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# Chapter 9 - PowerPoint PPT Presentation

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:.

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### 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

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

9.3

### The Gamma Function

For n > 1, integrate by parts: dg = e−xdxf = xn−1

Idea: extend the factorial to non-integral arguments.

9.4

rdθ

r

dx

dy

area = rdrdθ

area = dxdy

9.4

Use Pythagorean distance to define spheres:

Consequently, their volume depends proportionally on rn

converting to polar coordinates

9.5

r2 t

just verify

by substitution

9.5

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

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

### Chebyshev’s Inequality axis:

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

Variance axis:

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

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

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