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BU EECE 522 - CRLB for Vector Parameter Case

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13.7 CRLB for Vector Parameter CaseVector Parameter:[]Tpθθθ!21=θIts Estimate:[]Tpθθθˆˆˆˆ21!=θAssume that estimate is unbiased: {}θθ =ˆEFor a scalar parameter we looked at its variance…but for a vector parameter we look at its covariance matrix:{}[][]θCθθθθθˆˆˆˆvar =−−=TEFor example: for θ = [x y z]T=)ˆvar()ˆ,ˆcov()ˆ,ˆcov()ˆ,ˆcov()ˆvar()ˆ,ˆcov()ˆ,ˆcov()ˆ,ˆcov()ˆvar(ˆzyzxzzyyxyzxyxxθC2Fisher Information MatrixFor the vector parameter case… Fisher Info becomes the Fisher Info Matrix (FIM) I(θ) whose mnthelement is given by:[][]pnmpEmnmn,,2,1,,);(ln)(2…=∂∂∂−=θθθxθIEvaluate at true value of θ3The CRLB Matrix Then, under the same kind of regularity conditions, the CRLB matrix is the inverse of the FIM:)(1θI−=CRLBSo what this means is:nnnnn][][ )(1ˆ2ˆθICθ−≥=θσDiagonal elements of Inverse FIM bound the parameter variances, which are the diagonal elements of the parameter covariance matrix=)ˆvar()ˆ,ˆcov()ˆ,ˆcov()ˆ,ˆcov()ˆvar()ˆ,ˆcov()ˆ,ˆcov()ˆ,ˆcov()ˆvar(ˆzyzxzzyyxyzxyxxθC )(1333231232221131211θI−=bbbbbbbbb(!)4More General Form of The CRLB Matrixdefinite-semi positive is)(1ˆθICθ−−0θICθ≥−−)(1ˆMathematical Notation for this is:(!!)Note: property #5 about p.d. matrices on p. 573 states that (!!) ⇒ (!)5CRLB Off-Diagonal Elements InsightLet θ = [xeye]Trepresent the 2-D x-y location of a transmitter (emitter) to be estimated.Consider the two cases of “scatter plots” for the estimated location:exˆeyˆexˆeyˆexexeyeyeyˆσeyˆσexˆσexˆσEach case has the same variances… but location accuracy characteristics are very different. ⇒ This is the effect of the off-diagonal elements of the covariance Should consider effect of off-diagonal CRLB elements!!!Not In Book6CRLB Matrix and Error EllipsoidsNot In BookAssume[]Teeyxˆˆˆ=θis 2-D Gaussian w/ zero meanand a cov matrixθCˆOnly For ConvenienceThen its PDF is given by:()()−=−θCθCθθθˆˆ21exp21ˆ1ˆˆTNpπQuadratic Form!!(recall: it’s scalar valued)So the “equi-height contours” of this PDF are given by the values of θˆsuch that:kT=θAθˆˆSome constanteasefor :Let1ˆACθ=−Note: A is symmetric so a12= a21…because any cov. matrix is symmetric and the inverse of symmetric is symmetric7What does this look like? kyayxaxaeeee=++22212211ˆˆˆ2ˆAn Ellipse!!! (Look it up in your calculus book!!!)Recall: If a12= 0, then the ellipse is aligned w/ the axes & the a11 and a22 control the size of the ellipse along the axesNote: a12= 0 ⇒=⇒=−2211ˆ22111ˆ100100aaaaθθCCeduncorrelat areˆ&ˆeeyx⇒Note: a12≠ 0 correlated areˆ&ˆeeyx⇒=2ˆˆˆˆˆ2ˆˆeeeeeeyxyyxxσσσσθC8exˆexˆeyˆeduncorrelat areˆ&ˆeeyxifeyˆexˆexˆeyˆcorrelated areˆ&ˆeeyxifeyˆexˆ2~σeyˆ2~σexˆ2~σeyˆ2~σexˆ2~σeyˆ2~σexˆ2~σeyˆ2~σNot In BookError Ellipsoids and CorrelationChoosing k ValueFor the 2-D case… k = -2 ln(1-Pe)where Pe is the prob. that the estimate will lie inside the ellipseSee posted paper by Torrieri9Ellipsoids and Eigen-StructureConsider a symmetric matrix A & its quadratic form xTAxkT=Axx⇒ Ellipsoid: ork=xAx ,Principle Axes of Ellipse are orthogonal to each other…and are orthogonal to the tangent line on the ellipse:x1x2Theorem: The principle axes of the ellipsoid xTAx = k are eigenvectors of matrix A.Not In Book10Proof: From multi-dimensional calculus: gradient of a scalar-valued function φ(x1,…, xn) is orthogonal to the surface:x1x2Tnnxxxxgrad∂∂∂∂==∂∂=∇=φφφφφ!…11)()(),,(xxxxDifferent NotationsSee handout posted on Blackboard on Gradients and Derivatives11∑∑∑∑∂∂=∂∂⇒==ikjijijkijjiijTxxxaxxxa)(xAx)x(φφProduct rule:#$%#$%jkikkjikikijkikjixxxxxxxxxδδ∂∂+∂∂=∂∂≠===01)(For our quadratic form function we have:(♣)(♣♣)Using (♣♣) in (♣) gives:jjkjjiikjjjkkxaxaxax∑∑∑=+=∂∂2φBy Symmetry:aik= akiAnd from this we get:AxAxxx2)( =∇T12x1x2Since grad ⊥ ellipse, this says Ax is ⊥ ellipse:xAxk=xAx ,When x is a principle axis, then x and Ax are aligned:x1x2xAxk=xAx ,xAxλ=Eigenvectors are Principle Axes!!!< End of Proof >13Theorem: The length of the principle axis associated with eigenvalue λiis ikλ/Proof: If x is a principle axis, then Ax = λx. Take inner productof both sides of this with x:xxxAx ,,λ==#$%kλλkk=⇒==xxxx#$%2,< End of Proof >Note: This says that if A has a zero eigenvalue, then the error ellipse will have an infinite length principle axis ⇒NOT GOOD!!So… we’ll require that all λi > 0⇒ must be positive definiteθCˆ14Application of Eigen-Results to Error EllipsoidsThe Error Ellipsoid corresponding to the estimator covariance matrix must satisfy:θCˆkT=−θCθθˆˆ1ˆNote that the error ellipse is formed using the inversecovThus finding the eigenvectors/values ofshows structure of the error ellipse 1ˆ−θCRecall: Positive definite matrix A and its inverse A-1have the • same eigenvectors• reciprocal eigenvaluesThus, we could instead find the eigenvalues ofand then the principle axes would have lengthsset by its eigenvalues not inverted )(1ˆθICθ−=Inverse FIM!!15Illustrate with 2-D case:kT=−θCθθˆˆ1ˆv1& v2λ1& λ2Eigenvectors/values for (not the inverse!)θCˆv11ˆθ2ˆθv21λk2λk16The CRLB/FIM EllipseWe can re-state this in terms of the FIM…Once we find the FIM we can:• Find the inverse FIM• Find its eigenvectors… gives the Principle Axes• Find its eigenvalues… Prin. Axis lengths are then Can make an ellipse from the CRLB Matrix… instead of the Cov. MatrixThis ellipse will be the smallest error ellipse that an unbiased estimator can


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BU EECE 522 - CRLB for Vector Parameter Case

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