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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.

WeakLearner
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    $r(t+s)$ and $r(t)$ are jointly normal. Calculate their correlation by using the expression for $r(t)$ you found. Use the result for bivariate normal random variables that gives you the expected value of one of the two variables when the other one is known. – Calculon Sep 08 '14 at 06:56
  • calculating $E[r_t r_{t+s}]$ is quite tedious though, is there no way of solving this problem directly by incorporating the information we are conditioning on? – WeakLearner Sep 08 '14 at 07:28
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    One way to do that would be to use the Markov property of $r(t)$. $r(t+s)$ is then given by taking $r(t)$ as your initial condition and let the SDE evolve for a duration of $s$. In your formula for $r(t+s)$ use $r(t)$ as $r_0$ and $s$ for $t+s$. That might be less tedious than my previous answer. – Calculon Sep 08 '14 at 08:17

1 Answers1

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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 $r(t)$.

mookid
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  • I get the results now, so incorporating the information up to time t is the same as having the initial condition as r(t) and just evaluating it over the increment over t? – WeakLearner Sep 08 '14 at 09:11
  • indeed, because the sde does not depend on $t$ directly. – mookid Sep 08 '14 at 09:12
  • Is there a way of incorporating the information up to time t straight into the expectation of the $dW_t$ term, similar to the attempt outlined in my initial question?? – WeakLearner Sep 08 '14 at 09:46
  • edited, with a fix in the time in the $dW_u$. – mookid Sep 08 '14 at 10:03