The solution to$~~~~ dr_t=\alpha(\mu-r_t)dt+\sigma dW_t $ is given by:
$$
r_t=r_0e^{-\alpha t} +\mu(1-e^{-\alpha t})+\sigma \int_0^t e^{-\alpha (t-s)}dW_s
$$
I have been able to show that:
$$
r_t\sim N(~~r_0e^{-\alpha t} +\mu(1-e^{-\alpha t})~,~\frac{\sigma^2}{2 \alpha}(1-e^{-2\alpha t})~~)
$$
I am trying to find the conditional expectation and variance of $r(t+s)$ given $r(t)$
The final result should be:
$$
r_{t+s}~|~r_t\sim N(~~\mu+(r_t-\mu)e^{-\alpha s}~~,~\frac{\sigma^2}{2 \alpha}(1-e^{-2\alpha s})~~)
$$
I'm a little confused about incorporating the information provided by knowing $r_t$ into the expectation, that is, I'm having difficulty when the problem boils down to figuring out:
$$
E~[\int_0^{t+s}e^{-\alpha(t+s-s)}dW_s~|~r_t]=E~[\int_0^{t}e^{-\alpha(t+s-s)}dW_s~|~r_t]+E~[\int_t^{t+s}e^{-\alpha(t+s-s)}dW_s~|~r_t]
$$
I don't seem to get the right result after taking this step.
These are both statements made but not proven in "Interest Rate Models – Cairns", any help would be appreciated.
Best Answer
Hint: the SDE verified by $r(s+t)$ is the same as $r(t)$, hence $$ r(t+s) = r(t)e^{-\alpha s} + \mu (1-e^{-\alpha s}) +\sigma \int_0^se^{-\alpha (s-u)}dW_{t+u} $$as in your first expression, with initial condition $r(t)$.