2.13 Change of Variables Formula

2.13  Change of Variables Formula

The most basic arithmetic operation is addition. Integral calculus generalizes this operation with the definite integral, which is a generalized sum. Definite integrals will play an important role in our discussions of value-at-risk (VaR). Indeed, the task of calculating a portfolio’s VaR is largely one of valuing a definite integral.

Definite integrals can often be simplified through a judicious change of variables. If an integral is to be valued using numerical techniques, a change of variables may be essential to avoid a singularity or to convert an unbounded region of integration into one that is bounded. The notion of changing variables is as old as addition itself. It is easier to count eggs in dozens than to count them individually.

For a one-dimensional integral over an interval [a, b], an invertible, continuously differentiable change of variables x = g(u) yields

[2.166]

For example

[2.167]

Generalizing to multiple dimensions, consider integrable function f : n and invertible, continuously differentiable change of variables g : nn. Then

[2.168]

where Ω  n, and |Jg(u)| is the determinant of the Jacobian of g. Consider integral:

[2.169]

where the region Ω is indicated in Exhibit 2.21.

Exhibit 2.21: The region Ω is bounded by the four lines x1 = 0, x2 = 0, x2 = 1 – x1, and x2 = 2 – x1.

Integral [2.169] is difficult to solve directly, but consider the change of variables x = g(u) defined by

[2.170]

The Jacobian determinant of g is

[2.171]

so, by [2.168], the integral becomes

[2.172]

Region g-1(Ω) is indicated in Exhibit 2.22. It has twice the area of the original region, so it should come as no surprise that the Jacobian determinant introduces an offsetting scaling factor of 1/2 into integral [2.172].

Exhibit 2.22: The region g–1(Ω) is bounded by the four lines u2 = 1, u2 = 2, u2 = u1, and u2 = –u1.

Region g-1(Ω) has a convenient shape, which allows us to represent the integral as

[2.173]

This is easily valued as 0.8146.

Exercises
2.22

Use change of variables x = g(u) = to value the integral

[2.174]

Solution

2.23

Consider the integral

[2.175]

  1. Solve the integral directly.
  2. Solve the integral using the change of variables x = g(u) defined by

[2.176]

(Hint: Sketch the regions of integration Ω and g–1(Ω).)

Solution