TF ErrorAna StatInference

Statistical Inference

Frequentist -vs- Bayesian Inference

When it comes to testing a hypothesis, there are two dominant philosophies known as a Frequentist or a Bayesian perspective.

The dominant discussion for this class will be from the Frequentist perspective.

frequentist statistical inference

Statistical inference is made using a null-hypothesis test; that is, ones that answer the question Assuming that the null hypothesis is true, what is the probability of observing a value for the test statistic that is at least as extreme as the value that was actually observed?

The relative frequency of occurrence of an event, in a number of repetitions of the experiment, is a measure of the probability of that event. Thus, if nt is the total number of trials and nx is the number of trials where the event x occurred, the probability P(x) of the event occurring will be approximated by the relative frequency as follows:

[math]P(x) \approx \frac{n_x}{n_t}.[/math]

Bayesian inference.

Statistical inference is made by using evidence or observations to update or to newly infer the probability that a hypothesis may be true. The name "Bayesian" comes from the frequent use of Bayes' theorem in the inference process.

Bayes' theorem relates the conditional probability|conditional and marginal probability|marginal probabilities of events A and B, where B has a non-vanishing probability:

[math]P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}\,\! [/math].

Each term in Bayes' theorem has a conventional name:

• P(A) is the prior probability or marginal probability of A. It is "prior" in the sense that it does not take into account any information about B.
• P(B) is the prior or marginal probability of B, and acts as a normalizing constant.
• P(A|B) is the conditional probability of A, given B. It is also called the posterior probability because it is derived from or depends upon the specified value of B.
• P(B|A) is the conditional probability of B given A.

Bayes' theorem in this form gives a mathematical representation of how the conditional probabability of event A given B is related to the converse conditional probabablity of B given A.

Example

Suppose there is a school having 60% boys and 40% girls as students.

The female students wear trousers or skirts in equal numbers; the boys all wear trousers.

An observer sees a (random) student from a distance; all the observer can see is that this student is wearing trousers.

What is the probability this student is a girl?

The correct answer can be computed using Bayes' theorem.

[math] P(A) \equiv[/math] probability that the student observed is a girl = 0.4

[math]P(B) \equiv[/math] probability that the student observed is wearing trousers = 60+20/100 = 0.8

[math]P(B|A) \equiv[/math] probability the student is wearing trousers given that the student is a girl

[math]P(A|B) \equiv[/math] probability the student is a girl given that the student is wearing trousers

[math]P(B|A) =0.5[/math]

[math]P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.5 \times 0.4}{0.8} = 0.25.[/math]

Method of Maximum Likelihood

The principle of maximum likelihood is the cornerstone of Frequentist based hypothesis testing and may be written as

The best estimate for the mean and standard deviation of the parent population is obtained when the observed set of values are the most likely to occur;ie: the probability of the observing is a maximum.

Least Squares Fit to a Line

Applying the Method of Maximum Likelihood

Our object is to find the best straight line fit for an expected linear relationship between dependent variate [math](y)[/math] and independent variate [math](x)[/math].

If we let [math]y_0(x)[/math] represent the "true" linear relationship between independent variate [math]x[/math] and dependent variate [math]y[/math] such that

[math]y_o(x) = A + B x[/math]

Then the Probability of observing the value [math]y_i[/math] with a standard deviation [math]\sigma_i[/math] is given by

[math]P_i = \frac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{1}{2} \left ( \frac{y_i - y_0(x_i)}{\sigma_i}\right)^2}[/math]

assuming an experiment done with sufficiently high statistics that it may be represented by a Gaussian parent distribution.

If you repeat the experiment [math]N[/math] times then the probability of deducing the values [math]A[/math] and [math]B[/math] from the data can be expressed as the joint probability of finding [math]N[/math] [math]y_i[/math] values for each [math]x_i[/math]

[math]P(A,B) = \Pi \frac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{1}{2} \left ( \frac{y_i - y_0(x_i)}{\sigma_i}\right)^2}[/math]

[math]= \left ( \frac{1}{\sigma \sqrt{2 \pi}}\right )^N e^{- \frac{1}{2} \sum \left ( \frac{y_i - y_0(x_i)}{\sigma_i}\right)^2}[/math] = Max

The maximum probability will result in the best values for [math]A[/math] and [math]B[/math]

This means

[math]\chi^2 = \sum \left ( \frac{y_i - y_0(x_i)}{\sigma_i}\right)^2 = \sum \left ( \frac{y_i - A - B x_i }{\sigma_i}\right)^2[/math] = Min

The min for [math]\chi^2[/math] occurs when the function is a minimum for both parameters A & B : ie

[math]\frac{\partial \chi^2}{\partial A} = \sum \frac{ \partial}{\partial A} \left ( \frac{y_i - A - B x_i }{\sigma_i}\right)^2=0[/math]

[math]\frac{\partial \chi^2}{\partial B} = \sum \frac{ \partial}{\partial B} \left ( \frac{y_i - A - B x_i }{\sigma_i}\right)^2=0[/math]

