## Multivariate Affine Asset Pricing Model

As the dimension of the state process increases, the no-arbitrage PDE becomes increasingly difficult to apply in practice. There are some cases, however, for which this so-called curse of dimensionality can be avoided. The most important case is the affine asset pricing model, which has been widely applied, especially in modeling interest rate and futures price term structure and in valuing European options.

Suppose that the d-dimensional risk-neutral state process S can be described by an affine diffusion, which takes the form dS = (a + AS)dt + Cdiag (Vb+BS^ dz, where a and b are d x 1 and A, B and C are d x d (the ^ operator is applied element by element). Furthermore, the risk free interest rate is an affine function of the state, ro + rS and the log of the terminal value of the asset is an affine function of the state given by h0 + hS (where r0 and h0 are both scalar and where r and h are both 1 x d).

Expressing the value of the asset as a function of S and the time-to-maturity t = T — t, it can be verified that:

To satisfy the terminal condition 00(0) = h0 and 0(0) = h. To verify that the proposed value of V satisfies the no-arbitrage condition, substitute it into (10.2):

(r0 + rS)V = — (00 (t )+ 0' (t )S )V + 0 (t )(a + AS)V

V is a common term that can be divided out and the remaining expression is affine in S. (10.5) is therefore satisfied when3

The d+1 coefficient functions 00(t) and 0(t) are thus solutions to a system of ordinary differential equations, which are easily solved, even when d is quite large.

The Vasicek bond pricing model discussed on page 325 is a one-dimensional example with b =1, B = 0, r0 = 0 and r = 1. A more interesting example is the Longstaff-Schwartz two-dimensional model, in which A is diagonal, C = I, b = 0, B = I and r0 = 0. The two factors in this model are independent mean square root processes:

The instantaneous interest rate is a linear function of these two factors. Due to its separability, the ODEs to be solved are

01 (t ) = Aii0i(T)+ 2 02 (t ) — ri, with 0i(0) = 0, for i = 1, 2, and

3We use the facts that trace(xyz) = trace(zxy) and, when x and y are vectors, diag(x)y diag(y)x.

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