Two distributions of probability and multiple distributions

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Two distributions of probability and multiple distributions

This chapter uses the two-item distribution and the multi-item distribution formula in probability theory, and gives a brief explanation here.

An event is bound to occur, saying that it is 100% to appear. 100%=1, so the meaning of 100% appears is the probability P= 1.

That is, the probability of occurrence of the inevitable event is 1.

Two item distributions

If you toss a coin, the probability of a positive upward ending is 0.5. The probability of a negative upward outcome is also 0.5. Then the probability of a positive or negative upward event is 0.5+0.5=1, or caricatures.

If a coin is thrown two times, the probability multiplication theorem of the independent event is both positive (negative) and 0.5x0.5=0.25. The first one is that the second is the inverse of the probability of occurrence is also 0.5x0.5=0.25. The first counter-second positive occurrence probability is also 0.5x0.5=0.25. So the probability of a positive and reverse is the first two cases and, that is, 0.25+0.25=2x0.25=0.5. Their total value is still 1. The list is as follows:

Two probability of a positive	The probability of a positive and a reverse	Probability of two tails
0.25	2x0.25=0.5	0.25

　

Notice that in algebra

(a+b) 2=a++ +b2,

While in a=0.5,b=0.5, there is

12= (0.5+0.5) 2=0.25+2x0.5x0.5+0.25=1

This shows that the probability of the occurrence of the various endings of the two-coin toss can be obtained by the square expansion of the two-item formula. In this case, the probability of the occurrence of the various endings of the n -th coin can also be obtained by the expansion of the two-term n -th-square.

For example n= 3 o'clock, there (note 0.5x0.5x0.5=0.125)

13= (0.5+0.5) 3=0.125+3x0.125+3x0.125+0.125=

0.125+0.375+0.375+0.125=1

-Up 4 probability values in 4 items 0.125,0.375,0.375,0.125 correspond to 3 positive, 2 Positive 1 anti,1 positive 2 anti and 3 Reverse, four kinds of endings.

Notice the Newton formula for the expansion of the two-polynomial:

( a + b ) N=an + nan-1b + ... +[ n !/ m ! ( n - m )!] ( a n - m b m ) + b n

put a , Span style= "font-size:medium;" >b equals N The + 1 item, in its general term, represents the n-m Front m the occurrence probability of an event. That is, the probability distribution of this type of problem (such as a coin toss) can be represented by a Newton formula with a two-item expansion. And that's why this type of probability distribution is called the two-item distribution.

if a , Span style= "font-size:medium;" >b is not equal to 0.5 , so long as a The probability of the occurrence of the event is P substituting, b The occurrence probability of the event is (1- p ) substituting, the formula is still correct, ( a + b still =1 ).

so for a , Span style= "font-size:medium;" >b Two endings of random events, if P , b event probability is 1- P , then a event appears n-m  M The probability of the occurrence of the time (corresponding to a composite event) P should be ( here p is uppercase )

P =[n!/m! ( N-m)!] [pn-m(1-p)m]

Notice the symmetry of the above formula, which can also be written as

P =[n!/m! ( N-m)!] [pm(1-p)n-m]

It is the probability formula for the occurrence of random events of the so-called two-item distribution, and also the general term of Newton's two-term expansion in the case of the variable as the corresponding probability value. It is the origin of this chapter's formula (11.3) .

In addition, when p=0.5, it is obvious that [pm(1-p)N- m] always equals 1/(2)n, notes [P+ (1-P)]n= 1, so the two-item formula expands N the total value of each coefficient of the + 1 item should be equal to 2n. that

There is no pin the equation, so this coefficient and formula are independent of P 's specific values. In general probability books, there are more than two distributions.

Multi-item Distribution

By generalizing the two-item distribution formula, we get a multi-item distribution (it is seldom introduced in the general probability book, but it is involved in thermodynamics).

A random experiment if there is K A possible ending a1 , a2 , ... , ak , their probability distributions are p1 , p2 , , , then in the total results of N Secondary sampling, appears N1 times, a2 appears Span style= "font-size:medium;" > N2 times, , ak appears Span style= "font-size:medium;" > nk times the probability of the occurrence of this event P has the following formula:

This is the probability formula for multiple distributions. It is obvious that it is a polynomial distribution because it is a special type of polynomial expansion.

We know that in algebra, when k variables and the n -th-square expansion (p1+ p2+...+ pk )n is a polynomial whose general term is the value given by the preceding formula. If this k - variable happens to be the probability of the occurrence of various outcomes, then the probability of the total value of these probabilities corresponds to an inevitable event. and the probability of an inevitable event equals 1, so the above polynomial becomes

(p1+ p2+...+ PK ) n =1n=1

That is, the value of the polynomial at this time is equal to 1.

because ( p1 +  p2 +...+  pk  ) N value equals 1 . We also think that it represents an inevitable event carried out by n  Secondary sampling probability ( =1 , inevitable event). And when this polynomial can be expanded into many items, the aggregate value of these items equals 1 prompts us that these items are incompatible events ( n the corresponding probability of the sampling. That is, the polynomial expansion of each item is a special event occurrence probability. So we put the expansion of the pass as a1 appears N1 times, a2 appears N2 times, ... , ak appears nk Probability of occurrence of such events. This gives you the previous formula.

If the probability of the occurrence of individual events P1,p2,...,PK are equal, that is p1=p2=...=pk=p(Note that this is the lowercase p ),

Notice the p1+p2+...+pk =1, get p1= p2 =...=pk =p=1/k .

By substituting this value for the expansion of the polynomial, the sum of the individual items of the expansion satisfies the following formula:

∑[ n!/(n1! ) N2!... NK!)] (1/k) N=1

i.e. ∑[ n!/(n1! ) N2!... NK!)] =kN

The sum above all the possible positive integer values of each ni , but requires that the aggregate value of each ni equals N . That

N1 +n2+... nk=N

In thermodynamics, when discussing the possible number of microscopic states of matter, we often use additional ideas to elicit n!/(n1!). N2!... NK!) Expression and called it a thermodynamic probability. It is a much larger number than astronomical, and it is not appropriate to call it probability. But in thermodynamics the probability of the occurrence of each microscopic state is equal, which corresponds to the p1= p2 =...=pk =p=1/k, which we discussed earlier, so

[N!/(n1! N2!... NK!)] (1/KN)

It really has the meaning of mathematical probability. In other words, the thermodynamic odds in physics [n!/(n1! N2!... NK!)] Multiplying (1/KN) is the probability of being defined in mathematics (having a normalized nature).

Two-item distributions and distributions of probabilities