Sect. 1.5: Probability Distribution for Large N

• 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: (Δ*n1)/<n1> = (q½)(pN)½

as N increases, the mean value increases N, & the

relative width decreases (N)-½

N = 20, p = q = ½

• Now, 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, 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.

• 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 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. However, this is a very general result. If one starts with ANY discrete probability distribution & takes the limit of LARGE N, one will obtain a Gaussian or Normal Distribution. This is called

The Central Limit Theorem or

The Law of Large Numbers.

Sect. 1.6: Gaussian Probability Distributions• 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 be treated as continuous. In this case, |P(m+2) – P(m)| << P(m) & discrete values of P(m) get closer & closer together.

Now, lets ask: What is the probability that, after N steps, the particle is in the range x to x + dx? 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 Gaussian Distribution form:

P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2] Here: 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 any

discrete distribution

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 of these 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-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]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

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 < )

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 table)

<(Δx)2> = σ2 = 4Npqℓ2

0 2 4 6 8 10

x

0.00

0.05

0.10

0.15

0.20

0.25

fxComparison of Binomial & Gaussian Distributions

Dots: BinomialCurve: Gaussian The same mean &

the same width

The width of a Gaussian is 2σ

2σ

Areas under portions of a Gaussian

Sect. 1.7: Probability Distributions Involving Several Variables

• Consider a statistical description of a situation with more than one variable: For 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

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

• We must have: ∑i = 1 M ∑j = 1 N P(ui,vj) = 1

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

Pu(ui) ≡ ∑j = 1 N P(ui,vj)

• Pv(vj) ≡ Probability that v = vj independent of value u = ui

Pv(vj) ≡ ∑i = 1 M P(ui,vj)

• Of course, ∑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: P(ui,vj) ≡ Pu(ui)Pv(vj)

General Discussion of Mean Values:• If F(u,v) = any function of u,v, it’s mean value is given by:

<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, 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 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 (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 of the distribution 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:

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

• 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

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 proper probability:

W(Bx)dBx= 1 (limits: –B ≤ Bx ≤ B)

The Poisson Probability Distribution

Simeon Denis Poisson • "Researches on the probability of

criminal and civil verdicts" 1837 

• Looked at the form of the binomial distribution

when the number of trials is large. 

• He derived the cumulative Poisson distribution as the

limiting case of the binomial when

the chance of success tends to zero.

The Poisson Probability Distribution

Simeon Denis Poisson • "Researches on the probability of criminal and civil verdicts" 1837 

• Looked at the form of the binomial distribution

when the number of trials is large.  • He derived the cumulative Poisson

distribution as the

limiting case of the binomial

whenthe chance of success

tends to zero.

Simeon Denis “Fish”!

• Poisson Distribution: Approximation to binomial distribution for the special case when the average number of successes is very much smaller than the possible number i.e. µ <<

n because p << 1. – It is important for the study of such phenomena as radioactive decay. This

distribution is NOT necessarily symmetric! Data are usually bounded on one side and not the other. An advantage of this distribution is that σ2 = μ

µ = 1.67σ = 1.29

µ = 10.0σ =

Another Useful Probability Distribution: The Poisson Distribution

• The Poisson Distribution models counts: If events happen at a constant rate over time, the Poisson distribution

gives the probability of X number of events occurring in a time T.

• This distribution tells us the

probability of all possible numbers of counts, from 0 to infinity.

• If X= # of counts per second, then the Poisson probability that X = k (a particular count) is:

• Here, λ ≡ the average number of counts per second. !)(

k

ekXp

k

Mean and Variance for the Poisson Distribution

• It’s easy to show that:

The Mean

The Variance &

Standard Deviation

2

For a Poisson Distribution, the variance and mean are the

same!

Terminology: A “Poisson Process”

• The Poisson parameter can be given as the mean number of events that occur in a defined time period OR, equivalently, can be given as a rate, such as = 2 events per month. must often be multiplied by a time t in a physical process (called a “Poisson Process”)

!

)()(

k

etkXP

tk

μ = t σ = t

More on the Poisson Distribution

Example1. If calls to your cell phone are a Poisson process with a constant rate = 2 calls per hour, what is the probability that, if you forget to turn your phone off in a 1.5 hour movie, your phone rings during that time?

Answer: If X = # calls in 1.5 hours, we want

P(X ≥ 1) = 1 – P(X = 0)

P(X ≥ 1) = 1 – .05 = 95% chance

2. How many phone calls do you expect to get during the movie?

<X> = t = 2(1.5) = 3Editorial comment: People at the

movie will not be very happy with you!!

05.!0

)3(

!0

)5.1*2()0( 3

30)5.1(20

eee

XP

31

Conditions required for the

Poisson Distribution to hold:

– The rate is a constant, independent of time.– Two events never occur at exactly the same time.

– Each event is independent --- the occurrence of one event does not make the next event more or less likely to happen.

Example

= (5 defects/hour)*(0.25 hour) = 1.25

p(x) = (xe-)/(x!) x = given number of defects

P(x = 0) = (1.25)0e-1.25)/(0!)

= e-1.25 = 0.287

= 28.7%

• A production line produces 600 parts per hour with an average of 5 defective parts an hour. If you test every part that comes off the line in 15 minutes, what is the probability of finding no defective parts (and incorrectly concluding that your process is perfect)?