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

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

For the 1 Dimensional Random Walk ProblemWe’ve found: The Probability Distribution is Binomial:

WN(n1) = [N!/(n1!n2!)]pn1qn2

Mean number of steps to the right: <n1> = NpDispersion in n1: <(Δn1)2> = Npq

Relative Width:(Δ*n1)/<n1> = (q½)(pN)½

for N increasing, mean value increases N, & relative width decreases (N)-½

N = 20 p = q = ½

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 <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.

• Do a Taylor’s 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 it’s average value <n1> = Np.

• It also shows that the width is equal to the value of the width <(Δn1)2> = 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 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]

• 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 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, and

• The 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 Distribution• In the limit of a large number of steps in the random

walk, N (>>1), the Binomial Distribution becomes 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

P(m) = [2πNpq]-½exp[-(m – N{p – q})2/(8Npq)]

We can express this in terms of x = mℓ. As N >> 1, x can betreated as continuous. In this case, |P(m+2) – P(m)| << P(m)& discrete values of P(m) getcloser & closer together.

• Now, ask: What is the probability that, afterN 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 dx is dx = 2ℓ.

• After some math, we obtain the standard form of the

Gaussian (Normal) DistributionP(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]

μ ≡ N(p – q)ℓ ≡ mean value of xσ ≡ 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 oflarge numbers for all discrete distributions!

P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]μ ≡ N(p – q)ℓ σ ≡ 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, doesP(x)dx = 1? (limits - < x < )

P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]dx = (2π)-½σ-1exp[-y2/2σ2]dy (y = x – μ)

= (2π)-½σ-1 [(2π)½σ] (from a table)

P(x)dx = 1

P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]μ ≡ N(p – q)ℓ σ ≡ 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-1yexp[-y2/2σ2]dy + μ exp[-y2/2σ2]dyyexp[-y2/2σ2]dy = 0 (odd function times even function)

exp[-y2/2σ2]dy = [(2π)½σ] (from a table)

<x> = μ ≡ N(p – q)ℓ

P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]

μ ≡ N(p – q)ℓ σ ≡ 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-1y2exp[-y2/2σ2]dy (y = x – μ)

= (2π)-½σ-1(½)(π)½σ(2σ2)1.5 (from a table)

<(Δx)2> = σ2 = 4Npqℓ2

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x

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fxComparison of Binomial &

Gaussian Distributions

Dots = BinomialCurve = Gaussian

with the same mean μ & the same width σ

Some Well-known & Potentially Useful Properties of Gaussians

Gaussian Width = 2σ

2σ

P(x) =

Areas Under Portions of a Gaussian Distribution

Two Graphs with the Same Informationin Different Forms

Again, Two Forms ofthe Same Information

Areas 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, v• The possible values of u are: u1,u2,u3,…uM

• The possible values of v are: v1,v2,v3,…vM

P(ui,vj) ≡ Probability that u = ui, & v = vj

SIMULTANEOUSLY• We must have:

∑i = 1 M ∑j = 1 N P(ui,vj) = 1

P(ui,vj) ≡ Probability that u = ui, & v = vj

SIMULTANEOUSLY∑i = 1 M ∑j = 1 N P(ui,vj) = 1

• Let Pu(ui) ≡ Probability that u = ui independent of the value v = vj

So, Pu(ui) ≡ ∑j = 1 N P(ui,vj)

• Similarly, let Pv(vj) ≡ Probability that

v = vj independent of value u = ui So, Pv(vj) ≡ ∑i = 1 M P(ui,vj)

• Of course, it must also be true that∑i = 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 Values• If F(u,v) = any function of u,v, it’s mean value is:

<F(u,v)> ≡ ∑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:

<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

of v, 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.

Sect. 1.8: Comments on Continuous Probability Distributions

• Everything we’ve 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)> ≡ 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 ≤ b2

• The probability of finding u in the range u to u + du AND v in the range v to v + dv is

P(u,v) ≡ P(u,v)dudv P(u,v) ≡ Probability Density function

• Normalization:

P(u,v)dudv = 1(limits a1 ≤ u ≤ a2, b1 ≤ v ≤ b2)

• Mean values:

<G(u,v)> ≡ G(u,v)P(u,v)dudv

Functions of Random VariablesAn important, often occurring problem is:

• 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 φ in the range φ to φ + dφ?

• Answer using essentially the “Chain Rule” of differentiation, but take the absolute value to make sure that probability W ≥ 0:

W(φ)dφ ≡ P(u)|du/dφ|dφCaution!!

φ(u) may not be a single valued function of u!

• Equally Likely The probability of finding θ between θ & θ + dθ is: P(θ)dθ ≡ (dθ/2π)

Question• What 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 dθ.

Also, dBx = |Bsinθ|dθ

• 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]½ so 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

• W diverges 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)