One-dimensional array [x1, x2 ... xn], a required value of X, such that:
F(X) = (X-x1)2+(X-x2)2+...(X-xn)2 = min
F(X) = nX2 - 2 * (x1+x2+....+xn) + x12 + x22 + ...+xn2 = min
Derivative of X, when dF / dX = 0, F (X) has a minimum value;
2nX - 2 (x1+x2+....+xn) = 0
Then, X = (x1 + x2 + .... + xn) / n
Thus, in the case of one-dimensional, seeking the least squares parameter X, and averaging the same;
Using a matrix method, first create equations:
X - x1 = 0
...
X - xn = 0
That is, equations:
A n-*. 1 X-B =, equivalent to [1,1, ......] T X-= [X1, X2 ... Xn] T ;
A T AX = A T b
Similarly solve for: X = (x1 + x2 + .... + xn) / n
Application: in one dimension, an array [2,2,2,2,2,10] like this, to find the values therein lone
First find X = Mean = 3.333
Sqrt error = [[(2-3.333) 2 + (2-3.333) 2 + (2-3.333) 2 + (2-3.333) 2 + (2-3.333) 2 + (10-3.333) 2 ] / ( 6-1) ] = 3.1622
If the observed value is an one-dimensional array for a distance, assuming a normal distribution N [[mu], [sigma] 2 ], it should be close to the digital U 2, but in fact is not known, when a large sample size, and usually X error instead of
| X n- - U |> 2σ probability is as l - 95.449974%;
| X n- - U |> 3σ of probability as l - 99.730020%;
So, based on this principle, | 10 - 3.3333 | ≈ 2σ, belongs to a small probability event, I think that 10 is a solitary value;