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Sect. 1.5: Probability Distributions for Large N (Continuous Probability Distributions)

Sect. 1.5: Probability Distributions for Large N (Continuous Probability Distributions). We’ve found that, for the one-dimensional Random Walk Problem , the probability distribution is the Binomial Distribution : W N (n 1 ) = [N!/(n 1 !n 2 !)]p n1 q n2 Here, q = 1 – p, n 2 = N - n 1.

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Sect. 1.5: Probability Distributions for Large N (Continuous Probability Distributions)

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  1. Sect. 1.5: Probability Distributions for Large N(Continuous Probability Distributions)

  2. We’ve found that, for the one-dimensional Random Walk Problem, the probability distribution is the Binomial Distribution: • WN(n1) = [N!/(n1!n2!)]pn1qn2 • Here, q = 1 – p, n2 = N - n1 The Relative Width is (Δ*n1)/<n1> = (q½)(pN)½ for N increasing, the mean value increases  N, & the relative width decreases (N)-½ N = 20, p = q = ½

  3. Now, imagine Ngetting larger & larger. Based on what we just said, the relative width of WN(n1) gets smaller & smaller & the mean value <n1> gets larger & larger. • If N is VERY, VERY large, we can treat W(n1) as a continuous function of a continuous variable n1. • For N large, it’s convenient to look at the natural log ln[W(n1)] of W(n1), rather than the function itself. • Now, do a Taylor’s series expansion of ln[W(n1)] about value of n1 where W(n1) has a maximum. • Detailed math (in the text) shows that this value of n1 is it’s average value <n1> = Np. • It also shows that the width is equal to the value of the width <(Δn1)2> = Npq.

  4. For N VERY, VERY large, treat W(n1) as a continuous function of n1. For N large, look at ln[W(n1)], rather than the function itself. • Do a Taylor’s series expansion of ln[W(n1)] about the n1 for W(n1) = its maximum. Detailed math shows that this value of n1 is it’s mean <n1> = Np. It also shows that the width is equal to <(Δn1)2> = Npq. • For ln[W(n1)], use Stirling’s Approximation (Appendix A-6) for logs of large factorials. Stirling’s Approximation • If n is a large integer, the natural log of it’s factorial is approximately: ln[n!] ≈ n[ln(n) – 1]

  5. In this large N, large n1 limit, the Binomial Distribution W(n1) becomes (shown in detail in the text): W(n1) = Ŵexp[-(n1 - <n1>)2/(2<(Δn1)2>)] Here, Ŵ = [2π <(Δn1)2>]-½ • This is called the Gaussian Distribution or the Normal Distribution. We’ve found that <n1> = Np, <(Δn1)2> = Npq. • The reasoning which led to this for large N & continuous n1 limit started with the Binomial Distribution. But this is a very general result. Starting with ANYdiscrete probability distribution & taking the limit of LARGE N, will result in the Gaussian or Normal Distribution. This is called The Central Limit Theorem or The Law of Large Numbers.

  6. Sect. 1.6:The Gaussian Probability Distribution • In the limit of a large number of steps in the random walk, N (>>1), the Binomial Distributionbecomes a Gaussian Distribution: W(n1) = [2π<(Δn1)2>]-½exp[-(n1 - <n1>)2/(2<(Δn1)2>)] <n1> = Np, <(Δn1)2> = Npq • Recall that n1 = ½(N + m),where the displacement x = mℓ & that <m> = N(p – q). We can use this to convert to the probability distribution for displacement m, in the large N limit (after algebra): P(m) = [2π<(Δm)2>]-½exp[-(m- <m>)2/(2<(Δm)2>)] <m> = N(p – q), <(Δm)2> = 4Npq

  7. P(m) = [2πNpq]-½exp[-(m – N{p – q})2/(8Npq)] We can express this in terms of x = mℓ.As N  >> 1, xcan be treated as continuous. In this case, |P(m+2) – P(m)| << P(m) & discrete values of P(m) get closer & closer together. • Now, ask: What is the probability that, after • N steps, the particle is in the range x to x + dx? • Let the probability distribution for this≡ P(x). • Then, we have:P(x)dx = (½)P(m)(dx/ℓ). • The range dx contains (½)(dx/ℓ) possible values • of m, since the smallest possible dxis dx = 2ℓ.

  8. After some math, we obtain the standard form of the Gaussian (Normal) Distribution P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2] μ ≡ N(p – q)ℓ ≡mean value of x σ ≡ 2ℓ(Npq)-½ ≡width of the distribution NOTE: The generality of the arguments we’ve used is such that a Gaussian Distribution occurs in the limit of large numbers for all discrete distributions!

  9. P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2] μ ≡ N(p – q)ℓσ ≡ 2ℓ(Npq)-½ • Note:To deal with Gaussian distributions, you need to get used to doing integrals with them! Many are tabulated!! • Is P(x)properly normalized? That is, does P(x)dx = 1? (limits -  < x < ) P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]dx = (2π)-½σ-1exp[-y2/2σ2]dy (y = x – μ) = (2π)-½σ-1 [(2π)½σ] (from a table) P(x)dx = 1

  10. P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2] μ ≡ N(p – q)ℓσ ≡ 2ℓ(Npq)-½ • Compute the mean value of x (<x>): <x> = xP(x)dx = (limits -  < x < ) xP(x)dx = (2π)-½σ-1xexp[-(x – μ)2/2σ2]dx = (2π)-½σ-(y + μ)exp[-y2/2σ2]dy(y = x – μ) = (2π)-½σ-1yexp[-y2/2σ2]dy + μ exp[-y2/2σ2]dy yexp[-y2/2σ2]dy = 0(odd function times even function) exp[-y2/2σ2]dy = [(2π)½σ] (from a table) <x> = μ ≡ N(p – q)ℓ

