# Probability Adding and Multiplying Probabilities Permutations

## Transcription

Probability Adding and Multiplying Probabilities Permutations

Probability Adding and Multiplying Probabilities Permutations There is only 1 one-letter word containing the letter A: A There are 2 two-letter words containing the letters AB: AB BA There are 6 three-letter words containing the letters AB: ABC BAC CAB ACB BCA CBA There are 24 different four-letter words containing the four letters ABCD: 1 ABCD BACD CABD DABC ABDC BADC CADB DACB ACBD BCAD CBAD DBAC ACDB BCDA CBDA DBCA ADBC BDAC CDAB DCAB ADCB BDCA CDBA DCBA For N -letter words, there are N ! possible permutations: N choices for the first letter, N − 1 for the second, and so on: N ! = N · (N − 1) · (N − 2) · · · The number of two-letter words that can be made with the four letters in ABCD is written 4 P2 , which is 12: ABCD BACD CABD DABC ABDC BADC CADB DACB In general N Pk N P2 ACBD BCAD CBAD DBAC ACDB BCDA CBDA DBCA ADBC BDAC CDAB DCAB ADCB BDCA CDBA DCBA → → → → AB BA CA DA AC BC CB DB AD BD CD DC = N · (N − 1), and = N · (N − 1) · (N − 2) · · · (N − k + 1) N · (N − 1) · (N − 2) · · · (N − k + 1) · (N − k) · (N − k − 1) · · · (1) = (N − k) · (N − k − 1) · · · (1) N! = (N − k)! 2 If order is not important, we are looking for combinations rather than permutations: N Ck = N Pk k Pk = N! (N − k)!k! This is the same as dividing N items into two groups, with N1 items in the first group, and N2 in the second (with, of course, N1 + N2 = N .) Again with order irrelevant: N CN1 = N! N! = (N − N1 )!N1 ! N1 !N2 ! If we have 5 objects, for example, that we want to divide into groups of 3 and 2, we have 5 C3 = 5!/3!2! = 10 combinations : abc:de ade:bc abd:ce bcd:ae abe:cd bce:ad Multinomial coefficients (x + y)1 = x+y 1! 1! = x+ y 1!0! 0!1! 3 acd:be bde:ac ace:bd cde:ab Note that 5 C3 = 5 C2 . = 1 C1 x + 1 C0 y (x + y)2 = x2 + 2xy + y 2 2! 2 2! 2! 2 = x + xy + y 2!0! 1!1! 0!2! = 2 C2 x2 + 2 C1 xy + [2 C0 y 2 (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 3! 3 3! 2 3! 3! 3 = x + x y+ xy 2 + y 3!0! 2!1! 1!2! 0!3! = 3 C3 x2 + 3 C1 x2 y + 3 C2 xy 2 + 3 C0 y 2 In general, we can write (x + y)N = N X N CN1 x N1 N −N1 y N1 =0 Why does this work? (x + y)6 = x6 x5 y x4 y 2 (x + y)·(x + y)·(x + y)·(x + y)·(x + y)·(x + y) a · b · c · d · e · f abcdef: abcde:f, abcdf:e, abcef:d, etc. (6 C5 = 6 terms) abcd:ef, abce:df, abcf:de, etc. (6 C4 = 15 terms) We extend this to the task of dividing N items into three groups, with N1 , N2 , and N3 elements, respectively. We already know that there are N CN1 ways to form the first group; then, we have N − N1 elements left to apportion. There are thus N −N1 CN2 possible second groups, which, once formed, uniquely determine the remaining N3 (= N −N1 −N2 ) elements in the third group. The total number of possible combinations is N CN1 ·N −N1 CN2 = N! (N − N1 )! N! · = (N − N1 )!N1 ! (N − N1 − N2 )!N2 ! N1 !N2 !N3 ! For r groups, we have N! N1 !N2 !N3 ! · · · Nr ! which turn out to be the coefficients in a multinomial expansion: !∗ N N N X X X N! N r (x1 + x2 + · · · + xr ) = ··· xN1 xN2 · · · xN r N1 !N2 !N3 ! · · · Nr ! 1 2 N1 =0 N2 =0 Nr =0 where the asterisk calls out the requirement that N1 + N2 + · · · + Nr = N . 4 Stirling’s approximation Stirling’s approximation for ln N !: ln N ! = ln [N · (N − 1) · (N − 2) · · ·] = ln N + ln(N − 1) + ln(N − 2) + · · · + ln 1 = N X ln x 1 Z ≈ N ln xdx 1 N ≈ (x ln x − x)|1 ≈ (N ln N − N ) − (1 ln 1 − 1) ≈ N ln N − N + 1 ≈ N ln N − N This approximation, ln N ! = N ln N − N , is quite poor for small values of N , but becomes increasingly accurate as N increases: Means and Moments The sum of all possible probabilities is always equal to 1: X pj = 1 j Or, it is certain that some event is going to occur. The mean value of a measurement is given by the sum of all of the measurements, divided by the total number of measurements. 5 So Aavg X Nj = j X = N Aj p j Aj j That is, Aavg = X pj Aj = p19 · 19 + p20 · 20 + . . . j We rename the average age the expectation value of A, denoted hAi. If, for some reason, we wanted to tally up the average value of the square of A, we would denote this hA2 i, and computing it is straightforward: hA2 i = p19 · 192 + p20 · 202 + . . . 6 In general, if we have any function of j, we can compute its expectation value using X hf (j)i = f (j)pj j It is alway true that: hAi2 ≤ hA2 i The two are equal only when all values of A are the same. Any difference between hxi2 and hx2 i gives a standard deviation: D E 2 σx2 = (x − hxi) X 2 (xj − hxi) pj = j = X = X x2j − 2hxixj + hxi2 pj j j x2j pj − 2 X j 7 hxixj pj + X j hxi2 pj = X 2 x − 2hxi2 + hxi2 pj = 2 x − 2hxi2 + hxi2 2 x − hxi2 j = and the standard deviation is p p σx = σx2 = hx2 i − hxi2 With continuous variables Prob(x, x + dx) = ρ(x)dx and we can extend to definite (finite-length) intervals by integrating b Z Pab = ρ(x)dx a We have Z +∞ ρ(x)dx = 1 −∞ and +∞ Z hxi = hx2 i = xρ(x)dx −∞ Z +∞ x2 ρ(x)dx −∞ Z +∞ hf (x)i = f (x)ρ(x)dx −∞ σx = p hx2 i − hxi2 8