Spline Function/Examples/Cubic Spline

Example of Spline Function

 * How to find the coefficients $a$, $b$, $c$, and $d$ in the cubic spline function $ax^3 + bx^2 + cx + d$

Assume that an auto insurance company pays out losses incurred in a given year according to the table below:


 * $\begin{array} {|r|r|}

\text {Time in Months } & \text {Percent Paid } \\ \hline 0 & 0 \\ 12 & 40 \\ 24 & 70 \\ 36 & 90 \\ 48 & 100 \\ \end{array} $

We wish to create a model that will interpolate the amount of loss paid at any point in time between $0$ and $48$ months.

One potential model we could employ is the Cubic Spline.

For our cubic spline, we will need to solve for $4$ coefficients for $4$ separate cubic equations.

In total, we are solving for $16$ variables.

We proceed as follows:

We now have $16$ equations and $16$ unknowns.

By Simultaneous Linear Equations has Unique Solution iff Rank of Matrix of Coefficients equals Number of Columns, we have a unique solution.


 * $\mathbf A \mathbf x = \begin{bmatrix}

0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 12^3 & 12^2 & 12 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 12^3 & 12^2 & 12 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 24^3 & 24^2 & 24 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 24^3 & 24^2 & 24 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 36^3 & 36^2 & 36 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 36^3 & 36^2 & 36 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 48^3 & 48^2 & 48 & 1 \\ 432 & 24 & 1 & 0 & -432 & -24 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1728 & 48 & 1 & 0 & -1728 & -48 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3888 & 72 & 1 & 0 & -3888 & -72 & -1 & 0 \\ 72 & 2 & 0 & 0 & -72 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 144 & 2 & 0 & 0 & -144 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 216 & 2 & 0 & 0 & -216 & -2 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 288 & 2 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} a_0 \\ b_0 \\ c_0 \\ d_0 \\ a_1 \\ b_1 \\ c_1 \\ d_1 \\ a_2 \\ b_2 \\ c_2 \\ d_2 \\ a_3 \\ b_3 \\ c_3 \\ d_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 40 \\ 40 \\ 70 \\ 70 \\ 90 \\ 90 \\ 100 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \mathbf B$

To obtain our solution, we must multiply the left inverse matrix of matrix A to our $16 \times 1$ Matrix B


 * $\mathbf A \mathbf x = \mathbf B$
 * $\mathbf A^{-1} \mathbf A \mathbf x = \mathbf A^{-1}\mathbf B$
 * $\mathbf x = \mathbf A^{-1}\mathbf B$

Running the matrix calculation above through a spreadsheet, we get:


 * $\begin{bmatrix} a_0 \\ b_0 \\ c_0 \\ d_0 \\ a_1 \\ b_1 \\ c_1 \\ d_1 \\ a_2 \\ b_2 \\ c_2 \\ d_2 \\ a_3 \\ b_3 \\ c_3 \\ d_3 \end{bmatrix} =

\begin{bmatrix} -0.0012 \\ -0.0000 \\ 3.5119 \\ 0.0000 \\ 0.0004 \\ -0.0595 \\ 4.2262 \\ -2.8571 \\ -0.0004 \\ 0.0000 \\ 2.7976 \\ 8.5714 \\ 0.0012 \\ -0.1786 \\ 9.2262 \\ -68.5714 \end{bmatrix}$

Our $4$ Cubic equations are then:

As a simple example, at $20$ months, $61.16\text{%}$ of the losses incurred will have been paid.


 * $0.0004 \paren{20}^3 - 0.0595 \paren{20}^2 + 4.2262 \paren{20} - 2.8571 = 61.16 \text{%}$