1. Section 1.3: A General Discussion of Mean Values

2. P(u1), P(u2), P(u3),….., P(uM-1), P(uM)
The Binomial Distribution is only one example of a discrete probability distribution.
Now, we’ll have a brief discussion of a
General Distribution.
Most of the following is valid for ANY discrete probability 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)

3. ū ≡ <u> ≡ (S2/S1) u1,u2,u3,…,uM-1,uM with probabilities
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 the ratio of two sums:
ū ≡ <u> ≡ (S2/S1)
Here, S1 ≡ ∑iP(ui) or
S1 ≡ P(u1) + P(u2) + P(u3) +…+ P(uM-1) + P(uM)
For a properly normalized distribution, we must have: S1 = ∑iP(ui) = 1.
We’ll assume this from now on.

4. Note: μ ≡ Standard notation for the
ū ≡ <u> ≡ (S2/S1)
As just stated, S1 = 1, so
ū ≡ <u> ≡ S2
Here, S2 ≡ ∑iuiP(ui) or
S2 ≡ u1P(u1) + u2P(u2) + u3P(u3) +
…+ uM-1P(uM-1) + uMP(uM)
Note: μ ≡ Standard notation for the
mean value.
So, μ ≡ ū ≡ <u>

5. 1. The Mean Value, μ ≡ ū ≡ <u>
Sometimes, ū is called the 1st moment of P(u).
If F(u) is any function of u, the mean value of F(u) is:
<F> ≡ ∑iF(ui)P(ui)
Some simple mean values that are useful for describing any probability distribution P(u):
1. The Mean Value, μ ≡ ū ≡ <u>
This is a measure of the central value of u about which the various values of ui are distributed.
Consider the quantity Δu ≡ 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!

6. <(Δu)2> ≥ 0, or <u2> ≥ (<u>)2
Now, look at (Δu)2 = (u - <u>)2 (square of the deviation from the mean). It’s mean value is:
σ2  <(Δu)2> = <(u - <u>)2> = <u2 -2uū – (ū)2>
σ2 = <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 Dispersion or Variance or the 2nd Moment of P(u) about the mean.
σ2 ≡ <(Δu)2> is a measurelof the spread of the u values about
the mean ū. Note that <(Δu)2> = 0 if & only if ui = ū for all i.
It can easily be shown that,
<(Δu)2> ≥ 0, or <u2> ≥ (<u>)2
Note: σ2 ≡ Standard notation for the variance.

7. <(Δu)n> ≡ <(u - <u>)n>
Could also define the nth moment of P(u) about the mean:
<(Δu)n> ≡ <(u - <u>)n>
In Physics this is rarely used beyond n = 2 & almost never beyond n = 3 or 4.
NOTE: From math: 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 determine P(u).
Math Theorem
In order to uniquely determine a distribution P(u), we need to know ALL moments of it. That is we need all moments for
n = 0,1,2,3….  .

8. Section 1.4 Mean Values for the Random Walk Problem
Also we’ll discuss a few math “tricks” for doing discrete sums! We’ve found that: For N steps, the probability of making n1 steps 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.

9. ∑(n1 = 0N) WN(n1) = 1? ∑(n1 = 0N) WN(n1) = 1.
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, so
∑(n1 = 0N) WN(n1) = 1.

10. = ∑(n1 = 0N) n1[N!/(n1!(N-n1)!] pn1qN-n1 (1)
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)
Do this sum by looking it up in a table OR we 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, treat p & q as “arbitrary”, continuous variables, ignoring that p + q =1.
NOTE, if p is a continuous variable, then clearly:
n1pn1 ≡ p[(pn1)/p]

11. n1pn1 ≡ p[(pn1)/p] μ = <n1> = Np
<n1> ≡ ∑(n1 = 0N) n1WN(n1)
= ∑(n1 = 0N) n1[N!/(n1!(N-n1)!] pn1qN-n1 (1)
Temporarily, treat p & q as “arbitrary”, continuous variables. If p is a continuous variable, then:
n1pn1 ≡ p[(pn1)/p]
Use this in (1) (interchanging sum & 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
In our special case (p + q) = 1, (p + q)N-1 = 1, so
μ = <n1> = Np

12. The mean number of steps to <n1> + <n2> = N(p + q) = N
Summary: For the Binomial Distribution, 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!

13. What is the mean displacement, <x> = <m>ℓ?
Question 2:
What is the mean displacement,
<x> = <m>ℓ?
Clearly, m = n1 – n2,
so <m> = <n1> - <n2> = N( p – q)
& <x> = <m> ℓ = N( p – q)ℓ
If p = q = ½, <m> = 0 so,
<x> = <m>ℓ = 0

14. σ2 = <(Δn1)2> = <(n1)2> - (<n1>)2
Question 3:
What is the dispersion (variance)?
σ2 = <(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:
σ2 = <(Δn1)2> = <(n1)2> - (<n1>)2

15. σ2 = <(Δn1)2> = <(n1)2> - (<n1>)2
We’ve just seen that the mean
 = <n1> = Np
So, we first need to calculate <(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

16. <(n1)2> = (Np)2 + Npq = (<n1>)2 + Npq
After math (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
σ is the dispersion or variance of the binomial distribution. The root mean square (rms) deviation from the mean is defined in general as:
 [<(Δn1)2>]½.
For the binomial distribution, this is
σ = [Npq]½
Note that σ  The Distribution Width
σ2 = <(Δn1)2> = Npq

17. Summary: For the Binomial Distribution
Dispersion or variance in number of steps to right:
Root mean square (rms) deviation from the mean:
σ  [<(Δn1)2>]½ = [Npq]½
and σ  Distribution Width
Again note that: <n1> = Np. So, the relative width of the distribution is:
(σ/μ) = [Npq]½(Np) = (q½)(pN)½
If p = q, this is: (σ/μ) = 1(N)½ = (N)-½
 As N increases, the mean value increases
 N but the relative width decreases  (N)-½
σ2 = <(Δn1)2> = Npq

18. Question 4: <x2> = <(Δm)2>ℓ2 = <(m - <m>)2>ℓ2
What is the dispersion
<x2> = <(Δm)2>ℓ2 = <(m - <m>)2>ℓ2
in the net displacement?
Or, 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: l
If p = q = ½, <(Δm)2> = N
<(Δm)2> = 4Npq

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