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# Chapter 9

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Chapter 9 Mathematical Preliminaries Stirling s Approximation n n e n n n 2 n log n log k Let I log xdx x log x x 1 n log n n 1 k 1 Fig 9 2 1 n 1 by trapezoid rule 1 1 I log1 log 2 log n 1 log n 2 2 1 n log n n log n 1 log n n n e n n e n 2 Fig 9 2 2 1 1 by midpoint formula I log 2 log 3 log n 1 log n 8 2 7 1 1 n n n log n n 1 log n log n n e n e 8 n 8 2 take antilogs n n n n e take antilogs 7 8 n C where 2 4 e C e 2 7 9 2 Binomial Bounds Show the volume of a sphere of radius n in the n dimensional unit hypercube is n n n V n 1 2nH 2 1 2 n Assuming 0 since the terms are reflected about n 2 the terms grow monotonically and bounding the last by Stirling gives n n n ne n n C n n n n n n n C n n e n n C n n n n n e nn 1 e n C 1 n n n n n n n 1 1 n e ne n n n 1 n 1 n 1 1 C 2 nH 2 C for some C 1 1 1 n 2 n log 1 1 log 1 log 1 H 9 3 n n n 1 termwise by a geometric series bound n n 1 1 n n 1 3 2 1 with ratios the 1 n n 1 n n 2 k n 2 n 1 n 1 1 actual ratios between te rms So 1 2 k 0 1 1 1 k n n C C 1 nH 2 nH 2 nH 2 2 2 2 k n k 0 1 k 0 n 1 2 1 as n n 2 N b n n n 1 n nH 2 1 n 2 2 2 2 k 0 k k 0 k 9 3 The Gamma Function Idea extend the factorial to non integral arguments by convention Let n e x x n 1dx 0 For n 1 integrate by parts dg e xdx f xn 1 n x n 1 x e n 1 e x x n 2 dx 10 0 n n 1 n 1 1 e dx e 0 2 1 3 2 x x 0 1 4 3 n n 1 9 4 1 2 x t 2 1 x t2 1 e x dx e 2tdt t 2 0 0 2 t2 2 e dt e dt t called the error integral 0 1 1 x y x2 y2 e dx e dy e dxdy 2 2 dr dx dy rd r area rdrd 2 2 2 r2 r2 e e rdr d 2 2 0 0 0 area dxdy 1 2 9 4 N Dimensional Euclidean Space Use Pythagorean distance to define spheres 2 1 2 n x x r Consequently their volume depends proportionally on rn 4 n Vn r Cn r C1 2 C2 C3 3 n e x12 x n2 n n t2 1 x i2 e dt e dxi i 1 2 n 2 dx dx e r dVn r dr C e r nr n 1dr 1 n n 0 2 dr converting to polar coordinates 2 0 9 5 r2 t dr n nC t t 2 1 n n Cn e t t dt e t dt 2 0 0 C n n n n Cn Cn 1 1 2 2 2 2 2 n 1 2 1 2 1 2 n n 2 2 Cn Cn 2 n n 1 2 just verify by substitution 2 00000 3 14159 3 4 3 4 18879 4 2 3 4 93420 5 8 2 15 5 26379 6 3 6 5 16771 7 16 3 105 4 72477 8 4 24 4 05871 2k k k 0 From table on page 177 of book 9 5 Interesting Facts about N dimensional Euclidean Space Cn 0 as n Vn r 0 as n for a fixed r Volume approaches 0 as the dimension increases n n n Vn r Cn r Cn r 1 1 1 n Vn r Cn r r as n Almost all the volume is near the surface as n end of 9 5 Angle between the vector 1 1 1 and each coordinate axis 1 cos n length of projection along axis As n cos 0 2 length of entire vector For large n the diagonal line is almost perpendicular to each axis x y By cos x y definition What about the angle between random vectors x and y of the form 1 1 1 n n uniform for uniformly 1 1 1 random products distributed random vectors k 1 k 1 0 n n n Hence for large n there are almost 2n random diagonal lines which are almost perpendicular to each other end of 9 8 Chebyshev s Inequality Let X be a discrete or continuous random variable with p xi the probability that X xi The mean square is x2 p x 0 x p x x p x dx E X 2 i 2 i 2 i x p x dx p x dx P X 2 x 2 2 x E X2 P X 2 Chebyshev s inequality 9 7 Variance The variance of X is the mean square about the mean value of X E X x p x dx a So variance via linearity is V X E X a 2 E X 2 2a E X a 2 a 2 2 2 2 2 E X a E X E X Note V 1 0 V c 0 V cX c V X 9 7 The Law of Large Numbers Suppose X and Y are independent random variables with E X a E Y b V X 2 V Y 2 because of independence Then E X a Y b E X a E Y b 0 0 0 And V X Y E X a Y b 2 E X a 2 2E X a Y b E Y b 2 V X V Y 2 2 Consider n independent trials for X called X1 Xn The expectation of their average is as expected X 1 X n E X 1 E X n E E X a n n 9 8 The variance of their average is remember independence 2 2 X X V X V X n 1 n 1 n V 2 2 n n n n So what is the probability that their average A is not close to the mean E X a Use Chebyshev s inequality E A a P A a 2 2 V A 2 2 n 2 Let n Weak Law of Large Numbers The average of a large enough number of independent trials comes …

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