If [math]\sigma_i = \sigma[/math]

All variances are the same (weighted fits don't make this assumption)

Then

[math]\frac{\partial \chi^2}{\partial A} = \frac{1}{\sigma^2}\sum \frac{ \partial}{\partial A} \left ( y_i - A - B x_i \right)^2=\frac{-2}{\sigma^2}\sum \left ( y_i - A - B x_i \right)=0[/math]

[math]\frac{\partial \chi^2}{\partial B} = \frac{1}{\sigma^2}\sum \frac{ \partial}{\partial B} \left ( y_i - A - B x_i \right)^2=\frac{-2}{\sigma^2}\sum x_i \left ( y_i - A - B x_i \right)=0[/math]

or

[math]\sum \left ( y_i - A - B x_i \right)=0[/math]

[math]\sum x_i \left( y_i - A - B x_i \right)=0[/math]

The above equations represent a set of simultaneous of 2 equations and 2 unknowns which can be solved.

[math]\sum y_i = \sum A + B \sum x_i[/math]

[math]\sum x_i y_i = A \sum x_i + B \sum x_i^2[/math]

[math]\left( \begin{array}{c} \sum y_i \\ \sum x_i y_i \end{array} \right) = \left( \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array} \right)\left( \begin{array}{c} A \\ B \end{array} \right)[/math]

The Method of Determinants

for the matrix problem:

[math]\left( \begin{array}{c} y_1 \\ y_2 \end{array} \right) = \left( \begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array} \right)\left( \begin{array}{c} x_1 \\ x_2 \end{array} \right)[/math]

the above can be written as

[math]y_1 = a_{11} x_1 + a_{12} x_2[/math]

[math]y_2 = a_{21} x_1 + a_{22} x_2[/math]

solving for [math]x_1[/math] assuming [math]y_1[/math] is known

[math]a_{22} (y_1 = a_{11} x_1 + a_{12} x_2)[/math]

[math]-a_{12} (y_2 = a_{21} x_1 + a_{22} x_2)[/math]

[math]\Rightarrow a_{22} y_1 - a_{12} y_2 = (a_{11}a_{22} - a_{12}a_{21}) x_1[/math]

[math]\left| \begin{array}{cc} y_1 & a_{12}\\ y_2 & a_{22} \end{array} \right| = \left| \begin{array}{cc} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right| x_1[/math]

or

[math]x_1 = \frac{\left| \begin{array}{cc} y_1 & a_{12}\\ y_2 & a_{22} \end{array} \right| }{\left| \begin{array}{cc} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right| }[/math] similarly [math]x_2 = \frac{\left| \begin{array}{cc} y_1 & a_{11}\\ y_2 & a_{21} \end{array} \right| }{\left| \begin{array}{cc} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right| }[/math]

Solutions exist as long as

[math]\left| \begin{array}{cc} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right| \ne 0[/math]

Apply the method of determinant for the maximum likelihood problem above

[math]A = \frac{\left| \begin{array}{cc} \sum y_i & \sum x_i\\ \sum x_i y_i & \sum x_i^2 \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}[/math]

[math]B = \frac{\left| \begin{array}{cc} N & \sum y_i\\ \sum x_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}[/math]

If the uncertainty in all the measurements is not the same then we need to insert [math]\sigma_i[/math] back into the system of equations.

[math]A = \frac{\left| \begin{array}{cc} \sum\frac{ y_i}{\sigma_i^2} & \sum\frac{ x_i}{\sigma_i^2}\\ \sum\frac{ x_i y_i}{\sigma_i^2} & \sum\frac{ x_i^2}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|} \;\;\;\; B = \frac{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{ y_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i y_i}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|}[/math]

Uncertainty in the Linear Fit parameters

As always the uncertainty is determined by the Taylor expansion in quadrature such that

[math]\sigma_P^2 = \sum \left [ \sigma_i^2 \left ( \frac{\partial P}{\partial y_i}\right )^2\right ][/math] = error in parameter P: here covariance has been assumed to be zero

By definition of variance

[math]\sigma_i^2 \approx s^2 = \frac{\sum \left( y_i - A - B x_i \right)^2}{N -2}[/math] : there are 2 parameters and N data points which translate to (N-2) degrees of freedom.

