Calculating probability and the partition function

The probability of a certain property is defined to be the fraction of members of the ensemble having that property

$$P_i~=~\frac{n_i}{n}$$ (1)

where i is the property of interest, n_i is the number of members of the ensemble having property i, and n is the total number of members of the ensemble.

Properties i and j are said to be mutually exclusive if no member of the ensemble can have both properties i and j (for example a woman couldn't be 25 and 40 years old at the same time). In this case we would have

$$P_{i~\mbox{\small {or}}~ j}~=~\frac{n_{i~\mbox{\small {or}}~j}}{n}~=~\frac{n_{i}~+~n_{j}}{n}~=~P_i~+~P_j~~.$$ (2)

A simple extension of the above equation is

$$P_0~+~P_1~+~P_2~+...=\sum_i P_i~=\sum_i~\frac{n_{i}}{n}~=~\frac{n}{n}~=~1~~.$$ (3)

The meaning of the above equation is that $\sum_i P_i$ is the probability that the member has a value for the property (e.g. age) and that they all should have some value, the sum must yield the probability of a certainty, which is unity [3]. A probability that satisfies the above equation is said to be normalized, and this equation is often called the normalization condition. Often the physical situation determines only that P_i is proportional to some function of i [2]

P_i = af(i)

where a is the normalization constant, and

$$P~=~\sum_i P_i~=~a \sum_i f(i)~=~1 ~~.$$

Thus one has

$$a~=~\frac{1}{\sum_i f(i)}$$

and

$$P_i~=~\frac{f(i)}{\sum_i f_i}~~.$$ (4)

Example (1):

Suppose each system is a rolled die, and i indicates the number of dots showing upward. Suppose f(i) is the number of dots showing upward. Since the die is correctly balanced,

P₁ = P₂ = P₃ = P₄ = P₅ = P₆ = 1/6   .

We could calculate the function f_i from equation (4),

$$f(i)~=~\sum_i P_i f_i~=~\frac{1}{6}(1)~+~\frac{1}{6}(2)~+~\fr... ...frac{1}{6}(4)~+~\frac{1}{6}(5)~+~\frac{1}{6}(6)~=~3\frac{1}{2}$$

and this means that if one rolles a million dies, one would get a total score of about $3\frac{1}{2}$ million [2].

Actually, one could see from the above example that the correct way of writing equation (4) is

$$\overline{f(i)}~=~\overline{f}~=~\sum_i P_i f_i~~.$$ (5)

A generalized form of equation (5) is

$$\overline{f_{i,j,k,...}}~=~\overline{f}~=~\sum_{i,j,k,...} P_{i,j,k,...} f_{i,j,k,...}~~.$$ (6)

The function $\sum_i f_i$ or its generalized form $\sum_{i,j,k,...} f_{i,j,k,...}$ is called the partition function in statistical mechanics.

Example(2):

Consider an ensemble consisting of 10 cats with ages 1,2,2,3,3,3,4,4,5,6; thus the probabilities generated are

$$P_1~=~\frac{f(1)}{\sum_i f(1)}~=~0.1,~P_2~=~0.2,~P_3~=~0.3,~P_4~=~0.2,~P_5~=~0.1,~P_6~=~0.1~~.$$

The average age could be found by

$$\overline{f}~={\sum P_i f_i}~=\frac{\sum n_i f_i}{n}~=~0.1+0.4+0.9+0.8+0.5+0.6=\frac{33}{10}=3.3 ~~.$$

The average of the square of the age is

$$\overline{f^2}=\sum_i P_i f_i^2=(0.1)(1^2)+(0.2)(2^2)+(0.3)(3^2)+(0.2)(4^2)+(0.1)(5^2)+(0.1)(6^2)=12.9~~.$$

It is evident that the average of the squares is larger than the square of the average, which would be in this case 3.3²=10.89.

The deviation is defined as

$$\delta f_i~:=~f_i~-~\bar{f}~~.$$ (7)

The ensemble average of the deviation is zero

$$\delta f_i=\sum_i P_i(f_i~-~\bar{f}) ~=~\sum_i P_i f_i~-~\bar f \sum_i P_i~=~\bar f~-~\bar f~=~0 ~~.$$

To obtain a useful measure of the average deviation of f from $\bar f$,the signs of positive and negative deviations must somehow be omitted to ensure that they will not cancel. One could average the absolute value of the deviation $\vert\delta f\vert$ to find the so called mean deviation. It is however more common to use the average of the square of the deviation:

$$\overline {\delta f^2} \sum_i P_i (f_i~-~\bar f)^2=\sum_i P_i f_i^2-2\sum_i P_i f_i \bar f+\sum_i P_i \bar f^2=\overline{f^2}~-~2\bar f^2~+~\bar f^2$$ =

$$\overline {f^2}~-~\bar f^2~~.$$ = (8)

This quantity is called the mean square deviation or the variance.The left hand side of the above equation is positive so the right hand side should also be positive and that $\overline{f^2}$ is always bigger than $\bar f^2$ as we have seen in example number two.

The square root of $\overline{f^2}$ is called the root mean square deviation, or the standard deviation $\sigma$:

$$\sigma~=~\sqrt{\overline{\delta f^2}}~=~\sqrt{\overline{f^2}-\bar f^2}$$ (9)

and a good measure of how the values of f for the members of the ensemble are spread out around$\bar f$ is to examine the ratio ${\sigma}/{\bar f}$, a dimensionless measure that is very small for tightly clustered f and becomes larger as the values of f are less clustered around the mean.

Further measures for the distribution spread are the moments about the mean $\mu_n$ defined as

$$\mu _n~=~\sum_i P_i~ (\delta f_i)^n~~.$$ (10)

We know already that $\mu_0$ is 1, $\mu_1$ is 0, and $\mu_2$ is the variance.[3]