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Chapter 9

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Slide 1Slide 2Binomial BoundsSlide 6Slide 7Slide 8N – Dimensional Euclidean SpaceSlide 10Interesting Facts about N-dimensional Euclidean SpaceSlide 12Slide 13VarianceThe Law of Large NumbersSlide 16Chapter 9Mathematical PreliminariesStirling’s ApproximationFig.9.2-1by trapezoid ruletakeantilogsFig.9.2-2by midpoint formulatakeantilogs9.2. . . .1log)log(logLet . log!log111nnnxxxxdxIknnnnknnI log21)1log(2log1log21 !!log1log21log nenennnnnnnnnennn2~!nnI log21)1log(3log2log81 !!loglog21811log87nenennnnnnnn7.24.2 where!87eCeCnennnnBinomial BoundsShow the volume of a sphere of radius λn in the n-dimensional unit hypercube is:Assuming 0  λ  ½ (since the terms are reflected about n/2) the terms grow monotonically, and bounding the last by Stirling gives:9.3)(22211)(nHnnnnnV  ' somefor 2)1()1(1)1()1(1)()()!()!(!)(2111)()1(12CnCnCnnnCeeennnCnnennCnenCnennnnnnnnHnnnnnnnnnnnnnnnnnnn    ).()1log()1(log)1(log1H9.3N.b.)()(0)(002222 as 1211212,211111-1 So rms).between te ratios actual (the 112232111 ratioswith series geometric aby termwise111 boundnHnHkknHnkkknnCnCknnnnnnnnnnnnnnn   nknnHnnkknkn012022 2~212The Gamma FunctionIdea: extend the factorial to non-integral arguments.by conventionFor n > 1, integrate by parts: dg = e−xdx f = xn−19.4.)(Let 01 dxxennx0201)1(1)( dxxenexnnxxn1)1()1()1()(00xxedxennn)!1()(!3)4(!2)3(1)2(  nndxdyarea = dxdy9.4rdr rdθarea = rdrdθ  dxdyedyedxedtedtetdttedxxeyxyxttttxx222222222121integral)error the(called2212100021020 02222rredrdre21N – Dimensional Euclidean SpaceUse Pythagorean distance to define spheres:Consequently, their volume depends proportionally on rnconverting to polar coordinates9.5rxxn221342)(321 CCCrCrVnnnniixntnndxedtei122221  010122221)(drnreCdrdrrdVedxdxenrnnrnxxnjust verifyby substitutionr2  t9.5nCn1 2 2.000002 π 3.141593 4π/3 4.188794 π2/2 4.934205 8π2/15 5.263796 π3/6 5.167717 16π3/105 4.724778 π4/24 4.05871… … …2k πk/k! → 0From table on page 177 of book.  122220121022121nCnnCdttenCdrdttteCnnntntnnn22212nnnCnnCInteresting Facts aboutN-dimensional Euclidean SpaceCn → 0 as n → ∞ Vn(r) → 0 as n → ∞ for a fixed rVolume approaches 0 as the dimension increases! Almost all the volume is near the surface (as n → ∞)end of 9.5nrrCrCrCrVrVnnnnnnnnn as111)()()(What about the angle between random vectors, x and y, of the form (±1, ±1, …, ±1)? Hence, for large n, there are almost 2n random diagonal lines which are almost perpendicular to each other!end of 9.8By definition:length of projection along axislength of entire vectorFor large n, the diagonal line is almost perpendicular to each axis!Angle between the vector (1, 1, …, 1) and each coordinate axis:As n → ∞: cos θ → 0,  θ → π/2.0)1()1)(1( vectorsrandomddistributeuniformlyfor 1products randomuniform 1  nnnnknkyxyxcosn1cos Chebyshev’s InequalityLet X be a discrete or continuous random variable with p(xi) = the probability that X = xi. The mean square is x2 × p(x) ≥ 0Chebyshev’s inequality9.7  ||)()()()(222222XPdxxpdxxpxdxxpxxpxXExxiii  22||XEXP VarianceThe variance of X is the mean square about the mean value of X,So variance, (via linearity) is: 9.7Note: V{1} = 0 → V{c} = 0 & V{cX} = c²V{X}Also: V{X − b} = V{X}  adxxpxXE )(         .2)(22222222XEXEaXEaaXEaXEaXEXVThe Law of Large NumbersLet X and Y be independent random variables, with E{X} = a, E{Y} = b, V{X} = σ2, V{Y} = τ2. Then E{(X − a) (∙ Y − b)} = E{X − a} ∙ E{Y − b} = 0 0 = 0∙And V{X + Y} = V{X − a + Y − b} = E{(X − a + Y − b)2} = E{(X − a)2} + 2E{(X − a)(Y − b)} + E{(Y − b)2} = V{X} + V{Y} = σ2 + τ2because of independence9.8Consider n independent trials for X; called X1, …, Xn.The expectation of their average is (as expected!):    aXEnXEXEnXXEnn 11So, what is the probability that their average A is not close to the mean E{X} = a? Use Chebyshev’s inequality:Let n → ∞Weak Law of Large Numbers: The average of a large enough number of independent trials comes arbitrarily close to the mean with arbitrarily high probability.9.8The


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