Review Sheet for GEOS 567 at UA
Review Sheet for GEOS 567 at UA
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Geosciences 567 CHAPTER 2 RMRGZ CHAPTER 2 REVIEW OF LINEAR ALGEBRA AND STATISTICS 21 Introduction In discrete inverse methods matrices and linear transformations play fundamental roles So do probability and statistics This review chapter then is divided into two parts In the first we will begin by reviewing the basics of matrix manipulations Then we will introduce some special types of matrices Hermitian orthogonal and semiorthogonal Finally we will look at matrices as linear transformations that can operate on vectors of one dimension and return a vector of another dimension In the second section we will review some elementary probability and statistics with emphasis on Gaussian statistics The material in the first section will be particularly useful in later chapters when we cover eigenvalue problems and methods based on the length of vectors The material in the second section will be very useful when we consider the nature of noise in the data and when we consider the maximum likelihood inverse 22 Matrices and Linear Transformations Recall from the first chapter that by convention vectors will be denoted by lower case letters in boldface ie the data vector d while matrices will be denoted by upper case letters in boldface ie the matrix G in these notes 221 Review of Matrix Manipulations Matrix Multiplication If A is an N X M matrix as in Nrows by M columns and B is an M X L matrix we write theN X L product C of A and B as C 2 AB 21 We note that matrix multiplication is associative that is ABC ABC 22 but in general is not commutative That is in general AB BA 23 12 Geosciences 567 CHAPTER 2 RMRGZ In fact if AB exists then the product BA only exists if A and B are square In Equation 21 above the ijth entry in C is the product of the ith row of A and the jth column of B Computationally it is given by c 2611ka 24 One way to form C using standard FORTRAN code would be DO 3001 1 N DO 300 J 1 L CIJ 00 DO 300 K 1 M 300 CIJ CIJ AIKBKJ 25 A special case of the general rule above is the multiplication of a matrix G N X M and a vectorm MX 1 d G m 113 NX1NgtltMMgtlt1 In terms of computation the vector 1 is given by d G m 26 The Inverse Ufa Matrix The mathematical inverse of the M X M matrix A denoted A4 is defined such that AA 1 A1A 1M 27 Where IM is the M X M identity matrix given by 1 0 0 0 E 0 28 0 0 1 MxM 13 Geosciences 567 CHAPTER 2 RMRGZ A 1 is the matrix Which When either pre or postmultiplied by A returns the identity mat1ix Clearly since only square mat1ices can both pre and postmultiply each other the mathematical inverse of a mat1ix only exists for square mat1ices A useful theorem follows concerning the inverse of a product of matiices Theorem If A B C D 29 NgtltN NgtltN NgtltN NgtltN Then A l if it exists is given by A1 D 1 c1 B1 210 Proof AA 1 BCDD1C1B1 BC DD 1 c IB 1 BC I c lB l B cc 1 B l BB4 1 211 Similarly A 1A D1C1B1BCD I QED The Transpose and Trace Ufa Matrix The transpose of a mat1ix A is Wiitten as AT and is given by AT A 212 That is you interchange rows and columns The transpose of a product of mat1ices is the product of the transposes in reverse order That is ABT BTAT 213 14 Geosciences 567 CHAPTER 2 RMRGZ Just about everything we do With real matrices A has an analog for complex matrices In the complex case Wherever the transpose of a matrix occurs it is replaced by the complex conjugate transpose of the matrix denoted A That is if Aij aij biJi 214 mml AUCUdg 21 Where Ci aji 216 and di bji 217 that is A i a b i 218 Finally the trace of A is given by M trace A 2a 23919 11 Hermitian Matrices A matrix A is said to be Hermitian if it is equal to its complex conjugate transpose That is if A A 220 If A is a real matrix this is equivalent to A 2 AT 221 This implies that A must be square The reason that Hermitian matrices Will be important is that they have only real eigenvalues We Will take advantage of this many times When we consider eigenvalue and shifted eigenvalue problems later 222 Matrix Transformations Linear Transformations A matrix equation can be thought of as a linear transformation Consider for example the original matrix equation dGm 113 15 Geosciences 567 CHAPTER 2 RMRGZ where dis an N X 1 vector m is an M X 1 vector and G is an N X M matrix The matrix G can be thought of as an operator that operates on an M dimensional vector m and returns an N dimensional vector 1 Equation 113 represents an explicit linear relationship between the data and model parameters The operator G in this case is said to be linear because if m is doubled for example so is d Mathematically one says that G is a linear operator if the following is true If 1 Gm and f Gr then 1 f Gm r 222 In another way to look at matrix multiplications in the by now familiar Equation 113 d Gm 113 the column vector 1 can be thought