# Conditional distribution of process $W$ given $\{W_1 = y\}$ is Gaussian.

Suppose that $$X=(X_t)_{t \in [0,1]}$$ is a continuous Gaussian process, for which $$\mathbb{E}(X_t) = 0$$ for all $$t \in [0,1]$$ and $$Cov(X_s,X_t) = s(1-t)$$ for all $$0 \leq s \leq t \leq 1$$. Let $$Y \sim N(0,1)$$ be a random variable independent of $$X$$. Define the stochastic process $$W = (W_t)_{t \in [0,1]}$$ by setting $$W_t = X_t + tY$$.

Show that the conditional distribution of the process $$W$$ given $$\{W_1 =y\}$$ is Gaussian, and calculate their mean and covariance functions. As a hint it is given how to make sense of conditioning on a set of zero probability:

The event $$\{W_1 =y \}$$ has zero probability, but it is natural to define conditioning on this event by the following limiting procedure. For any $$\varepsilon >0$$, the event $$\{|W_1 - y| < \varepsilon\}$$ has positive probability, so we can consider the conditional distribution of the process $$W$$ given the event $$\{|W_1 - y| < \varepsilon\}$$, and then take the (weak) limit of this conditional distribution as $$\varepsilon \downarrow 0$$.

I have already shown that W is a Gaussian process. Now my idea was to calculate $$\frac{\mathbb{P}(\{W \in B\} \cap \{|W_1 - y| < \varepsilon\})}{\mathbb{P} \{|W_1 - y| < \varepsilon\}}$$, but I do not seem to come to the right conclusion. Is this even the right approach? Further I do not see where I have to use weak convergence here. Thank you very much in advance for any help.

You can finesse this by showing that the distribution of $$(X_t+ty)_{t\in[0,1]}$$ as $$y$$ varies over $$\Bbb R$$, (call it $$Q_y$$) is a regular conditional distribution of $$W$$ given $$W_1$$. Namely, $$y\mapsto Q_y$$ is suitably measurable and $$P[W\in B, W_1\in C]=\int_{C}Q_y(B)\varphi(y)\,dy,\qquad(*)$$ where $$\varphi$$ is the standard normal density. As $$X_t+ty$$ is clearly Gaussian, this will meet your needs. To see (*), just compute: Becaue $$Y$$ is independent of $$X$$, \eqalign{ P[W\in B, W_1\in C] &=P[(X_t+tY)_{t\in[0,1]}\in B, Y\in C]\cr &=\int_C P[(X_t+ty)_{t\in[0,1]}\in B]\cdot P[Y\in dy]\cr }
• Thank you very much for your answer. There are a few things that are not clear to me. If I do it as you said, I would get somthing like $\frac{\int_{-\varepsilon+y \leq Y \leq \varepsilon +y} Q_y(B) \varphi(y) \ dy}{\mathbb{P}[\{-\varepsilon + y \leq Y \leq \varepsilon y\}]}$. Wouldn't this blow up as $\varepsilon \downarrow 0$. Further I still do not understand where weak convergence is used here. – Rupert R Mar 24 at 8:09