Chapter 1 : Introduction to Statistical Methods

1. Chapter 1:Introduction to Statistical Methods “The true logic of this world is in the calculus of probabilities”. James Clerk Maxwell

2. Introductory Remarks • This course: The Physics of systems containing HUGE numbers (~ 1023) of particles: Solids, liquids, gases, E&M radiation (photons), …. • Challenge:Describe a system’s Macroscopic characteristics from a Microscopic theory. Classical:Newton’s Lawsneed to solve 1023coupled differential equations of motion (ABSURD!!) Quantum:Schrödinger’s Equationneed to solve for 1023particles (ABSURD!!)

3.  Use a Statistical description of such a system.  Talk about Probabilities & Average Properties We are NOTconcerned with the detailed behavior of individual particles. • Definitions: Microscopic:~ Atomic dimensions ~ ≤ a fewÅ Macroscopic:Large enough to be “visible” in the “ordinary” sense

4. Definitions • An Isolated System is in Equilibrium when it’s Macroscopic parameters are time-independent. This is the usual case in this course! • But, note! Even if it’s Macroscopic parameters are time-independent, a system’s Microscopic parameters can still vary with time!

5. The Random Walk & The Binomial Distribution Section 1.1:Elementary Statistical Concepts; Examples • Some math preliminaries (& methods) for the next few lectures. • To treat statistical physics problems, we must first know something about the mathematics of Probability & Statistics The following should be a review! (?) • Keep in mind:Whenever we want to describe a situation using probability & statistics, we must consider an assembly of a large number N (in principle, N ∞)of “similarly prepared systems”.

6. This assembly is called an ENSEMBLE (“Ensemble” = the French word for assembly). • The Probability of an occurrence of a particular event is DEFINED with respect to this particular ensemble & is given by the fraction of systems in the ensemble characterized by the occurrence of this event. • Example: In throwing a pair of dice, we can give a statistical description by considering that a very large number N of similar pairs of dice are thrown under similar circumstances. Alternatively, we could imagine the same pair of dice thrown N times under similar circumstances. The probability of obtaining two 1’s is then given by the fraction of these experiments in which two 1’s is the outcome.

7. Note that this probability depends strongly on the nature of the ensemble to which we are referring. • Reif’s flower seed example (p. 5). • To quantitatively introduce probability concepts, we use a specific, simple example, which is actually much more general than you first might think. • The example is called The Random Walk Problem

8. One-Dimensional Random Walk In it’s simplest, idealized form, the random walk problem can be viewed as in the figure above: A drunk starts out from a lamp post on a street. Each step he takes is of equal length ℓ. The man is SO DRUNK, that the direction of each step (right or left) is completely independent of the preceding step. The probability of stepping to the right is p & of stepping to the left is q = 1 – p. In general, q ≠ p. The x axis is along the sidewalk, the lamp post is at x = 0. Each step is of length ℓ, so his location on the x axis must be x = mℓ where m = a positive or a negative integer.

9. Question: After N steps, what is the probability that the man is at a specific location x = mℓ (m specified)?To answer, we first consider an ensemble of a large numberN of drunk men starting from similar lamp posts. Or repeat this with the same drunk man walking on the sidewalk N times. • This can be easily generalized to 2 dimensions. See figure to the right:

10. The 2 dimensional random walk corresponds to aPHYSICS problemof adding N, 2 dimensional vectors of equal length (see figure)& random directions & asking: What is the probability that the resultant has a certain magnitude & direction?

11. Physical Examplesto which theRandom Walk Problemapplies • Magnetism Natoms, each with magnetic moment μ. Each has spin ½. By Quantum Mechanics, each magnetic moment can point either “up” or “down”. If these are equally likely, what is the Net magnetic moment of the N atoms? • Diffusion of a Molecule of |Gas A molecule travels in 3 dimensions with a mean distance ℓbetween collisions. How far is it likely to have traveled after Ncollisions? Answer using Classical Mechanics.

12. The Random Walk Problem • The Random Walk Problemillustrates some fundamental results ofProbability Theory. • The techniques used arePowerful & General. They are used repeatedly throughout Statistical Mechanics. So, it’s very important to spend some time on this problem & understand it!

13. Section 1.2: 1 Dimensional Random Walk • Forget the drunk, let’s get back to Physics! Think of a particle moving in 1 dimension in steps of length ℓ, with probability p of stepping to the right & q = 1 – p of stepping to the left. After N steps, the particle is at position: x = mℓ (- N ≤m≤N). Let n1≡ # of steps to the right (of N), n2≡ # of steps to the left. Clearly, N = n1+ n2 (1) Clearly also, x ≡ mℓ = (n1- n2)ℓ or, m = n1- n2 (2) Combining (1) & (2) gives m = 2n1– N (3) Thus, if N is odd, so is m and if N is even, so is m.