of as a weighted sum of the columns of G with the weighting factors being the elements in m That is dm1g1 m2g2 mMgM 223 where m m1 m2 mMT 224 and gi 81139s 8213 sgNilT 225 is the ith column of G Also if GA B then the above can be used to infer that the first column of B is a weighted sum of the columns of G with the elements of the first column of A as weighting factors etc for the other columns of B Each column of B is a weighted sum of the columns of G Next consider dT GmT 226 or dT mT GT 227 1gtltN 1gtltM MgtltN The row vector dT is the weighted sum of the rows of GT with the weighting factors again being the elements in m That is 16 Geosciences 567 CHAPTER 2 RMRGZ dT mg ngg mMgATI 228 Extending this to ATGT 2 BT 229 we have that each row of BT is a weighted sum of the rows of GT with the weighting factors being the elements of the appropriate row of AT In a long string of matrix multiplications such as ABC 2 D 230 each column of D is a weighted sum of the columns of A and each row of D is a weighted sum of the rows of C Orthogonal Transformations An orthogonal transformation is one that leaves the length of a vector unchanged We can only talk about the length of a vector being unchanged if the dimension of the vector is unchanged Thus only square matrices may represent an orthogonal transformation Suppose L is an orthogonal transformation Then if LX y 231 where L is N X N and X y are both N dimensional vectors Then XTX yTy 232 where Equation 232 represents the dot product of the vectors with themselves which is equal to the length squared of the vector If you have ever done coordinate transformations in the past you have dealt with an orthogonal transformation Orthogonal transformations rotate vectors but do not change their lengths Properties of orthogonal transformations There are several properties of orthogonal transformations that we will wish to use First if L is an N X N orthogonal transformation then LTL 2 IN 233 This follows from yTy LXTLX 17 Geosciences 567 CHAPTER 2 RMRGZ XTLTLX 234 but yTy XTX by Equation 232 Thus LTL 2 IN QED 235 Second the relationship between L and its inverse is given by L 1 2 LT 236 and L 2 LT 1 237 These two follow directly from Equation 235 above Third the determinant of a matrix is unchanged if it is operated upon by orthogonal transformations Recall that the determinant of a 3 X 3 matrix A for example where A is given by an an an A all an an 238 a a a 32 is given by det A 11111122 133 11231132 412 121 133 11231131 a13a21a32 11221131 239 Thus if A is an M X M matrix and Lis an orthogonal transformations and if A LALT 240 it follows that det A det A 241 Fou h the trace of a matrix is unchanged if it is operated upon by an orthogonal transformation where trace A is defined as M trace A 2a 23942 11 18 Geosciences 567 CHAPTER 2 RMRGZ That is the sum of the diagonal elements of a matrix is unchanged by an orthogonal transformation Thus trace A trace A 243 Semiorthogonal Transformations Suppose that the linear operator L is not square but N X M N 75 M Then L is said to be semiorthogonal if and only if LTL 1M but LLT 1N Ngt M 244 or LLT 2 IN but LTL 75 IM M gt N 245 Where IN and IM are the N X N and M X M identity matiices respectively A matiix cannot be both orthogonal and semiorthogonal OIthogonal mat1ices must be square and semi01thogonal mat1ices cannot be square Fuithermore if L is a square N X N matiix and LTL 2 IN 235 then it is not possible to have LLT 1N 246 223 Matrices and Vector Spaces The columns or rows of a matiix can be thought of as vectors For example if A is an N X M matiix each column can be thought of as a vector in N space because it has N entiies Conversely each row of A can be thought of as being a vector in M space because it has M entiies We note that for the linear system of equations given by Gm d 113 Where G is N X M m is M X 1 and dis N X 1 that the model parameter vector m lies in M space along With all the rows of G While the data vector lies in N space along With all the columns of G In general we Will think of the M X 1 vectors as lying in model space While the N X 1 vectors lie in data space Spanning a Space 19 Geosciences 567 CHAPTER 2 RMRGZ The notion of spanning a space is important for any discussion of the uniqueness of solutions or of the ability to fit the data We first need to introduce definitions of linear independence and vector orthogonality A set on M vectors vi i 1 M in M space the set of all M dimensional vectors is said to be linearly independent if and only if a1v1 a2v2 anM 0 247 where 111 are constants has only the trivial solution al 0 i 1 M This is equivalent to saying that an arbitrary vector 5 in M space can be written as a linear combination of the vi i 1 M That is one can find al such that for an arbitrary vector 5 s a1v1 a2v2 anM 248 Two vectors 1 and s in M space are said to be orthogonal to each other if their dot or inner