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 $X$, $$ \eqalign{ P[W\in B, W_1\in C] &=P[(X_t+tY)_{t\in[0,1]}\in B, Y\in C]\cr &=\int_C P[(X_t+ty)_{t\in[0,1]}\in B]\cdot P[Y\in dy]\cr } $$