14. A Fundamental Assumption is that successive steps are statistically independent • Let p≡the probability of stepping to the right and q = 1 – p≡the probability of stepping to the left. • Since each step is statistically independent, the probability of agiven sequenceof n1steps to the right followed by n2steps to the left is given by multiplying the respective probabilities for each step: p·p·p·p·p· · · · · · · p·p· ···· · ···· · q·q·q·q·q·q·q·q·q·q· · · q·q ≡ pn1qn2  n1 factors  n2 factors  • But, also, clearly, there are MANY different possible ways of taking N steps so n1 are to right & n2 are to left!

15. The # of distinct possibilities is the SAME as counting the # of distinct ways we can place N objects, n1of one type & n2of another in N = n1+ n2 places: 1st place: Can be occupied any one of N ways 2nd place: Can be occupied any one of N - 1 ways 3rd place: Can be occupied any one of N - 2 ways · · (N – 1)th place: Can be occupied only 2 ways Nth place: Can be occupied only 1way All available places can be occupied in: N(N-1)(N-2)(N-3)(N-4)·····(3)(2)(1) ≡ N! Ways N! ≡ “N-Factorial”

16. Note However!This analysis doesn’t take into account the fact that there areonly 2 distinguishablekinds of objects:n1 of the1st type& n2 of the2ndtype.All n1!possible permutations of the1st typeof object lead toexactly the sameN! possible arrangements of the objects. Similarly, alln2!possible permutations of the2ndtypeof objectalsolead toexactly the sameN! arrangements.  So, we need to divide the result by n1!n2! So, the#of distinct ways in which Nobjects can be arranged with n1of the 1st type & n2 of the 2ndtypeis ≡ N!/(n1!n2!) This is the sameas the # of distinct ways of taking N steps, with n1to the right & n2 to the left.

17. In Summary: The probability WN(n1) of taking N steps, with n1 to the right & n2=N -n1 to the left is WN(n1) = [N!/(n1!n2!)]pn1qn2or WN(n1) = [N!/{n1!(N – n1)!]}pn1(1-p)n2 • Often, this is written as WN(n1) = N pn1qn2 n1 • This probability distribution is called the Binomial Distribution. • This is because the Binomial Expansion has the form (p + q)N = ∑(n1 = 0N)[N!/[n!(N–n1)!]pn1qn2 Remember that q = 1- p

18. We really want the probabilityPN(m) that x = mℓafter N steps. This really the same as WN(n1)if we change notation: PN(m) = WN(n1).But m = 2n1– N,so n1= (½)(N + m)& n2= N - n1 = (½)(N - m).So the probabilityPN(m) that x = mℓafter N steps is: PN(m) = {N!/([0.5(N + m)]![0.5(N – m)!]}p0.5(N+m)(1-p)0.5(N-m) For the common case of p = q = ½, this is: PN(m) = {N!/([0.5(N + m)]![0.5(N – m)!]}(½)N This is the usual form of the Binomial Distribution which is probably the most elementary (discrete) probability distribution.

19. Possible Step Sequences • As a trivial example, suppose that p = q = ½, N = 3 steps: P3(m) = {3!/[0.5(3+m)!][0.5(3-m)!](½)3 So P3(3) = P3(-3) = (3!/[3!0!](⅛) = ⅛ P3(1) = P3(-1) = (3!/[2!1!](⅛) = ⅜

20. Note: The envelope of the histogram is a bell-shaped curve. The significance of this is that, after N random steps, the probability of a particle being a distance of N steps away from the start is very small & the probability of it being at or near the origin is relatively large: • As another example, suppose that: p = q = ½, N = 20. P20(m) = {20!/[0.5(20 + m)!][0.5(20 - m)!](½)3 • Doing this gives the histogram results in the figure P20(20) = [20!/(20!0!)](½)20 P20(20)  9.5  10-7 P20(0) = [20!/(10!)2](½)20 P20(0)  1.8  10-1

21. Sect. 1.3: General Discussion of Mean Values

22. The Binomial Distribution is only one example of a probability distribution. Now, we’ll begin a discussion of a General Distribution. • Most of the following results are valid forANYprobability distribution, • Let u = a variable which can take on any of M discrete values: u1,u2,u3,…,uM-1,uM with probabilities P(u1),P(u2),P(u3),…..,P(uM-1),P(uM) • The Mean (average) value of u is defined as: ū≡ <u> ≡ (S2/S1) where S1 ≡ P(u1) + P(u2) + P(u3) +…..+ P(uM-1) + P(uM)≡ ∑iP(ui) S2 ≡ u1P(u1) + u2P(u2) + u3P(u3) +…..+ uM-1P(uM-1) + uM-1P(uM) ≡ ∑iuiP(ui) For a properly normalized distribution,S1 = ∑iP(ui) = 1. We assume this from now on.