product with each other is zero That is if 1 r s quotsllcosg 0 249 a I The dot product of two vectors is also given by where Bis the angle between the vectors and squot are the lengths of r and s respectively T T 250 M space is spanned by any set of M linearly independent M dimensional vectors Rank Ufa Matrix The number of linearly independent rows in a matrix which is also equal to the number of linearly independent columns is called the rank of the matrix The rank of matrices is defined for both square and nonsquare matrices The rank of a matrix cannot exceed the minimum of the num ber of rows or columns in the matrix ie the rank is less than or equal to the minimum of N M If an M X M matrix is an orthogonal matrix then it has rank M The M rows are all linearly independent as are the M columns In fact not only are the rows independent for an orthogonal matrix they are orthogonal to each other The same is true for the columns If a matrix is semiorthogonal then the M columns orN rows if N lt M are orthogonal to each other We will make extensive use of matrices and linear algebra in this course especially when we work with the generalized inverse Next we need to turn our attention to probability and statistics 20 Geosciences 567 CHAPTER 2 RMRGZ 23 Probability and Statistics 231 Introduction We need some background in probability and statistics before proceeding very far In this review section I will cover the material from Menke39s book using some material from other math texts to help clarify things Basically what we need is a way of describing the noise in data and estimated model parameters We will need the following terms random variable probability distribution mean or expected value maximum likelihood variance standard deviation standardized normal variables covariance correlation coe icients Gaussian distributions and con dence intervals 232 De nitions Part 1 Random Variable A function that assigns a value to the outcome of an experiment A random variable has well defined properties based on some distribution It is called random because you cannot know beforehand the exact value for the outcome of the experiment One cannot measure directly the true properties of a random variable One can only make measurements also called realizations of a random variable and estimate its properties The birth weight of baby goslings is a random variable for example Probability Density Function The true properties of a random variable b are specified by the probability density function Pb The probability that a particular realization of b will fall between b and b db is given by Pbdb Note that Menke uses d where Iuse b His notation is bad when one needs to use integrals Pb satisfies 1 1207 db 251 which says that the probability of b taking on some value is 1 Pb completely describes the random variable b It is often useful to try and find a way to summarize the properties of Pb with a few numbers however Mean or Expected Value The mean value Eb also denoted ltbgt is much like the mean of a set of numbers that is it is the balancing point of the distribution Pb and is given by M 27 Pb db 252 Maximum Likelihood This is the point in the probability distribution Pb that has the highest likelihood or probability It may or may not be close to the mean Eb ltbgt An important point is that for Gaussian distributions the maximum likelihood point and the mean Eb ltbgt 21 Geosciences 567 CHAPTER 2 RMRGZ are the same The graph below after Figure 23 p 23 Menke illustrates a case where the two are different Pb The maximum likelihood point bML of the probability distribution Pb for data 17 gives the most probable value of the data In general this value can be different from the mean datum ltbgt which is at the balancing point of the distribution Variance Variance is one measure of the spread or width of Pb about the mean Eb It is given by 02 may ltbgt2Pb db 253 Computationally for L experiments in which the kth experiment gives bk the variance is given by L 02 Lzay k ltbgt2 254 L l k1 Standard Deviation Stande deviation is the positive square root of the variance given by a 255 Covariance Covariance is a measure of the correlation between errors If the errors in two observations are uncorrelated then the covariance is zero We need another definition before proceeding 22 Geosciences 567 CHAPTER 2 RMRGZ Joint Density Function Pb The probability that 171 is between 271 and 171 dbl that 172 is between 172 and 172 d172 etc If the data are independent then 1 00 P071 P072 quot39Pbn 256 If the data are correlated then Pb will have some more complicated form Then the covariance between 271 and 172 is defined as covb1 1 fa lt b1 gtb2 lt b gtPbalb1 db db 257 In the event that the data are independent this reduces to 40d 4 com b L La lt b gtgtltb lt b gtPbPbgt db db 258 0 The reason is that for any value of 171 ltb1gt b2 ltb2gt is as likely to be positive as negative