Assume that $W_s,W_t$ are Standard Brownian motions with $s<t$. Find the following
$$E[W_s | W_t]=? $$
The hint is to use projection method. If I understand correctly, we have the following property by projection
$$E[W_sZ]=E[YZ] $$ for $Y=E[W_s|W_t]$ and $Z$ is some random variable that is measurable under the same filtration that generates $W_t$. How to use this property and proceed? Any hint is appreciated.
P.S. I do not understand the solution given here and would prefer to understand using projection method (if at all possible).
Best Answer
The main idea is that we're going to guess that $Y=\mathbb{E}[W_s | W_t] = \beta W_t$ where $\beta$ is a (non-random) constant that depends only on $s$ and $t$. Then the projection method tells us (letting $Z = W_t$) that
\begin{align*} \mathbb{E}[W_s W_t] &= \mathbb{E}[Y W_t] \\ &= \beta \mathbb{E}[W_t^2]. \end{align*}
Now we know $\mathbb{E}[W_s W_t] = s$ and $\mathbb{E}[W_t^2]=t$, so the equation simplifies to $s = \beta t$ and hence $\beta = \frac st$. Therefore we conclude that $\mathbb{E}[W_s | W_t] = \beta W_t = \frac st W_t$.