The least square fit ( assuming equal [math]\sigma[/math]) has the following solution for the parameters A & B as

[math]A = \frac{\left| \begin{array}{cc} \sum y_i & \sum x_i\\ \sum x_i y_i & \sum x_i^2 \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|} \;\;\;\; B = \frac{\left| \begin{array}{cc} N & \sum y_i\\ \sum x_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}[/math]

uncertainty in A

[math]\frac{\partial A}{\partial y_j} =\frac{\partial}{\partial y_j} \frac{\sum y_i \sum x_i^2 - \sum x_i \sum x_i y_i }{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}[/math]

[math] = \frac{(1) \sum x_i^2 - x_j\sum x_i }{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}[/math] only the [math]y_j[/math] term survives

[math] = D \left ( \sum x_i^2 - x_j\sum x_i \right)[/math]

Let

[math]D \equiv \frac{1}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}=\frac{1}{N\sum x_i^2 - \sum x_i \sum x_i }[/math]

[math]\sigma_A^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial A}{\partial y_j}\right )^2\right ][/math]

[math]= \sum_{j=1}^N \sigma_j^2 \left ( D \left ( \sum x_i^2 - x_j\sum x_i \right) \right )^2[/math]

[math] = \sigma^2 D^2 \sum_{j=1}^N \left ( \sum x_i^2 - x_j\sum x_i \right )^2[/math] : Assume [math]\sigma_i = \sigma[/math]

[math] = \sigma^2 D^2 \sum_{j=1}^N \left ( \sum x_i^2\right )^2 + \left (x_j\sum x_i \right )^2 - 2 \left ( \sum x_i^2 x_j \sum x_i \right )[/math]

[math] = \sigma^2 D^2\sum x_i^2\left [ \sum_{j=1}^N \left ( \sum x_i^2\right ) + \sum_{j=1}^N x_j^2 - 2 \sum x_i \sum_{j=1}^N x_j \right ][/math]

[math] = \sigma^2 D^2\sum x_i^2\left [ N \left ( \sum x_i^2\right ) - 2 \sum x_i \sum_{j=1}^N x_j + \sum_{j=1}^N x_j^2 \right ][/math]

[math] \sum x_i \sum_{j=1}^N x_j \approx \sum_{j=1}^N x_j^2[/math] Both sums are over the number of observations [math]N[/math]

[math] = \sigma^2 D^2\sum x_i^2\left [ N \left ( \sum x_i^2\right ) - 2 \sum_{j=1}^N x_j^2 + \sum_{j=1}^N x_j^2 \right ][/math]

[math] = \sigma^2 D^2\sum x_i^2 \frac{1}{D}[/math]

[math] \sigma_A^2= \sigma^2 \frac{\sum x_i^2 }{N\sum x_i^2 - \left (\sum x_i \right)^2}[/math]

[math] \sigma_A^2= \frac{\sum \left( y_i - A - B x_i \right)^2}{N -2} \frac{\sum x_i^2 }{N\sum x_i^2 - \left (\sum x_i \right)^2}[/math]

If we redefine our origin in the linear plot so the line is centered a x=0 then

[math]\sum{x_i} = 0[/math]

[math]\Rightarrow \frac{\sum x_i^2 }{N\sum x_i^2 - \left (\sum x_i \right)^2} = \frac{\sum x_i^2 }{N\sum x_i^2 } = \frac{1}{N}[/math]

or

[math] \sigma_A^2= \frac{\sum \left( y_i - A - B x_i \right)^2}{N -2} \frac{1}{N} = \frac{\sigma^2}{N}[/math]

Note

The parameter A is the y-intercept so it makes some intuitive sense that the error in the Y -intercept would be dominated by the statistical error in Y

uncertainty in B

[math]B = \frac{\left| \begin{array}{cc} N & \sum y_i\\ \sum x_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|} [/math]

[math]\sigma_B^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial B}{\partial y_j}\right )^2\right ][/math]

[math]\frac{\partial B}{\partial y_j} =\frac{\partial}{\partial y_j} \frac{\left| \begin{array}{cc} N & \sum y_i\\ \sum x_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|}= \frac{\partial}{\partial y_j} D \left ( N\sum x_i y_i -\sum x_i \sum y_i \right )[/math]

[math]= D \left ( N x_j - \sum x_i \right) [/math]

[math]\sigma_B^2 = \sum_{j=1}^N \left [ \sigma_j^2 D^2 \left ( N x_j - \sum x_i \right)^2 \right ][/math]

[math]= \sigma^2 D^2 \sum_{j=1}^N \left [ \left ( N x_j - \sum x_i \right)^2 \right ][/math] assuming [math]\sigma_j = \sigma[/math]

[math]= \sigma^2 D^2 \sum_{j=1}^N \left [ \left ( N^2 x_j^2 - 2N x_j \sum x_i + \sum x_i^2\right) \right ][/math]

[math]= \sigma^2 D^2 \left [ \left ( N^2 \sum_{j=1}^Nx_j^2 - 2N \sum x_i \sum_{j=1}^N x_j + \sum x_i^2 \sum_{j=1}^N \right) \right ][/math]

[math]= \sigma^2 D^2 \left [ \left ( N^2 \sum x_i^2 - 2N \sum x_i \sum_{j=1}^N x_j + N \sum x_i^2 \right) \right ][/math]

[math]= N \sigma^2 D^2 \left [ \left ( N \sum x_i^2 - 2 \sum x_i \sum_{j=1}^N x_j + \sum x_i^2 \right) \right ][/math]

