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Suppose $X_t$ satisfies the SDE $dX_t = a \; dt + dW_t$, where $a$ is a constant and $X_0 = 0$. What is the distribution of $X_3$ given that $X_1 = -1, X_4 = 2$ and $X_5 = 1$? Give the type of distribution and the parameters that describe the conditional distribution of $X_3$ completely.

OK I am a bit embarrassed that I do not remember how to do this. Here is what I have started on: one can see by integrating that $X_t$ is distributed as $N(at, t)$, so each $X_t$ is Gaussian. I know that $X_3$ conditioned on $X_1, X_4, X_5$ is a one dimensional Gaussian, but what is the easiest approach to finding the conditional mean and variance? I remember being able to do this given a covariance matrix, but I suspect that's not really necessary here. Any help would be greatly appreciated.

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Given that SDE $\mathrm{d} X_t = a \mathrm{d} t + \mathrm{d} W_t$ with the initial condition $X_0 = 0$, the process $X_t$ is Gaussian, meaning that the joint distribution of $(X_1, X_3, X_4, X_5)$ has the normal distribution with mean vector $(a, 3a, 4a, 5a)$, and the covariance matrix $\Sigma_{t,s} = \min(t,s)$, i.e.: $$ \Sigma = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 3 & 3 & 3 \\ 1 & 3 & 4 & 4 \\ 1 & 3 & 4 & 5 \\ \end{pmatrix} $$

The conditional normal of a multi-normal vector is also normal. Using notation's from the wikipedia page: $$ \hat{\Sigma}_{11} = \begin{pmatrix} 3 \end{pmatrix} \qquad \hat{\Sigma}_{22} = \begin{pmatrix} 1 & 1 & 1\\ 1 & 4 & 4\\ 1 & 4 & 5 \end{pmatrix} \quad \hat{\Sigma}_{12} = \begin{pmatrix} 1 & 3 & 3 \end{pmatrix} $$ Notice that $$ \hat{\Sigma}_{22}^{-1} =\frac{1}{3} \left( \begin{array}{rrr} 4 & -1 & 0 \\ -1 & 4 & -3 \\ 0 & -3 & 3 \end{array}\right) $$

Conditional distribution of $X_3$ given $X_1 = x_1$, $X_4 = x_4$ and $X_5 = x_5$ is normal with mean $$ \begin{eqnarray} \mathbb{E}(X_3 &|& X_1=x_1, X_4=x_4, X_5=x_5) \\&=& \mathbb{E}(X_3) + \hat{\Sigma}_{12} \hat{\Sigma}_{22}^{-1} \begin{pmatrix} x_1 - \mathbb{E}(X_1) & x_4 - \mathbb{E}(X_4) & x_5 - \mathbb{E}(X_5) \end{pmatrix}^t \\ &=& 3 a + \left( -3 a + \frac{x_1}{3} + \frac{2 x_4}{3} \right) = \frac{x_1 + 2 x_4}{3} \end{eqnarray} $$ The conditional variance: $$ \mathbb{Var}(X_3 | X_1=x_1, X_4=x_4, X_5=x_5) = \hat{\Sigma}_{11} - \hat{\Sigma}_{12} \hat{\Sigma}_{22}^{-1} \hat{\Sigma}_{12}^t = \frac{2}{3} $$ Thus the conditional distribution of $X_3$ results to be $\mathcal{N}\left( \frac{x_1 + 2 x_4}{3}, \sqrt{\frac{2}{3}} \right) = \mathcal{N}\left(1,\sqrt{\frac{2}{3}}\right)$.

As you see, the conditional distribution of $X_3$ only depends on the conditional value of $X_4$, but not $X_5$.

Indeed, the conditional Brownian motion $X_t$ for $1 \leqslant X_t \leqslant 4$, conditioned on $X_1 = x_1$ and $X_4 = x_4$ is known as the Brownian bridge process.


My original incorrect response:

Remember that $X_t$ is an independent increment process, such that $X_t - X_s \sim X_{t-s}$, that is $X_3$, given $X_1 = -1$ is equal in distribution to $X_3 = X_1 + (X_3-X_1) \sim -1 + X_2$, that is it is normal with variance $t$ and mean $-1 + 2 a$.

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Thank you Sasha –  David Dec 19 '11 at 22:23
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But you're ignoring $X_4$. The conditional distribution of $X_3$ given observed values of $X_1$, $X_4$, and $X_5$ should not depend on $X_5$, but it should depend on $X_1$ and $X_4$. –  Michael Hardy Dec 20 '11 at 3:25
    
@Bob I am afraid my original response is entirely incorrect per Michael's comment. I will write a new one momentarily. –  Sasha Dec 20 '11 at 5:48
    
@MichaelHardy Thank you for alerting me to the embarrassing mistake in my post. I have corrected it now. –  Sasha Dec 20 '11 at 6:30
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