ie the sum will average to zero The matrix cov b contains all of the covariances defined using Equation 257 in an N X N matrix Note also that the covariance of bl with itself is just the variance of bi In practical terms if one has an N dimensional data vector b that has been measured L times then the ijth term in cov b denoted cov blj is defined as 1 L covbU 2 bf 271be b1 259 1H where b is the value of the ith datum in b on the kth measurement of the data vector E is the mean or average value of bl for all L measurements also commonly written ltbigt and the L 1 term results from sampling theory Correlation Coef cients This is a normalized measure of the degree of correlation of errors It takes on values between 1 and 1 with a value of 0 implying no correlation The correlation coefficient matrix cor b is defined as covbU 00 cor my 260 where cov bl is the covariance matrix defined term by term as above for cov 171 172 and 03 039 are the standard deviations for the ith and jth observations respectively The diagonal terms of cor b i are equal to 1 since each observation is perfectly correlated with itself 23 Geosciences 567 CHAPTER 2 RMRGZ The figure below after Figure 28 page 26 Menke shows three different cases of degree of correlation for two observations b1 and b2 a b C I I I b2 b2 b2 I b b b Contour plots of Pb1 272 when the data are a uncorrelated b positively correlated c negatively correlated The dashed lines indicate the four quadrants of alternating sign used to determine correlation 233 Some Comments on Applications to Inverse Theory Some comments are now in order about the nature of the estimated model parameters We will always assume that the noise in the observations can be described as random variables Whatever inverse we create will map errors in the data into errors in the estimated model parameters Thus the estimated model parameters are themselves random variables This is true even though the true model parameters may not be random variables If the distribution of noise for the data is known then in principle the distribution for the estimated model parameters can be found by mapping through the inverse operator This is often very difficult but one particular case turns out to have a rather simple form We will see where this form comes from when we get to the subject of generalized inverses For now consider the following as magic If the transformation between data b and model parameters m is of the form m 2 Mb v 261 where M is any arbitrary matrix and v is any arbitrary vector then ltmgt Mltbgt v 262 and cov m M cov b MT 263 24 Geosciences 567 CHAPTER 2 RMRGZ 234 De nitions Part 2 Gaussian Distribution This is a particular probability distribution given by Pb 1 expb ltZbgt2 264 V 27039 2039 The figure below after Figure 210 page 29 Menke shows the familiar bell shaped curve It has the following properties Mean Eb ltbgt and Variance 0392 050 Pb 0 8 I 54321012345 b Gaussian distribution with zero mean and 039 1 for curve A and 039 2 for curve B Many distributions can be approximated fairly accurately especially away from the tails by the Gaussian distribution It is also very important because it is the limiting distribution for the sum of random variables This is often just what one assumes for noise in the data One also needs a way to represent the joint probability introduced earlier for a set of random variables each of which has a Gaussian distribution The joint probability density function for a vector b of observations that all have Gaussian distributions is chosen to be see Equation 210 of Menke page 30 detcov b712 131 27mm exp b lt b gtTcovb 1b lt b gt 265 25 Geosciences 567 CHAPTER 2 RMRGZ which reduces to the previous case in Equation 264 for N 1 and var bl 52 In statistics books Equation 265 is often given as Poo 2702 liblU2 exp 2b 11T 21 b um With this background it makes sense statistically at least to replace the original relationship b Gm 113 with ltbgt Gm 266 The reason is that one cannot expect that there is an m that should exactly predict any particular realization of b when b is in fact a random variable Then the joint probability is given by detcovb 12 2mm exp 7n Gm covb b Gm 267 Pb What one then does is seek an m that maximizes the probability that the predicted data are in fact close to the observed data This is the basis of the maximum likelihood or probabilistic approach to inverse theory Standardized Normal Variables It is possible to standardize random variables by subtracting their mean and dividing by the standard deviation If the random variable had a Gaussian ie normal distribution then so does the standardized random variable Now however the standardized normal variables have zero mean and standard deviation equal to one Random variables can be standardized by the following transformation S 268 where you will often see 2 replacing s in statistics books We will see when all