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$y(\infty)$

The results of section [*] apply and special expressions for the specification EFM(2) can be obtained. First, $Q\,b = Q\,D^-\,Q^-\,e_1 = ({A^\star}^\prime)^-\,e_1$ has a simple expression:
\begin{displaymath}
Q\,b = -({\kappa_r^\star}^-, {\kappa_x^\star}^-, {\kappa_y^\star}^-)^\prime =: -\kappa^-\quad .
\end{displaymath} (62)

This leads to
\begin{displaymath}
y(\infty) = {a^\star}^\prime\,\kappa^- - {1\over 2}\,{\kappa^-}^\prime\,V\,\kappa^- \quad .
\end{displaymath} (63)

The long yield $y(\infty)$ is dominated by the smallest mean reversion strength (under $P^\star$). For $\kappa_r^\star > \kappa_x^\star \gg \kappa_y^\star$ and $\lambda_y^1 \gg \kappa_y\,\alpha_y$ it scales with $\kappa_y^\star$ as
\begin{displaymath}
y(\infty) \sim \frac{\lambda^1_y}{\kappa_y^\star} - \frac{\sigma_y^2}{2\,{\kappa_y^\star}^2} \quad ,
\end{displaymath} (64)

i.e. the blow-up effect of $\kappa_y^\star\to 0$ can only be compensated by $\lambda^1_y, \sigma_y$ of the order of $\kappa_y^\star$.

Markus Mayer 2009-06-22