Would be grateful for any assistance. Below are the expected value and variance of the integral of the short rate under the Vasicek model:
$E\left[ \int_{0}^{t}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\frac{(r_{0}-b)}{a}(1-e^{-at})+bt$
$Var\left[ \int_{0}^{t}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\frac{\sigma^{2}}{2a^{3}}(2at-3+4e^{-at}-e^{-2at})$
But what if I would like to find the following:
$E\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} \\ Var\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} \\ where \;\;0\lt t_{1}\lt t_{2}$
Can I simply rewrite the above expressions as:
$E\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=E\left[ \int_{0}^{t_{2}}r(u)du - \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} = \frac{(r_{0}-b)}{a}(e^{-at_{1}}-e^{-at_{2}})+b(t_{2}-t_{1})$
$Var\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} = Var\left[ \int_{0}^{t_{2}}r(u)du-\int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\=Var\left[ \int_{0}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}+Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}-2Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\=Var\left[ \int_{0}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}-Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\= \frac{\sigma^{2}}{2a^{3}}(2a(t_{2}-t_{1})+4(e^{-at_{2}}-e^{-at_{1}})-e^{-2at_{2}}+e^{-2at_{1}})\\$