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

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

• For the 1 Dimensional Random Walk ProblemWeve found: The Probability Distribution is Binomial:WN(n1) = [N!/(n1!n2!)]pn1qn2Mean number of steps to the right: = NpDispersion in n1: = NpqRelative Width:(*n1)/ = (q)(pN)for N increasing, mean value increases N, & relative width decreases (N)-N = 20 p = q = q = 1 p n2 = N - n1

• Imagine N getting larger & larger. Based on what we just said, the relative width of WN(n1) gets smaller & smaller & the mean value 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, its convenient to look at the natural log ln[W(n1)] of W(n1), rather than the function itself. Do a Taylors series expansion of ln[W(n1)] about value of n1 where W(n1) has a maximum.Detailed math (in text) shows that this value of n1 is its average value = Np. It also shows that the width is equal to the value of the width = Npq.

• 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 Taylors series expansion of ln[W(n1)] about the n1 for W(n1) = its maximum. Detailed math shows that this value of n1 is its mean = Np. It also shows that the width is equal to = Npq.For ln[W(n1)], use Stirlings Approximation (Appendix A-6) for logs of large factorials. Stirlings ApproximationIf n is a large integer, the natural log of its factorial is approximately:ln[n!] n[ln(n) 1]

• In this large N, large n1 limit, the Binomial Distribution W(n1) becomes (shown in detail in the text): W(n1) = exp[-(n1 - )2/(2)] Here, = [2 ]- This is called the Gaussian Distribution or the Normal Distribution. Weve found that = Np, = 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 ANY discrete probability distribution & taking the limit of LARGE N, will result in the Gaussian or Normal Distribution. This is called The Central Limit Theoremor The Law of Large Numbers.

• One of the most important results of probability theory is The Central Limit Theorem:The distribution of any random phenomenon tends to be Gaussian or Normal if we average it over a large number of independent repetitions.This theorem allows us to analyze and predict the results of chance phenomena when we average over many observations.

• Related to the Central Limit Theorem is The Law of Large Numbers:As a random phenomenon is repeated a large number of times, the proportion of trials on which each outcome occurs gets closer and closer to the probability of that outcome, andThe mean of the observed values gets closer and closer to the mean of a Gaussian Distribution which describes the data.

• Sect. 1.6: The Gaussian Probability DistributionIn the limit of a large number of steps in the random walk, N (>>1), the Binomial Distribution becomes aGaussian Distribution: W(n1) = [2]-exp[-(n1 - )2/(2)] = Np, = NpqRecall that n1 = (N + m), where the displacement x = m & that = 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) = [2]-exp[-(m - )2/(2)] = N(p q), = 4Npq

• P(m) = [2Npq]-exp[-(m N{p q})2/(8Npq)] We can express this in terms of x = m. As N >> 1, x can betreated as continuous. In this case, |P(m+2) P(m)|
• After some math, we obtain the standard form of theGaussian (Normal) DistributionP(x)dx = (2)--1exp[-(x )2/22] N(p q) mean value of x 2(Npq)- width of the distribution

NOTE: The generality of the arguments weve used is such that a Gaussian Distribution occurs in the limit oflarge numbers for all discrete distributions!

• P(x)dx = (2)--1exp[-(x )2/22] N(p q) 2(Npq)- 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, doesP(x)dx = 1? (limits - < x < )P(x)dx = (2)--1exp[-(x )2/22]dx = (2)--1exp[-y2/22]dy (y = x ) = (2)--1 [(2)] (from a table)P(x)dx = 1

• P(x)dx = (2)--1exp[-(x )2/22] N(p q) 2(Npq)- Compute the mean value of x (): = xP(x)dx = (limits - < x < )xP(x)dx = (2)--1xexp[-(x )2/22]dx= (2)--(y + )exp[-y2/22]dy (y = x )= (2)--1yexp[-y2/22]dy + exp[-y2/22]dyyexp[-y2/22]dy = 0 (odd function times even function) exp[-y2/22]dy = [(2)] (from a table) = N(p q)