[math]= N \sigma^2 D^2 \left [ \left ( N \sum x_i^2 - \sum x_i^2 \right) \right ][/math]

[math] = N D^2 \sigma^2 \frac{1}{D} = ND \sigma^2[/math]

[math] \sigma_B^2= \frac{N \frac{\sum \left( y_i - A - B x_i \right)^2}{N -2}} {N\sum x_i^2 - \left (\sum x_i \right)^2}[/math]

Linear Fit with error

From above we know that if each independent measurement has a different error [math]\sigma_i[/math] then the fit parameters are given by

[math]A = \frac{\left| \begin{array}{cc} \sum\frac{ y_i}{\sigma_i^2} & \sum\frac{ x_i}{\sigma_i^2}\\ \sum\frac{ x_i y_i}{\sigma_i^2} & \sum\frac{ x_i^2}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|} \;\;\;\; B = \frac{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{ y_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i y_i}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|}[/math]

Weighted Error in A

[math]\sigma_A^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial A}{\partial y_j}\right )^2\right ][/math]

[math]A = \frac{\left| \begin{array}{cc} \sum\frac{ y_i}{\sigma_i^2} & \sum\frac{ x_i}{\sigma_i^2}\\ \sum\frac{ x_i y_i}{\sigma_i^2} & \sum\frac{ x_i^2}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|} [/math]

Let

[math]D = \frac{1}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right| }= \frac{1}{\sum \frac{1}{\sigma_i^2} \sum \frac{x_i^2}{\sigma_i^2} - \sum \frac{x_i}{\sigma_i^2} \sum \frac{x_i}{\sigma_i^2}}[/math]

[math]\frac{\partial A}{\partial y_j} = D\frac{\partial }{\partial y_j} \left [ \sum\frac{ y_i}{\sigma_i^2} \sum\frac{ x_i^2}{\sigma_i^2}- \sum\frac{ x_i}{\sigma_i^2} \sum\frac{ x_i y_i}{\sigma_i^2} \right ][/math]

[math]= \frac{ D}{\sigma_j^2} \sum\frac{ x_i^2}{\sigma_i^2}- \frac{ x_j}{\sigma_j^2}\sum\frac{ x_i}{\sigma_i^2} [/math]

[math]\sigma_A^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial A}{\partial y_j}\right )^2\right ] = \sum_{j=1}^N \left [ \sigma_j^2 D^2 \left ( \frac{ 1}{\sigma_j^2} \sum\frac{ x_i^2}{\sigma_i^2}- \frac{ x_j}{\sigma_j^2}\sum\frac{ x_i}{\sigma_i^2} \right )^2\right ][/math]

[math]= D^2 \sum_{j=1}^N \sigma_j^2 \left [ \frac{ 1}{\sigma_j^4} \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right)^2 - 2 \frac{ x_j}{\sigma_j^4} \sum\frac{ x_i^2}{\sigma_i^2}\sum\frac{ x_i}{\sigma_i^2} + \frac{ x_j^2}{\sigma_j^4} \left (\sum\frac{ x_i}{\sigma_i^2} \right) ^2 \right ][/math]

[math]= D^2 \left [ \sum_{j=1}^N \frac{ 1}{\sigma_j^2} \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right)^2 - 2 \sum_{j=1}^N \frac{ x_j}{\sigma_j^2} \sum\frac{ x_i^2}{\sigma_i^2}\sum\frac{ x_i}{\sigma_i^2} + \sum_{j=1}^N \frac{ x_j^2}{\sigma_j^2} \left (\sum\frac{ x_i}{\sigma_i^2} \right) ^2 \right ][/math]

[math]= D^2 \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right) \left [ \sum_{j=1}^N \frac{ 1}{\sigma_j^2} \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right)^2 - 2 \sum_{j=1}^N \frac{ x_j}{\sigma_j^2} \sum\frac{ x_i}{\sigma_i^2} + \sum_{j=1}^N \frac{ x_j^2}{\sigma_j^2} \left (\sum\frac{ 1}{\sigma_i^2} \right) \right ][/math]

[math]= D^2 \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right) \left [ \sum_{j=1}^N \frac{ 1}{\sigma_j^2} \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right)^2 - \left( \sum \frac{ x_j}{\sigma_j^2} \right)^2 \right ][/math]

[math]= D \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right) = \frac{ \left ( \sum\frac{ x_i^2}{\sigma_i^2}\right) }{\sum \frac{1}{\sigma_i^2} \sum \frac{x_i^2}{\sigma_i^2} - \sum \frac{x_i}{\sigma_i^2} \sum \frac{x_i}{\sigma_i^2}}[/math]

Compare with the unweighted error

[math] \sigma_A^2= \frac{\sum \left( y_i - A - B x_i \right)^2}{N -2} \frac{\sum x_i^2 }{N\sum x_i^2 - \left (\sum x_i \right)^2}[/math]