is said and done that most inverses represent a transformation to standardized variables followed by a simple inverse analysis and then a transformation back for the final solution ChiSquared Goodness of Fit Test A statistical test to see whether a particular observed distribution is likely to have been drawn from a population having some known form 26 Geosciences 567 CHAPTER 2 RMRGZ The application we will make of the chi squared test is to test whether the noise in a particular problem is likely to have a Gaussian distribution This is not the kind of question one can answer with certainty so one must talk in terms of probability or likelihood For example in the chi squared test one typically says things like there is only a 5 chance that this sample distribution does not follow a Gaussian distribution As applied to testing whether a given distribution is likely to have come from a Gaussian population the procedure is as follows One sets up an arbitrary number of bins and compares the number of observations that fall into each bin with the number expected from a Gaussian distribution having the same mean and variance as the observed data One quantifies the departure between the two distributions called the chi squared value and denoted 12 as zz x k obs in bin 139 expected in bin i2 269 1 expected in bin 139 where the sum is over the number of bins k Next the number of degrees of freedom for the problem must be considered For this problem the number of degrees is equal to the number of bins minus three The reason you subtract three is as follows You subtract 1 because if an observation does not fall into any subset of k 1 bins you know it falls in the one bin left over You are not free to put it anywhere else The other two come from the fact that you have assumed that the mean and standard deviation of the observed data set are the mean and standard deviations for the theoretical Gaussian distribution With this information in hand one uses standard chi squared test tables from statistics books and determines whether such a departure would occur randomly more often than say 5 of the time Officially the null hypothesis is that the sample was drawn from a Gaussian distribution If the observed value for If is greater than 12 called the critical If value for the 0 significance level then the null hypothesis is rejected at the 0 significance level Commonly 0 005 is used for this test although 0 001 is also used The 0significance level is equivalent to the 1001 00 confidence level ie 0 005 corresponds to the 95 confidence level Consider the following example where the underlying Gaussian distribution from which all data samples d are drawn has a mean of 7 and a variance of 10 Seven bins are set up with edges at 4 2 4 6 8 10 12 and18 respectively Bin widths are not prescribed for the chi squared test but ideally are chosen so there are about an equal number of occurrences expected in each bin Also one rule of thumb is to only include bins having at least five expected occurrences I have not followed the about equal number expected in each bin suggestion because I want to be able to visually compare a histogram with an underlying Gaussian shape However I have chosen wider bins at the edges in these test cases to capture more occurrences at the edges of the distribution Suppose our experiment with 100 observations yields a sample mean of 676 and a sample variance of 827 and 3 13 26 25 16 14 and 3 observations respectively in the bins from left to right Using standard formulas for a Gaussian distribution with a mean of 676 and a variance of 827 the number expected in each bin is 490 1198 2273 2710 2031 956 and 341 respectively The calculated using Equation 269 is 448 For seven bins the DOFs 27 Geosciences 567 CHAPTER 2 RMRGZ for the test is 4 and 25 949 for a 005 Thus in this case the null hypothesis would be accepted That is we would accept that this sample was drawn from a Gaussian distribution with a mean of 676 and a variance of 827 at the a 005 significance level 95 confidence level The distribution is shown below with a filled circle in each histogram at the number expected in that bin X2 448 25 O 20 9 15 Number 10 o 0 5 10 15 d It is important to note that this distribution does not look exactly like a Gaussian distribution but still passes the 22 test A simple non chi square analogy may help better understand the reasoning behind the chi square test Consider tossing a true coin 10 times The most likely outcome is 5 heads and 5 tails Would you reject a null hypothesis that the coin is a true coin if you got 6 heads and 4 tails in your one experiment of tossing the coin ten times Intuitively you probably would not reject the null hypothesis in this case because 6 heads and 4 tails is not that unlikely for a true coin In order to make an informed decision as we try to do with the chi square test you would need to quantify how