Given a number of measured values x₁, x₂, ... of a quantity X we can construct the frequency distribution by subdividing the range of X in intervals (bins) and counting the number of x_k that fall within each bin,

The frequency distribution if f_i=N_i / N and it satisfies

When the bins are very small we have a continuous distribution f(x), which satisfies ∫ f(x) dx=1.

Given a distribution we can compute several statistical estimators of the "true" value. The mean is X_mean=∑ x_i f_i. The mode is the value of x for which f(x) is maximum. It makes sense for unimodal distributions, ie, distribution with a single maximum. The median is the value of x such that X has equal probability to fall before or after it. We usually take the mean as the best estimate of the true value,

It is also important to have an estimate of the accuracy of the apparatus. The mean absolute deviation is S_abs=∫ |x-X_mean| f(x) dx. The mean square deviation is S_sqr=∫ (x-X_mean)²f(x) dx The standard deviation is its square root, S = S_sqr^1/2. The estimate of the presision of the measurement is not quite the standard deviation, because it must be compensated for number of independent values, thus there is N-1 instead of N,

The accuracy of the estimate of the "true" value is the standard error s(X_N) = S/√(N), or the adjusted standard error s'(X_N) = S_N/√(N).

The accuracy of the standard deviation gives the numer of significant figures. The relative accuracy is usually s(S_N)/S_N = (N-2)^-1/2.

Given Z=F(X), the statistical estimators of Z can be computed from those of X. In general

For example for a linear varaiable Z = a + b X + c Y, we have

The Least Squares is the procedure of minimization of the square errors, E = ∑ (x_k - X)². It leads to the mean, X_N, as value for X. It can be used to find the best straight line that interpolates a set of combined (X,Y) measures. Supposing that the variables are linearly related, Y = a X + b, the coefficients turn out

If the errors in x_k are negligeable, and the errors in a and b arise only from errors in y_k we have

where S(Y)=[Y²] -[Y]² - ([XY]-[X][Y])²/([X² - [X]²) The adjusted error have N-2 instead of N.

When we combine the estimates of two different experiments, X'+S' and X"+S", the means are weighted with the inverse of the square errors,

The distribution of random errors is binomial. The porbability of having an error E=m e out of n errors of equal amplitude e is

As m and n go to infinity (while E stays finite) this probability tends to exp(-E²/[2 S²] )/√(2 π S²). Therefore the probability of obtaining an error smaller than E is ψ(E/S) where ψ(y)=√(2/π) ∫₀^y exp(-y²/2) dy. The followin table contains some values of ψ.

   y   ψ(y)
 0.0   0.399
 0.5   0.352
 1.0   0.242
 1.5   0.129
 2.0   0.054
 2.5   0.017
 3.0   0.004

If the probability of having a certain event is p, the probability of having r events in n trials is the Poisson distribution (neglecting multiple occurrences of the event in a single trial)

Letting u=np fixed while n tends to infinity, we have P(r,u)=e^-u u^r/r! This has mode {u}, the greatest integer less than or equal to u, and mean u. The standard deviation is √(u). When n is large the Poisson distribution becomes a gaussian with mean np and standard deviation √(np).

Likelyhood and confidence

Is a process has distribution f(x) than the likelyhood of a set of results x_k is

When f(x) is gaussian the maximum likelyhood is obtained for X = (1/N) ∑ x_k, and S² = (1/N) ∑ (x_k - X)². If f(x) is Poisson u=(1/N) ∑ r_k and the standard deviation should be √(u): this is a check for the assumption of Poisson distribution.

The confidence is the probability that x lies in a certain interval. For the gaussian the confidence of the 1-sigma interval [X-S,X+S] is 0.683. That of the 3-sigma interval is 0.997 (0.99 confidence is achieved at 2.58 sigma). The central limit theorem assures that the mean X_N of N measures follows a gaussian distribution with standard deviation the standard error S_N.

We can approach the combination of two different measurement sets (assumed both gaussian with same mean X but different standard deviations) using the principle of maximum likelyhood. It results a cubic equation for X

where S'²=S_N² + (X'_N-X)² and a similar formula for S". This equation could be solved in general, but it is already useful in special limiting cases where simplifying assumptions can be made.

x is worse than x' if f(x) < f(x'). The backing of x is the probability of getting a value worse than X. Therefore B(x) = ∫ f(x) dx where the integral is extended to the values that are worse than x. If the backing of x is 1 than x is the mode. The backing of x tends to 0 as x tends to infinity. For a gaussian distribution B(x)=ψ([x-X]/S).

Given two measures x' and x" (each with its own gaussian distribution) we can define when a pair (x',x") is worse than another, and define the backing B(x',x") which turns out 1/(2 π) ∫ exp( - [x'²+x"²]/2 ) where the integral is over the region outside the circle of radius X. This generalizes to 2n measurements,

where G is the Gamma function and the integral ranges from X to infinity.

Now suppose to test a distribution and fix x₀ ... x_m with probability p_k of having x in the interval [x_k, x_k+1). If the outcome of a measure gives n_k in this intervals, and the total number of N is large enough that each Np_k>10, so that each n_k can be thought to belong to a gaussian distribution (Np_k, √(Np_k)), letting the chi-square

the backing is P_m-1(>X²). Here there is m-1 because there is one constraint among the x. Any other constraint that might come from particular situation of the measurements reduces the index of P by 1.

The chi-square test is a check to decide whether a set of measurements supports a certain distribution. It gives the confidence level for it. The following table has some values of P_n(>X²).

n      1.0   3.0   5.0  10.0 
X2
 1.0  0.32  0.80  0.96  1.00  
 5.0  0.02  0.17  0.41  0.89
10.0  0.00  0.02  0.07  0.44
15.0        0.00  0.01  0.13
20.0              0.00  0.03