Weighted Error in B

[math]\sigma_B^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial B}{\partial y_j}\right )^2\right ][/math]

[math]B = \frac{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{ y_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i y_i}{\sigma_i^2} \end{array}\right|}{\left| \begin{array}{cc} \sum \frac{1}{\sigma_i^2} & \sum \frac{x_i}{\sigma_i^2}\\ \sum \frac{x_i}{\sigma_i^2} & \sum \frac{x_i^2}{\sigma_i^2} \end{array}\right|}[/math]

[math]\frac{\partial B}{\partial y_j} = D\frac{\partial }{\partial y_j} \left [\sum \frac{1}{\sigma_i^2}\sum \frac{x_i y_i}{\sigma_i^2} - \sum \frac{ y_i}{\sigma_i^2}\sum \frac{x_i}{\sigma_i^2} \right ][/math]

[math]= \frac{ D}{\sigma_j^2} \sum\frac{ x_i}{\sigma_i^2}- \frac{1}{\sigma_j^2}\sum\frac{ x_i}{\sigma_i^2} [/math]

[math]\sigma_B^2 = \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{\partial B}{\partial y_j}\right )^2\right ][/math]

[math]= \sum_{j=1}^N \left [ \sigma_j^2 \left ( \frac{ D}{\sigma_j^2} \sum\frac{ x_i}{\sigma_i^2}- \frac{1}{\sigma_j^2}\sum\frac{ x_i}{\sigma_i^2} \right )^2\right ][/math]

[math]= D^2 \sum_{j=1}^N \left [ \frac{ 1}{\sigma_j^2} \left (\sum\frac{ x_i}{\sigma_i^2}\right)^2 - 2 \left (\sum\frac{x_i}{\sigma_i^2}\right)\frac{1}{\sigma_j^2}\sum\frac{ x_i}{\sigma_i^2} + \left (\frac{1}{\sigma_j}\sum\frac{ x_i}{\sigma_i^2} \right )^2\right ][/math]

[math]= D^2 \sum_{j=1}^N \frac{ 1}{\sigma_j^2} \left [ \left (\sum\frac{ x_i}{\sigma_i^2}\right)^2 - 2 \left (\sum\frac{x_i}{\sigma_i^2}\right)\sum\frac{ x_i}{\sigma_i^2} + \left (\sum\frac{ x_i}{\sigma_i^2} \right )^2\right ][/math]

[math]= D^2 \sum_{j=1}^N \frac{ 1}{\sigma_j^2} \left [ \left (\sum\frac{ x_i}{\sigma_i^2}\right)^2 - 1 \left (\sum\frac{x_i}{\sigma_i^2}\right)\sum\frac{ x_i}{\sigma_i^2} \right ][/math]

[math]= D \sum_{j=1}^N \frac{ 1}{\sigma_j^2}[/math]

[math]\sigma_B^2 = \frac{ \sum\frac{ 1}{\sigma_i^2}}{\sum \frac{1}{\sigma_i^2} \sum \frac{x_i^2}{\sigma_i^2} - \sum \frac{x_i}{\sigma_i^2} \sum \frac{x_i}{\sigma_i^2}}[/math]

Correlation Probability

Once the Linear Fit has been performed, the next step will be to determine a probability that the Fit is actually describing the data.

The Correlation Probability (R) is one method used to try and determine this probability.

This method evaluates the "slope" parameter to determine if there is a correlation between the dependent and independent variables , x and y.

The liner fit above was done to minimize \chi^2 for the following model

[math]y = A + Bx[/math]

What if we turn this equation around such that

[math]x = A^{\prime} + B^{\prime}y[/math]

If there is no correlation between [math]x[/math] and [math]y[/math] then [math]B^{\prime} =0[/math]

If there is complete correlation between [math]x[/math] and [math]y[/math] then

[math]\Rightarrow[/math]

[math]A = -\frac{A^{\prime}}{B^{\prime}}[/math] and [math]B = \frac{1}{B^{\prime}}[/math]

and [math]BB^{\prime} = 1[/math]

So one can define a metric BB^{\prime} which has the natural range between 0 and 1 such that

[math]R \equiv \sqrt{B B^{\prime}}[/math]

since

[math]B = \frac{\left| \begin{array}{cc} N & \sum y_i\\ \sum x_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum x_i\\ \sum x_i & \sum x_i^2 \end{array}\right|} = \frac{N\sum x_i y_i - \sum y_i \sum x_i }{N\sum x_i^2 - \sum x_i \sum x_i }[/math]

and one can show that

[math]B^{\prime} = \frac{\left| \begin{array}{cc} N & \sum x_i\\ \sum y_i & \sum x_i y_i \end{array}\right|}{\left| \begin{array}{cc} N & \sum y_i\\ \sum y_i & \sum y_i^2 \end{array}\right|} = \frac{N\sum x_i y_i - \sum x_i \sum y_i }{N\sum y_i^2 - \sum y_i \sum y_i }[/math]

