank(G)=p G=[Up U0] [Sp 0;0 0] [Vp V0]T = UpSpVpT G+=VpSp-1UpT = Generalized Inverse (pseudoinverse) m+=G+d=VpSp-1UpT d = pseudoinverse solution Since Vp-1=VpT and Up-1=UpT (A.6)

  22. SVD G=USVT U (m x m) orthogonal spanning data space V (n x n) orthogonal spanning model space S (m x n) eigenvalues along diagonal Rank(G)=p Theorem A.5: N(GT)+R(G)=Rm, i.e. p columns of Up form an orthonormal basis for R(G) columns of U0 form an orthonormal basis for N(GT) p columns of Vp form an orthonormal basis for R(GT) columns of V0 form an orthonormal basis for N(G)

  23. Properties of SVD G=USVT N(G) and N(GT) are trivial (only null vector): Up=U, Vp=V square orthogonal matrices, and UpT=Up-1 and VpT=Vp-1 G+=VpSp-1UpT = (UpSpVpT)-1 = G-1 (inverse for full rank matrix, m=n=p). Unique solution, data are fit exactly.

  24. Properties of SVD G=USVT 2. N(G) is nontrivial (model, V); N(GT) is trivial (data, U) UpT=Up-1 and VpTVp=Ip Gm+=GG+d=UpSpVpTVpSp-1UpTd=UpSpIpSp-1UpTd=d i.e. the data are fit exactly an LS solution, but nonuniquely due to the nontrivial model null space m=m++m0 =m++∑ iV.,I ||m||22=||m+||22 +∑ i ≥||m+||22 -> minimum length solution n i=p+1 n i=p+1

  25. Properties of SVD G=USVT 3. N(G) is trivial (model, V); N(GT) is nontrivial (data, U) Gm+=GG+d=UpSpVpTVpSp-1UpTd=UpUpTd = projection of d onto R(G), I.e. the point in R(G) that is closest to d, m+ is LS solution to Gm=d. If d is in R(G), m+ will be solution to Gm=d. Solution is unique but does not fit data exactly i.e. the data are fit exactly an LS solution, but nonuniquely due to the nontrivial model null space m=m++m0 =m++∑ iV.,I ||m||22=||m+||22 +∑ i ≥||m+||22 -> minimum length solution n i=p+1 n i=p+1

  26. Properties of SVD G=USVT 4. N(G) is nontrivial (model, V); N(GT) is nontrivial (data, U) p < (m,n) Gm+=GG+d=UpSpVpTVpSp-1UpTd=UpUpTd = projection of d onto R(G), I.e. the point in R(G) that is closest to d, m+ is LS solution to Gm=d. i.e. LS solution to minimum norm, as case 2)

  27. Properties of SVD G=USVT - Always exists - LS or minimum length - Properly accommodates the rank and dimensions of G - Nontrivial model null space m0 the is heart of the problem -> - Infinite # of solutions will fit the data equally well, since components of N(G) have no effect on data fit, I.e., selection of a particular solution requires a priori constraints (smoothing, minimum length) - Nontrivial data space are vectors d0 that have no influence on m+. If p<n, there are an infinite # of data sets that will produce the same model

  28. Properties of SVD - covariance/resolution Least squares solution is unbiased: min ∑ [di-(Gm)I]2/2 W=diag(1/1, 1/2, …,1/m) Gw=WG, dw=Wd, Gwm=dw mL2=[GwTGw]-1 GwTdw ||dw-Gwmw||22= ∑ [di-(Gm)I]2/2 E[mL2]=E[(GwTGw)-1 GwTdw] = (GwTGw)-1 GwT E[dw]= (GwTGw)-1 GwT dwtrue = (GwTGw)-1 GwT Gwmtrue = mtrue

  29. Properties of SVD - covariance/resolution Generalized inverse not necessarilyunbiased: E[m+]=E[G+d] = G+E[d]= G+Gmtrue = Rmmtrue Bias= E[m+]-mtrue =Rmmtrue-mtrue = (Rm-I)mtrue = VpVpT-VVTmtrue=-V0V0T mtrue I.e., as p increasees Rm->I Cov(mL2)=2(GTG)-1 Cov(m+)=G+[Cov(d)]G+T = G+G+T =VpSp-2VpT = ∑ V.,i V.,iT/i2 I.e., as p increases, Rm->I: bias decreases while variance increases…! P I=1

  30. Model resolution Rm -> I: increasing resolution Resolution test: multiply Rm onto a particular model, fx a spike model, with one element 1 and the rest 0, picks out the corresponding column of Rm Data resolution D+=Gm+ = GG+d = Rdd Rd=UpSpVpTVpSp-1UpT=UpUpT p=m -> Rd=I, d+=d p<m -> Rd<>I, m+ doesn’t fit data exactly Rm, Rd can be assessed during design of experiment

  31. Instabilitites of SVD Small eigenvalues -> m+ sensitive to small amounts of noise Small eigenvalues maybe indistinguishable from 0 Possible to remove small eigenvalues to stabilize solution -> Truncated SVD, TSVD Condition number cond(G)=s1/sk

More Related