• P(x)dx = (2)--1exp[-(x )2/22] N(p q) 2(Npq)- Compute the dispersion in x () = = (x )2P(x)dx (limits - < x < ) = xP(x)dx = (2)--1xexp[-(x )2/22]dx= (2)--1y2exp[-y2/22]dy (y = x ) = (2)--1()()(22)1.5 (from a table)

= 2 = 4Npq2

• Comparison of Binomial & Gaussian DistributionsDots = BinomialCurve = Gaussianwith the same mean & the same width

• Some Well-known & Potentially Useful Properties of Gaussians Gaussian Width = 22P(x) =

• Areas Under Portions of a Gaussian DistributionTwo Graphs with the Same Informationin Different Forms

• Again, Two Forms ofthe Same InformationAreas Under Portions of a Gaussian Distribution

• Sect. 1.7: Probability Distributions Involving Several Variables: Discrete or Continuous

• Consider a statistical description of a situation with more than one random variable: Example, 2 variables, u, vThe possible values of u are: u1,u2,u3,uMThe possible values of v are: v1,v2,v3,vMP(ui,vj) Probability that u = ui, & v = vjSIMULTANEOUSLY We must have:i = 1 M j = 1 N P(ui,vj) = 1

• P(ui,vj) Probability that u = ui, & v = vj SIMULTANEOUSLYi = 1 M j = 1 N P(ui,vj) = 1Let Pu(ui) Probability that u = ui independent of the value v = vjSo, Pu(ui) j = 1 N P(ui,vj)Similarly, let Pv(vj) Probability thatv = vj independent of value u = ui So, Pv(vj) i = 1 M P(ui,vj)Of course, it must also be true thati = 1 M Pu(ui) = 1 & j = 1 N Pv(vj) = 1

• In the special case that u & v are Statistically Independent or Uncorrelated:Then & only then can we write:P(ui,vj) Pu(ui)Pv(vj)

• A General Discussion of Mean ValuesIf F(u,v) = any function of u,v, its mean value is: i = 1 M j = 1 N P(ui,vj)F(ui,vj) If F(u,v) & G(u,v) are any 2 functions of u, v, we can easily show: = + If f(u) is any function of u & g(v) is any function of v, we can easily show: The only case when the inequality becomes an equality is if u & v are statistically independent.

• Sect. 1.8: Comments on Continuous Probability DistributionsEverything weve discussed for discrete distributions generalizes to continuous distributions in obvious ways.Let 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 (limits a1 u a2)Mean values: F(u)P(u)du.

• Consider two continuous random variables:u continuous random variable in range: a1 u a2 v continuous random variable in range: b1 v b2The probability of finding u in the range u to u + du AND v in the range v to v + dv isP(u,v) P(u,v)dudv P(u,v) Probability Density functionNormalization:P(u,v)dudv = 1(limits a1 u a2, b1 v b2)Mean values: G(u,v)P(u,v)dudv

• Functions of Random VariablesAn important, often occurring problem is:Consider a random variable u. Suppose (u) any continuous function of u. QuestionIf P(u)du Probability of finding u in the rangeu to u + du, what is the probability W()d of finding in the range to + d?Answer using essentially the Chain Rule of differentiation, but take the absolute value to make sure that probability W 0:W()d P(u)|du/d|dCaution!!(u) may not be a single valued function of u!

• Equally Likely The probability of finding between & + d is: P()d (d/2)QuestionWhat is the probability W(Bx)dBx that 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 d. Also, dBx = |Bsin|d

• So, we have:W(Bx)dBx = 2P()|d/dBx|dBx = ()-1dBx/|Bsin| Note also that: |sin| = [1 cos2] = [1 (Bx)2/B2] so finally, W(Bx)dBx = ()-1dBx[1 (Bx)2/B2]-, B Bx B = 0, otherwiseW not only has a maximum at Bx = B, it diverges there! It has a minimum at Bx = 0. So, it looks like W diverges at Bx = B, but it can be shown that its integral is finite. So that W(Bx) is a properly normalized probability: W(Bx)dBx= 1 (limits: B Bx B)

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