Thus

[math]R = \sqrt{ \frac{N\sum x_i y_i - \sum y_i \sum x_i }{N\sum x_i^2 - \sum x_i \sum x_i } \frac{N\sum x_i y_i - \sum x_i \sum y_i }{N\sum y_i^2 - \sum y_i \sum y_i } }[/math]

[math]= \frac{N\sum x_i y_i - \sum y_i \sum x_i }{\sqrt{\left( N\sum x_i^2 - \sum x_i \sum x_i \right ) \left (N\sum y_i^2 - \sum y_i \sum y_i\right) } }[/math]

Note

The correlation coefficient (R) CAN'T be used to indicate the degree of correlation. The probability distribution [math]R[/math] can be derived from a 2-D gaussian but knowledge of the correlation coefficient of the parent population [math](\rho)[/math] is required to evaluate R of the sample distribution.

Instead one assumes a correlation of [math]\rho=0[/math] in the parent distribution and then compares the sample value of [math]R[/math] with what you would get if there were no correlation.

The smaller [math]R[/math] is the more likely that the data are correlated and that the linear fit is correct.

[math]P_R(R,\nu) = \frac{1}{\sqrt{\pi}} \frac{\Gamma\left ( \frac{\nu+1}{2}\right )}{\Gamma \left ( \frac{\nu}{2}\right)} \left( 1-R^2\right)^{\left( \frac{\nu-2}{2}\right)}[/math]

= Probability that any random sample of UNCORRELATED data would yield the correlation coefficient [math]R[/math]

where

[math]\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t} dt[/math]

(ROOT::Math::tgamma(double x) )

[math]\nu=N-2[/math] = number of degrees of freedom = Number of data points - Number of parameters in fit function

Derived in "Pugh and Winslow, The Analysis of Physical Measurement, Addison-Wesley Publishing, 1966."

Least Squares fit to a Polynomial

Let's assume we wish to now fit a polynomial instead of a straight line to the data.

[math]y(x) = \sum_{j=0}^{n} a_j x^{n}=\sum_{j=0}^{n} a_j f_j(x)[/math]

[math]f_j(x) =[/math] a function which does not depend on [math]a_j[/math]

Then the Probability of observing the value [math]y_i[/math] with a standard deviation [math]\sigma_i[/math] is given by

[math]P_i(a_0,a_1, \cdots ,a_n) = \frac{1}{\sigma_i \sqrt{2 \pi}} e^{- \frac{1}{2} \left ( \frac{y_i - \sum_{j=0}^{n} a_j f_j(x_i)}{\sigma_i}\right)^2}[/math]

assuming an experiment done with sufficiently high statistics that it may be represented by a Gaussian parent distribution.

If you repeat the experiment [math]N[/math] times then the probability of deducing the values [math]a_n[/math] from the data can be expressed as the joint probability of finding [math]N[/math] [math]y_i[/math] values for each [math]x_i[/math]

[math]P(a_0,a_1, \cdots ,a_n) = \Pi_i P_i(a_0,a_1, \cdots ,a_n) =\Pi_i \frac{1}{\sigma_i \sqrt{2 \pi}} e^{- \frac{1}{2} \left ( \frac{y_i - \sum_{j=0}^{n} a_j f_j(x_i)}{\sigma_i}\right)^2} \propto e^{- \frac{1}{2}\left [ \sum_i^N \left ( \frac{y_i - \sum_{j=0}^{n} a_j f_j(x_i)}{\sigma_i}\right)^2 \right]}[/math]

Once again the probability is maximized when the numerator of the exponential is a minimum

Let

[math]\chi^2 = \sum_i^N \left ( \frac{y_i - \sum_{j=0}^{n} a_j f_j(x_i)}{\sigma_i}\right )^2[/math]

where [math]N[/math] = number of data points and [math]n[/math] = order of polynomial used to fit the data.

The minimum in [math]\chi^2[/math] is found by setting the partial derivate with respect tot he fit parameters [math]\left (\frac{\partial \chi}{\partial a_k} \right)[/math] to zero

[math]\frac{\partial \chi^2}{\partial a_k} = \frac{\partial}{\partial a_k}\sum_i^N \frac{1}{\sigma_i^2} \left ( y_i - \sum_{j=0}^{n} a_j f_j(x_i)\right )^2[/math]

[math]= \sum_i^N 2 \frac{1}{\sigma_i^2} \left ( y_i - \sum_{j=0}^{n} a_j f_j(x_i)\right ) \frac{\partial \left( - \sum_{j=0}^{n} a_j f_j(x_i) \right)}{\partial a_k}[/math]

[math]= \sum_i^N 2 \frac{1}{\sigma_i^2}\left ( y_i - \sum_{j=0}^{n} a_j f_j(x_i)\right ) \left ( - f_k(x_i) \right)[/math]

[math]= 2 \sum_i^N \frac{1}{\sigma_i^2}\left ( y_i - \sum_{j=0}^{n} a_j f_j(x_i)\right ) \left ( - f_k(x_i) \right)[/math]