23. Sometimes, ū is called the 1st moment of P(u). • If O(u) is any function of u, the mean value of O(u) is: Ō ≡ <O> ≡ ∑iO(ui)P(ui) • Some simple mean values that are useful for describing the probability distribution P(u): 1. The mean value,ū This is a measure of the central value of u about which the various values of ui are distributed. • Consider the quantityΔu ≡ u - ū(deviation from the mean).It’s mean is: <Δu> = <u - ū> = ū – ū = 0  The mean value of the deviation from the mean is always zero!

24. Now, lets look at(Δu)2 = (u - <u>)2(square of the deviation from the mean). It’s mean value is: <(Δu)2> = <(u - <u>)2>= <u2 -2uū – (ū)2> = <u2> - 2<u><u> – (<u>)2 = <u2> - (<u>)2 • This is called the “Mean Square Deviation”(from the mean). It is also called several different (equivalent!) other names: the Dispersionor the Variance or the 2nd Moment of P(u) about the mean. <(Δu)2> is a measure of the spread of theuvaluesabout the mean ū. NOTE that <(Δu)2> = 0 if & only if ui = ūfor all i. • It can easily be shown that,<(Δu)2> ≥ 0, or <u2> ≥ (<u>)2

25. We could also define the nth moment of P(u) about the mean: <(Δu)n> ≡ <(u - <u>)n> • This is rarely used beyond n = 2. Almost never beyond n = 3 or 4. • NOTE:A knowledge of the probability distribution function P(u) gives complete information about the distribution of the values of u. But, a knowledge of only a few moments, like knowing just ū & <(Δu)2>implies only partial, though useful knowledge of the distribution. A knowledge of only some moments is not enough to uniquely determineP(u). Math Theorem • In order to uniquely determine a distributionP(u),we need to knowALLmoments of it. That is we need all moments for n = 0,1,2,3….  .

26. Section 1.4: Calculation of Mean Values for the Random Walk Problem • Also we’ll discuss a few math “tricks” for doing discrete sums! • We’ve found: The probability in N steps of making n1 to the right & n2 = N - n1 to the left is the Binomial Distribution: WN(n1) = [N!/(n1!n2!)]pn1qn2 p =the probability of a step to the right,q = 1 – p =the probability of a step to the left. • First, lets verify normalization: ∑(n1 = 0N)WN(n1) = 1? • Recall the binomial expansion: (p + q)N = ∑(n1 = 0N)[N!/(n1!n2!)]pn1qn2 = ∑(n1 = 0N)WN(n1) • But, (p + q) = 1, so(p + q)N = 1& ∑(n1 = 0N)WN(n1) = 1.

27. Question 1:What is the mean number of steps to the right? <n1> ≡ ∑(n1 = 0N) n1WN(n1) = ∑(n1 = 0N) n1[N!/(n1!(N-n1)!]pn1qN-n1 (1) • We can do this sum by looking it up in a table ORwe can use a “trick” as follows. The following is a general procedure which usually works, even if it doesn’t always have mathematical “rigor”. • Temporarily, lets treat p & q as “arbitrary”, continuous variables, ignoring the fact that p + q =1. • NOTEthat, ifpis a continuous variable, then we clearly have: n1pn1≡ p[(pn1)/p] • Now, use this in (1) (interchanging the sum & the derivative): <n1> = ∑(n1 = 0N)[N!/(n1!(N-n1)!]n1pn1qn2 = ∑(n1 = 0N)[N!/(n1!(N-n1)!]p[(pn1)/p]qn2 =p[/p]∑(n1 = 0N) [N!/(n1!(N-n1)!]pn1qN-n1 =p[/p](p + q)N = pN(p + q)N-1 But, for our special case (p + q) = 1, (p + q)N-1 = 1, so <n1> = Np

28. Summary:The mean number of steps to the right is: <n1> = Np • We might have guessed this! Similarly, we can also easily show that The mean number of steps to the left is: <n2> = Nq • Of course, <n1> + <n2> = N(p + q) = N as it should! • Question 2: What is the mean displacement,<x> = <m>ℓ? Clearly, m =n1 – n2, so <m> = <n1> - <n2> = N( p – q) So, if p = q = ½, <m> = 0 so, <x> = <m>ℓ = = 0

