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D - 1 / 2 = 1 / λ 1 1 / λ 2 .
( x - x 0 ^ ) T Γ ^ - 1 ( x - x 0 ^ ) = ( x - x 0 ^ ) T U D - 1 U T ( x - x 0 ^ ) = D - 1 / 2 U T ( x - x 0 ^ ) 2 .

So we can set up a change of variable:

w = f ( x ) = W ( x - x 0 ^ ) , W = D - 1 / 2 U T .

Notice that

d w = d e t ( W ) d x = 1 λ 1 λ 2 d x = 1 det ( Γ ^ ) 1 / 2 d x .

This change of variable transforms the equiprobability ellipses of x into circles centered at the origin:

f ( B α ) = w R 2 | w < δ , δ = δ ( α ) > 0 .

We are now ready to evaluate the integral. We will first perform a change of variable from x to w , then change to polar coordinates.

p = B α π x d x = 1 2 π det Γ ^ 1 / 2 B α exp - 1 2 ( x - x 0 ^ ) T Γ ^ - 1 ( x - x 0 ^ ) d x = 1 2 π det Γ ^ 1 / 2 B α exp - 1 2 W ( x - x 0 ^ ) 2 d x = 1 2 π f ( B α ) exp - 1 2 w 2 d w = 0 δ exp - 1 2 r 2 r d r = 1 - exp - 1 2 δ 2 .

Given this, we can solve for δ :

δ = δ ( p ) = 2 log 1 1 - p .

This gives us a straightforward way to check whether or not a point in the sample, x j , lies in the credibility region. If

w j < δ ( p ) , w j = W ( x j - x 0 ^ )

holds, then the point is in the credibility ellipse.

Recall that this expression is for two dimensions. If we repeat the integral in different dimensions, we get a different expression relating δ and p . We are also interested in distributions in R , R 14 , and R 16 , so we repeat the calculations for those spaces. In R 14 and R 16 , we do not get a nice expression, so we approximate the values of δ for p = { 0 . 1 , 0 . 2 , , 0 . 9 } using a TI-89 calculator (MATLAB was unable to carry out the calculation symbolically). In R , using w = σ - 1 , we find that

δ = 2 erf - 1 p .

We can then plot the number of points inside a credibility ellipse against the credibility. A perfectly normal distribution will be a straight line from the origin to ( 1 , 1 ) . If a sample significantly deviates from this line, then this criterion suggests that the sample is not normal.

Distributions

Armed with these criteria for looking at the normality of a sample, we study the displacements of the nodes. First we look at the nodes individually. We look at how a single node moves under a particular load. We expect the sample to be normally distributed: under a fixed load, most of the displacements should be about the same, with a few outliers. Data Set A has 104 samples split evenly between two loads, so we have 14 samples in R 2 with 52 points each. [link] plots the number of points inside the credibility ellipse against the credibility for Node 4, Pattern 2. The red line is from the sample, the blue line is a randomly generated normally distributed sample with 52 points, and the black line is a reference line. The blue allows us to have an idea for how much we can expect a sample this size to deviate from the straight line. [link] is a histogram of the positions of the node. This plot suggests that the distribution of this node is normally distributed. All of the nodes, under both loads, give similar results (See [link] ). Because of this, we believe that the displacements are normally distributed, as we expected.

Node 4, Load 2
Normality of Individual Nodes [Load Pattern:Node]

Next we look at the nodes collectively, giving us two samples in R 14 with 52 points each. Because we believe that each node is normally distributed, we expect the combined displacements to be normally distributed. [link] is the plot from Load 1; [link] is the plot from Load 2. [link] suggests that the displacements of the nodes are indeed normally distributed.

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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