[math]= 2\sum_i^N \frac{1}{\sigma_i^2} \left ( - f_k(x_i) \right) \left ( y_i - \sum_{j=0}^{n} a_j f_j(x_i)\right ) =0[/math]

[math]\Rightarrow \sum_i^N y_i \frac{ f_k(x_i)}{\sigma_i^2} = \sum_i^N \frac{ f_k(x_i)}{\sigma_i^2} \sum_{j=0}^{n} a_j f_j(x_i)= \sum_{j=0}^{n} a_j \sum_i^N \frac{ f_k(x_i)}{\sigma_i^2} f_j(x_i)[/math]

You now have a system of [math]n[/math] coupled equations for the parameters [math]a_j[/math] with each equation summing over the [math]N[/math] measurements.

The first equation [math](f_1)[/math] looks like this

[math]\sum_i^N f_1(x_i) \frac{y_i}{\sigma_i^2} = \sum_i^N \frac{f_1(x_i)}{\sigma_i^2} \left ( a_1 f_1(x_i) + a_2 f_2(x_i) + \cdots a_n f_n(x_i)\right )[/math]

You could use the method of determinants as we did to find the parameters [math](a_n)[/math] for a linear fit but it is more convenient to use matrices in a technique referred to as regression analysis

Regression Analysis

The parameters [math]a_j[/math] in the previous section are linear parameters to a general function which may be a polynomial.

The system of equations is composed of [math]n[/math] equations where the [math]k^{\mbox{th}}[/math] equation is given as

[math]\sum_i^N y_i \frac{ f_k(x_i)}{\sigma_i^2} = \sum_{j=0}^{n} a_j \sum_i^N \frac{ f_k(x_i)}{\sigma_i^2} f_j(x_i)[/math]

may be represented in matrix form as

[math]\tilde{\beta} = \tilde{a} \tilde{\alpha}[/math]

where

[math]\beta_k= \sum_i^N y_i \frac{ f_k(x_i)}{\sigma_i^2} [/math]

[math]a_k = a_k [/math]

[math]\alpha_{kj} = \sum_i^N \frac{ f_k(x_i)}{\sigma_i^2} f_j(x_i)[/math]

or in matrix form

[math]\tilde{\beta}= ( \beta_1, \beta_2, \cdots , \beta_n) [/math] = a row matrix of order [math]n[/math]

[math]\tilde{a} =( a_1, a_2, \cdots , a_n)[/math] = a row matrix of the parameters

[math]\tilde{\alpha} = \left ( \begin{matrix} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1j} \\ \alpha_{21} & \alpha_{22}&\cdots&\alpha_{2j} \\ \vdots &\vdots &\ddots &\vdots \\ \alpha_{k1} &\alpha_{k2} &\cdots &\alpha_{kj}\end{matrix} \right )[/math] = a [math]k \times j = n \times n[/math] matrix

the object if to find the parameters [math]a_n[/math]

To find [math]a_n[/math] just invert the matrix

[math]\tilde{\beta} = \tilde{a} \tilde{\alpha}[/math]

[math]\left ( \tilde{\beta} = \tilde{a} \tilde{\alpha} \right) \tilde{\alpha}^{-1}[/math]

[math]\tilde{\beta}\tilde{\alpha}^{-1} = \tilde{a} \tilde{\alpha}\tilde{\alpha}^{-1}[/math]

[math]\tilde{\beta}\tilde{\alpha}^{-1} = \tilde{a} \tilde{1}[/math]

[math] \Rightarrow \tilde{a} =\tilde{\beta}\tilde{\alpha}^{-1} [/math]

Thus if you invert the matrix [math]\tilde{\alpha}[/math] you find [math]\tilde{\alpha}^{-1}[/math] and as a result the parameters [math]a_n[/math].

Matrix inversion

The first thing to note is that for the inverse of a matrix [math](\tilde{A})[/math] to exist its determinant can not be zero

[math]\left |\tilde{A} \right | \ne 0[/math]

The inverse of a matrix [math](\tilde{A}^{-1})[/math] is defined such that

[math]\tilde{A} \tilde{A}^{-1} = \tilde{1}[/math]

If we divide both sides by the matrix [math]\tilde {A}[/math] then we have

[math]\tilde{A}^{-1} = \frac{\tilde{1}}{\tilde{A} }[/math]

The above ratio of the unity matrix to matrix [math]\tilde{A}[/math] is always equal to [math]\tilde{A}^{-1}[/math] as long as both the numerator and denominator are multiplied by the same constant factor.

If we do such operations we can transform the ratio such that the denominator has the unity matrix and then the numerator will have the inverse matrix.

This is the principle of Gauss-Jordan Elimination.

