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Sect. 1.5: Probability Distributions for Large N: (Continuous Distributions)
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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
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• 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.
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• 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]
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• 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.
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• 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.
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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.
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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
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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.
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• 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!
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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
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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
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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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0 2 4 6 8 10
x
0.00
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fxComparison of Binomial &
Gaussian Distributions
Dots = BinomialCurve = Gaussian
with the same mean μ & the same width σ
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Some Well-known & Potentially Useful Properties of Gaussians
Gaussian Width = 2σ
2σ
P(x) =
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Areas Under Portions of a Gaussian Distribution
Two Graphs with the Same Informationin Different Forms
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Again, Two Forms ofthe Same Information
Areas Under Portions of a Gaussian Distribution
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Sect. 1.7: Probability Distributions Involving Several Variables: Discrete or Continuous
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• 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
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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
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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)
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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.
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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.
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• 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
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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!
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• 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
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• 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)