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I'm given a Ornstein-Uhlenbeck proces $V=(V_t)_{t\geq 0}$, i.e., $$ V_t = \frac{\sigma}{\sqrt{2\beta}}e^{-\beta t}B_{e^{2\beta t}}. $$

I'm told to prove that $V_{s+t}$, conditionated on $V_s=v$, follows a $$ N(e^{-\beta t}v,\frac{\sigma^2}{2\beta}(1-e^{-2\beta t}), $$ but I don't see why the expectation is not zero, since the expectation of the brownian motion is zero. Any hint, please?

R__
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    Hint: For fixed $s,t$ the joint distribution of $(V_t, V_{s +t})$ is bivariate Gaussian with known mean and covariance matrix. Now you just have to find the conditional distribution of the first component of a bivariate normal random vector given the second which is a straightforward calculation. – Stefan May 17 '21 at 14:18
  • very helpful, I didn't realise that! – R__ May 17 '21 at 16:52

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$$V_{t+s}=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}B_{e^{2\beta(t+s)}}$$ Let us find the moments. As the expectation of the increment $B_{e^{2\beta(t+s)}}-B_{e^{2\beta s}}$ is $0$ and if $V_s$ is known then $B_{e^{2\beta s}}$ is known: $$E[V_{t+s}|V_s]=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}E[B_{e^{2\beta(t+s)}}|V_s]=\frac{\sigma}{\sqrt{2\beta}}e^{-\beta(t+s)}B_{e^{2\beta s}}=e^{-\beta t}V_s$$ Also $$E[(B_{e^{2\beta(t+s)}}-B_{e^{2\beta s}})^2|V_s]=E[B_{e^{2\beta(t+s)}}^2|V_s]+B_{e^{2\beta s}}^2-2B_{e^{2\beta s}}E[B_{e^{2\beta(t+s)}}|V_s]$$ $$e^{2\beta(t+s)}-e^{2\beta s}=E[B_{e^{2\beta(t+s)}}^2|V_s]+B_{e^{2\beta s}}^2-2B_{e^{2\beta s}}^2=E[B_{e^{2\beta(t+s)}}^2|V_s]-B_{e^{2\beta s}}^2$$ Therefore $$E[V_{t+s}^2|V_s]=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}E[B_{e^{2\beta(t+s)}}^2|V_s]$$ $$E[V_{t+s}|V_s]^2=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}E[B_{e^{2\beta(t+s)}}|V_s]^2$$ Thus by using $\textrm{Var}[X]=E[X^2]-E[X]^2$ we get $$\textrm{Var}[V_{t+s}|V_s]=\frac{\sigma^2}{2\beta}e^{-2\beta(t+s)}\bigg(e^{2\beta(t+s)}-e^{2\beta s}+B_{e^{2\beta s}}^2-B_{e^{2\beta s}}^2\bigg)=\frac{\sigma^2}{2\beta}(1-e^{-2\beta t})$$

Snoop
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  • That's great! Thank you very much. One more question, how can I use this result to calculate $\lim_{t\to\infty}\mathbb{E}[(V_{s+t}-v)^2|V_s=v]$? I'm trying to expand the expresion and make use of the variance, but I get that this lim is equal to zero, which does not make sense in my context. Thanks! – R__ May 17 '21 at 17:32
  • You're welcome. Did you mean to divide that limit by $t$? – Snoop May 17 '21 at 17:50
  • Yes, sorry, divided by $t$. Maybe it's a typo and should be tending to $\infty$, but in the paper is written $t\to\infty$ and I get just zero (it should be $\sigma^2$). – R__ May 17 '21 at 17:51
  • Yes it is $\sigma^2$. You should post another question about that limit, though – Snoop May 17 '21 at 18:00
  • the typo is probably that it should be $t \to 0$ – Snoop May 17 '21 at 18:08