  11. P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2] μ ≡ N(p – q)ℓσ ≡ 2ℓ(Npq)-½ • Compute the dispersion in x (<(Δx)2>) <(Δx)2> = <(x – μ)2> = (x – μ)2P(x)dx (limits -  < x < ) <(Δx)2> = xP(x)dx = (2π)-½σ-1xexp[-(x – μ)2/2σ2]dx = (2π)-½σ-1y2exp[-y2/2σ2]dy(y = x – μ) = (2π)-½σ-1(½)(π)½σ(2σ2)1.5(from a table) <(Δx)2> = σ2 = 4Npqℓ2

  12. Comparison of Binomial & Gaussian Distributions Dots = Binomial Curve = Gaussian with the same mean μ & the same width σ

  13. Gaussian Width = 2σ 2σ P(x) =

  14. Areas Under Portions of a Gaussian Distribution Two Graphs with the Same Information in Different Forms

  15. Areas Under Portions of a Gaussian Again, Two Forms of the Same Information

  16. Sect. 1.7: Probability Distributions Involving Several Variables

  17. Consider a statistical description of a situation with more than one random variable: Example, 2 variables, u, v The possible values of u are: u1,u2,u3,…uM The possible values ofv are: v1,v2,v3,…vM Let P(ui,vj) ≡ Probability thatu = ui, &v = vjsimultaneously • We must have: ∑i = 1 M ∑j = 1 NP(ui,vj) = 1 • Pu(ui) ≡ Probability thatu = ui independent of value v = vjPu(ui) ≡ ∑j = 1 NP(ui,vj) • Pv(vj) ≡ Probability that v = vjindependent of value u = uiPv(vj) ≡ ∑i = 1 MP(ui,vj) • Of course, ∑i = 1 M Pu(ui) = 1 &∑j = 1 NPv(vj) = 1

  18. In the special casethat u, v are Statistically Independent or Uncorrelated: Then & only then: P(ui,vj) ≡ Pu(ui)Pv(vj) General Discussion of Mean Values: • If F(u,v)= any function ofu,v, it’s mean value is given by: <F(u,v)> ≡ ∑i = 1 M ∑j = 1 NP(ui,vj)F(ui,vj) • If F(u,v) & G(u,v)are any 2 functions ofu,v, can easily show: <F(u,v) + G(u,v)>= <F(u,v)> + <G(u,v)> • If f(u) is any function of u& g(v) is any function ofv, we can easily show: <f(u)g(v)>≠ <f(u)><g(v)> The only case when the inequality becomes an equality is if u& v are statistically independent.

  19. Sect. 1.8: Comments on Continuous Probability Distributions • Everything we’ve discussed for discrete distributions generalizes to continuous distributions in obvious ways. • u ≡ a continuous random variable in the range: a1≤ u ≤ a2 • The probability of finding u in the range u to u + du ≡ P(u) ≡ P(u)du P(u)≡Probability Density of the distribution function • Normalization: P(u)du= 1 (limitsa1≤ u ≤ a2) • Mean values: <F(u)> ≡F(u)P(u)du.

  20. Consider two continuous random variables: u ≡ continuous random variable in range: a1≤ u ≤ a2 v ≡ continuous random variable in range: b1≤ v ≤ b2 • The probability of finding u in the range u to u + duAND v in the range v to v + dvis P(u,v) ≡ P(u,v)dudv P(u,v) ≡Probability Density of the distribution function • Normalization: P(u,v)dudv= 1 (limitsa1≤ u ≤ a2, b1≤ v ≤ b2) • Mean values: <G(u,v)> ≡G(u,v)P(u,v)dudv

  21. Functions of Random Variables An important, often occurring problem: • Consider a random variable u. • Suppose φ(u)≡ any continuous function of u. Question • If P(u)du ≡Probability of finding u in the range u to u + du, what is the probability W(φ)dφ of finding φin the range φto φ + dφ? • Answer using essentially the “Chain Rule” of differentiation, but take the absolute value to make sure that W ≥ 0: W(φ)dφ ≡ P(u)|du/dφ|dφ Caution!! φ(u) may not be a single valued function of u!

  22. Equally Likely The probability of finding θbetween θ& θ + dθis: P(θ)dθ ≡ (dθ/2π) Question What is the probability W(Bx)dBxthat the x component of B lies between Bx & Bx + dBx? Clearly, we must have –B ≤ Bx ≤ B. Also, each value of dBx corresponds to 2 possible values of dθ. Also, dBx = |Bsinθ|dθ • Example: A 2-dimensional vector B of constant magnitude |B| is EQUALLY LIKELY to point in any direction θ in the x-y plane. Figures 

  23. So, we have: W(Bx)dBx = 2P(θ)|dθ/dBx|dBx = (π)-1dBx/|Bsinθ| Note also that:|sinθ| = [1 – cos2θ]½= [1 – (Bx)2/B2]½so finally, W(Bx)dBx = (π)-1dBx[1 – (Bx)2/B2]-½, –B ≤ Bx ≤ B = 0, otherwise W not only has a maximum at Bx = B,it diverges there! It has a minimum at Bx = 0. So, it looks like Wdiverges at Bx = B, but it can be shown that it’s integral is finite. So that W(Bx)is a properly normalized probability: W(Bx)dBx= 1 (limits:–B ≤ Bx ≤ B)

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