likely or unlikely a particular outcome is before accepting or rejecting the null hypothesis that it is a true coin For a true coin 5 heads and 5 tails has a probability of 0246 that is on average it happens 246 of the time while the probability of 6 heads and 4 tails is 0205 7 heads and 3 tails is 0117 and 8 heads and 2 tails is 0044 respectively A distribution of 7 heads and 3 tails does not look like 5 heads and 5 tails but occurs more than 10 of the time with a true coin Hence by analogy it is not too unlikely and you would probably not reject the null hypothesis that the coin is a true coin just because you tossed 7 heads and 3 tails in one experiment Ten heads and no tails only occurs on average one time in 1024 experiments or about 0098 of the time If you got 10 heads and 0 tails you d probably reject the null hypothesis that you are tossing a true coin because the outcome is very unlikely Eight heads and two tails occurs 44 of the time on average You might also reject the null hypothesis in this 28 Geosciences 567 CHAPTER 2 RMRGZ case but you would do so with less confidence or at a lower significance level In both cases however your conclusion will be wrong occasionally just due to random variations You accept the possibility that you will be wrong rejecting the null hypothesis 44 of the time in this case even if the coin is true The same is true with the chi square test That is at the a 005 significance level 95 confidence level with 252 greater than 25 you reject the null hypothesis even though you recognize that you will reject the null hypothesis incorrectly about 5 of the time in the presence of random variations Note that this analogy is a simple one in the sense that it is entirely possible to actually do a chi square test on this coin toss example Each time you toss the coin ten times you get one outcome x heads and 10 x tails This falls into the x heads and 10 x tails bin If you repeat this many times you get a distribution across all bins from 0 heads and 10 tails to 10 heads and 0 tails Then you would calculate the number expected in each bin and use Equation 269 to calculate a chi square value to compare with the critical value at the a significance level Now let us return to another example of the chi square test where we reject the null hypothesis Consider a case where the observed number in each of the seven bins defined above is now 2 17 13 24 26 9 and 9 respectively and the observed distribution has a mean of 728 and variance of 1028 The expected number in each bin for the observed mean and variance is 495 1032 1916 2440 2132 1278 and 702 respectively The calculatedjf is now 1077 and the null hypothesis would be rejected at the a 005 significance level 95 confidence level That is we would reject that this sample was drawn from a Gaussian distribution with a mean of 728 and variance of 1028 at this significance level The distribution is shown on the next page again with a filled circle in each histogram at the number expected in that bin X2 1077 25 I o 20 15 Number 0 10 39 d Con dence Intervals One says for example with 98 confidence that the true mean of a random variable lies between two values This is based on knowing the probability distribution 29 Geosciences 567 CHAPTER 2 RMRGZ for the random variable of course and can be very difficult especially for complicated distributions that include nonzero correlation coefficients However for Gaussian distributions these are well known and can be found in any standard statistics book For example Gaussian distributions have 68 and 95 confidence intervals of approximately 10 and 20 respectively T and F Tests These two statistical tests are commonly used to determine whether the properties of two samples are consistent with the samples coming from the same population The F test in particular can be used to test the improvement in the fit between predicted and observed data when one adds a degree of freedom in the inversion One expects to fit the data better by adding more model parameters so the relevant question is whether the improvement is significant As applied to the test of improvement in fit between case 1 and case 2 where case 2 uses more model parameters to describe the same data set the F ratio is given by F E E2D0F D0Fz 270 E DOF where E is the residual sum of squares and DOF is the number of degrees of freedom for each case If F is large one accepts that the second case with more model parameters provides a significantly better fit to the data The calculated F is compared to published tables with DOFI DOFZ and DOFZ degrees of freedom at a specified confidence level Reference T M Hearns P travel times in Southern California J Geophys Res 89 1843 1855 1984 The next section will deal with solving inverse problems based on length measures This will include the classic least squares approach 30
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