1 / 14

Chapter 9

Chapter 9. Mathematical Preliminaries. Stirling’s Approximation. Fig . 9.2-1. by trapezoid rule. take antilogs. Fig . 9.2-2. by midpoint formula. take antilogs. . 9.2. Binomial Bounds. Show the volume of a sphere of radius λ n in the n -dimensional unit hypercube is:.

lesley-levy
Download Presentation

Chapter 9

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. Chapter 9 Mathematical Preliminaries

  2. Stirling’s Approximation Fig. 9.2-1 by trapezoid rule take antilogs . . . . Fig. 9.2-2 by midpoint formula take antilogs  9.2

  3. Binomial Bounds Show 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

  4. N.b. 9.3

  5. by convention The Gamma Function For n > 1, integrate by parts: dg = e−xdxf = xn−1 Idea: extend the factorial to non-integral arguments. 9.4

  6. dr rdθ r dx dy area = rdrdθ area = dxdy 9.4

  7. N – Dimensional Euclidean Space Use Pythagorean distance to define spheres: Consequently, their volume depends proportionally on rn converting to polar coordinates 9.5

  8. r2 t just verify by substitution 9.5

  9. Interesting Facts aboutN-dimensional Euclidean Space Cn→ 0 as n → ∞  Vn(r) → 0 as n → ∞ for a fixed r Volume approaches 0 as the dimension increases! Almost all the volume is near the surface (as n→∞) end of 9.5

  10. Angle between the vector (1, 1, …, 1) and each coordinate axis: As n→ ∞: cos θ → 0,  θ → π/2. length of projection along axis What about the angle between random vectors, x and y, of the form (±1, ±1, …, ±1)? length of entire vector For large n, the diagonal line is almost perpendicular to each axis! By definition: Hence, for large n, there are almost 2n random diagonal lines which are almost perpendicular to each other! end of 9.8

  11. Chebyshev’s Inequality x2× p(x) ≥ 0 Let X be a discrete or continuous random variable with p(xi) = the probability that X = xi. The mean square is Chebyshev’s inequality 9.7

  12. Variance The variance of X is the mean square about the mean value of X, So variance, (via linearity) is: Note: V{1} = 0 → V{c} = 0 & V{cX} = c²V{X} 9.7

  13. The Law of Large Numbers Suppose X and Y are independent random variables, with E{X} = a, E{Y} = b, V{X} = σ2, V{Y} = τ2. because of independence Then E{(X− a) ∙ (Y − b)} = E{X − a} ∙ E{Y − b} = 0 ∙ 0 = 0And 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} = σ2 + τ2 Consider n independent trials for X; called X1, …, Xn. The expectation of their average is (as expected!): 9.8

  14. The variance of their average is (remember independence): So, 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. 9.8

More Related