Gauss-Jordan Elimination

If Gauss–Jordan elimination is applied on a square matrix, it can be used to calculate the inverse matrix. This can be done by augmenting the square matrix with the identity matrix of the same dimensions, and through the following matrix operations:

[math]\tilde{A} \tilde{1} \Rightarrow \tilde{A}^{-1} \tilde{ A} \tilde{1} \Rightarrow \tilde{1} \tilde{A}^{-1} . [/math]

If the original square matrix, [math]A[/math], is given by the following expression:

[math] \tilde{A} = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}. [/math]

Then, after augmenting by the identity, the following is obtained:

[math] \tilde{A}\tilde{1} = \begin{bmatrix} 2 & -1 & 0 & 1 & 0 & 0\\ -1 & 2 & -1 & 0 & 1 & 0\\ 0 & -1 & 2 & 0 & 0 & 1 \end{bmatrix}. [/math]

By performing elementary row operations on the [math]\tilde{A}\tilde{1}[/math] matrix until it reaches reduced row echelon form, the following is the final result:

[math] \tilde{1}\tilde{A}^{-1} = \begin{bmatrix} 1 & 0 & 0 & \frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\ 0 & 1 & 0 & \frac{1}{2} & 1 & \frac{1}{2}\\ 0 & 0 & 1 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{bmatrix}. [/math]

The matrix augmentation can now be undone, which gives the following:

[math] \tilde{1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\qquad \tilde{A}^{-1} = \begin{bmatrix} \frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{bmatrix}. [/math]

or

[math] \tilde{A}^{-1} =\frac{1}{4} \begin{bmatrix} 3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 3 \end{bmatrix}=\frac{1}{det(A)} \begin{bmatrix} 3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 3 \end{bmatrix}. [/math]

A matrix is non-singular (meaning that it has an inverse matrix) if and only if the identity matrix can be obtained using only elementary row operations.

Error Matrix

As always the uncertainty is determined by the Taylor expansion in quadrature such that

[math]\sigma_P^2 = \sum \left [ \sigma_i^2 \left ( \frac{\partial P}{\partial y_i}\right )^2\right ][/math] = error in parameter P: here covariance has been assumed to be zero

Where the definition of variance

[math]\sigma_i^2 \approx s^2 = \frac{\sum \left( y_i - \sum_{j=0}^{n} a_j f_j(x_i) \right)^2}{N -n}[/math] : there are [math]n[/math] parameters and [math]N[/math] data points which translate to [math](N-n)[/math] degrees of freedom.

Applying this for the parameter [math]a_k[/math] indicates that

[math]\sigma_{a_k}^2 = \sum \left [ \sigma_i^2 \left ( \frac{\partial a_k}{\partial y_i}\right )^2\right ][/math]

But what if there are covariances?

In that case the following general expression applies

[math]\sigma_{a_k,a_l}^2 = \sum_i^N \left [ \sigma_i^2 \frac{\partial a_k}{\partial y_i}\frac{\partial a_l}{\partial y_i}\right ][/math]

[math]\frac{\partial a_k}{\partial y_i} = \frac{\partial }{\partial y_i} \sum_j^n \beta_j \alpha_{jk}^{-1}[/math]

[math]= \frac{\partial }{\partial y_i} \sum_j^n \left( \sum_i^N f_j(x_i) \frac{y_i}{\sigma_i}\right) \alpha_{jk}^{-1}[/math]

[math]= \sum_j^n \left( f_j(x_i) \frac{1}{\sigma_i}\right) \alpha_{jk}^{-1}[/math] :only one [math]y_i[/math] in the sum over [math]N[/math] survives the derivative

similarly

[math]\frac{\partial a_l}{\partial y_i} = \sum_p^n \left( f_p(x_i) \frac{1}{\sigma_i}\right) \alpha_{pl}^{-1}[/math]

substituting

[math]\sigma_{a_k,a_l}^2 = \sum_i^N \left [ \sigma_i^2 \frac{\partial a_k}{\partial y_i}\frac{\partial a_l}{\partial y_i}\right ][/math]

[math]= \sum_i^N \left [ \sigma_i^2 \sum_j^n \left( f_j(x_i) \frac{1}{\sigma_i}\right) \alpha_{jk}^{-1}\sum_p^n \left( f_p(x_i) \frac{1}{\sigma_i}\right)\alpha_{pl}^{-1} \right ][/math]

notice that [math]\sigma_i[/math] cancels

[math]\sigma_{a_k,a_l}^2 = \sum_i^N \left [ \sum_j^n \left( f_j(x_i) \right) \alpha_{jk}^{-1}\sum_p^n \left( f_p(x_i) \right)\alpha_{pl}^{-1} \right ][/math]

[math]= \sum_j^n \left( \alpha_{jk}^{-1}\sum_p^n \sum_i^N f_j(x_i) f_p(x_i) \right)\alpha_{pl}^{-1}[/math]

[math]= \sum_j^n \left( \alpha_{jk}^{-1}\sum_p^n \alpha_{jp} \right)\alpha_{pl}^{-1}[/math]

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