Exercise 10.1 (Statistics in the two-factor Vasicek model). According to Example 4.7.3, $Y_1(t)$ and $Y_2(t)$ in $(10.2 .43)-(10.2 .46)$ are Gaussian processes.
(i) Show that
$$
\widetilde{\mathbb{E}} Y_1(t)=e^{-\lambda_1 t} Y_1(0),
$$
that when $\lambda_1 \neq \lambda_2$, then
$$
\tilde{\mathbb{E}} Y_2(t)=\frac{\lambda_{21}}{\lambda_1-\lambda_2}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) Y_1(0)+e^{-\lambda_2 t} Y_2(0),
$$
and when $\lambda_1=\lambda_2$, then
$$
\widetilde{\mathbb{E}} Y_2(t)=-\lambda_{21} t e^{-\lambda_1 t} Y_1(0)+e^{-\lambda_1 t} Y_2(0) .
$$
We can write
$$
Y_1(t)-\widetilde{\mathbb{E}} Y_1(t)=e^{-\lambda_1 t} I_1(t),
$$
when $\lambda_1 \neq \lambda_2$,
$$
Y_2(t)-\mathbb{E} Y_2(t)=\frac{\lambda_{21}}{\lambda_1-\lambda_2}\left(e^{-\lambda_1 t} I_1(t)-e^{-\lambda_2 t} I_2(t)\right)-e^{-\lambda_2 t} I_3(t),
$$
and when $\lambda_1=\lambda_2$,
$$
Y_2(t)-\tilde{\mathbb{E}} Y_2(t)=-\lambda_{21} t e^{-\lambda_1 t} I_1(t)+\lambda_{21} e^{-\lambda_1 t} I_4(t)+e^{-\lambda_1 t} I_3(t),
$$
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10 Term-Structure Models
where the ItƓ integrals
$$
\begin{aligned}
& I_1(t)=\int_0^t e^{\lambda_1 u} d \widetilde{W}_1(u), I_2(t)=\int_0^t e^{\lambda_2 u} d \widetilde{W}_1(u), \\
& I_3(t)=\int_0^t e^{\lambda_2 u} d \widetilde{W}_2(u), I_4(t)=\int_0^t u e^{\lambda_1 u} d \widetilde{W}_1(u),
\end{aligned}
$$
all have expectation zero under the risk-neutral measure $\widetilde{\mathbb{P}}$. Consequently, we can determine the variances of $Y_1(t)$ and $Y_2(t)$ and the covariance of $Y_1(t)$ and $Y_2(t)$ under the risk-neutral measure from the variances and covariances of $I_j(t)$ and $I_k(t)$. For example, if $\lambda_1=\lambda_2$, then