29. Question 3: What is the dispersion (or variance) <(Δn1)2> = <(n1 - <n1>)2> in the number of steps to the right? That is, what is the spread in n1 values about <n1>? • Our general discussion has shown that: <(Δn1)2> = <(n1)2> - (<n1>)2 Also we’ve just seen that<n1> = Np So, we first need to calculate the quantity <(n1)2> <(n1)2> = ∑(n1 = 0N)(n1)2 WN(n1) = ∑(n1 = 0N)(n1)2[N!/(n1!(N-n1)!]pn1qN-n1 (2) • Use a similar “trick” as we did before & note that: (n1)2pn1≡ [p(/p)]2pn1

30. <(Δn1)2> = Npq • After algebra (in the book) & using p + q = 1, we find: <(n1)2> = (Np)2 + Npq = (<n1>)2 + Npq • So, finally, using <(Δn1)2> = <(n1)2> - (<n1>)2 • This is the dispersion or variance of the binomial distribution. • The root mean square (rms) deviation from the mean is defined as: (Δ*n1)  [<(Δn1)2>]½(in general). For the binomial distribution this is (Δ*n1)= [Npq]½ The distribution width • Again note that: <n1> = Np.So, the relative width of the distribution is: (Δ*n1)/<n1> = [Npq]½(Np) = (q½)(pN)½ If p = q, this is: (Δ*n1)/<n1> = 1(N)½ = (N)-½ As N increases, the mean value increases N but the relative width decreases (N)-½

31. Question 4: What is the dispersion <(Δm)2> = <(m - <m>)2> in the net displacement? x = mℓ. What is the spread in m values about <m>)? • We had, m = n1 – n2 = 2n1 – N. So, <m> = 2<n1> – N. Δm = m - <m> = (2n1 – N) – (2<n1> - N) = 2(n1 – <n1>) = 2(Δn1) (Δm)2 = 4(Δn1)2. So, <(Δm)2> = 4<(Δn1)2> • Using <(Δn1)2> = Npq, this becomes: • If p = q = ½,<(Δm)2> = N <(Δm)2> = 4Npq

32. Summary: 1 Dimensional Random Walk Problem • Probability Distribution is Binomial: WN(n1) = [N!/(n1!n2!)]pn1qn2 • Mean number of steps to the right: <n1> = Np • Dispersion in n1: <(Δn1)2> = Npq Relative width:(Δ*n1)/<n1> = (q½)(pN)½ for N increasing, the mean value increases N, the relative width decreases (N)-½

33. Some General Comments about the Binomial Distribution

34. The Binomial Distribution applies to cases where there are only two possible outcomes: head or tail, success or failure, defective item or good item, etc. Requirements justifying the use of the Binomial Distribution 1.The experiment must consist ofn identical trials. 2.Each trial must result inonly one of two possible outcomes. 3.The outcomes of the trialsmust be statistically independent. 4.All trials must havethe same probabilityfor a particular outcome.

35. Common Notation for the Binomial Distribution r items of one type and (n – r) of a second type can be arranged in nCr ways. Here: ≡ nCris calledthe binomial coefficient In this notation, the probability distribution can be written: Wn(r) = nCr pr(1-p)n-r ≡ probability of finding r items of one type& n – r items of the other type. p = probability of a given item being of one type

36. Binomial Distribution Example Problem: A sample of n = 11 electric bulbs is drawn every day from those manufactured at a plant. The probabilities of getting defective bulbs are random and independent of previous results. The probability that a bulb is defective isp = 0.04. 1. What is the probability of finding exactly three defective bulbs in a sample? (Probability that r = 3?) 2. What is the probability of finding three or more defective bulbs in a sample? (Probability that r ≥ 3?)

37. Binomial Distribution, n = 11

38. Question 1: Probability of finding exactly three defective bulbs in a sample? P(r = 3 defective bulbs) = W11(r = 3) = 0.0076 Question 2: Probability of finding three or more defective bulbs in a sample? P(r ≥ 3 defective bulbs ) = 1- W11(r = 0) – W11(r = 1) – W11(r = 2) = 1 – 0.6382 - 0.2925 – 0.0609 = 0.0084

39. Binomial Distribution, Same Problem, Larger r

40. Binomial Distribution

41. Binomial Distribution

42. Binomial distribution with n=10, p=0.5

43. “Wandering Photon” Animation found on the Internet!

44. The “Wandering Photon” Photon Walks straight for a random length Stops with probabilityg Turns in a random direction with probability (1-g)

45. One Dimension

46. x After a random length x with probability gstop with probability (1-g )/2continue in each direction

47. x

48. x

49. x

50. x