TITLE: Conditional expectations and Brownian bridge
QUESTION [5 upvotes]: I'm currently working about Brownian Bridge and I have to compute the following expectation
\begin{equation} E[W_tW_s\vert W_T]\end{equation}
We consider here that the Brownian bridge is defined as a Brownian motion conditioned to hit 0 at time T, and for $(W_t)_{t≥0}$ a Brownian motion, we give that for $s≤ t ≤ 1$,
\begin{equation} E[W_s \vert W_t = x] = \frac{s}{t}x \text{  and   } Var[W_s \vert W_t = x] = \displaystyle\frac{s(t-s)}{t}\end{equation}
I found 2 different ways to do it that seems good, but I find a different result. Can somebody explain me why there is this difference, and if there's a mistake where it is ?
Method 1 (this is the result I'm supposed to find)
\begin{equation}E[W_tW_s\vert W_T = 0] = E[E(W_s W_t \vert W_t) \vert W_T] = E[W_t E[W_s \vert W_t]  \vert  W_T] = E[W_t^2 \frac{s}{t} \vert W_T] = \frac{s}{t} Var(W_t \vert W_T) = \frac{s}{t} t \frac{(T-t)}{T} = \displaystyle\frac{s(T-t)}{T}\end{equation}
Method 2 (not the same result but (I think) mathematically true)
\begin{equation}E[W_tW_s\vert W_T = 0] = E[(W_t - W_s)W_s + W_s^2 \vert W_T] = E[W_t - W_s \vert W_T] E[W_s \vert W_T] + E[W_s W_s\vert W_T] =  Var(W_s\vert W_T) = \displaystyle\frac{s(T-s)}{T}\end{equation}
Thanking you in advance !

REPLY [4 votes]: I am not convinced by the use (?) of a tower property in method 1, nor of the independence in method 2. Suppose $s<t<T$. Note
$$\begin{bmatrix}
W_T\\
W_t\\
W_s
\end{bmatrix}\sim \mathcal{N}(0,\Sigma),\,\quad \Sigma=\begin{bmatrix}
T&t&s\\
t&t&s\\
s&s&s
\end{bmatrix}$$
So we use the conditional density
$$(W_t,W_s)|W_T=w\sim \mathcal{N}(\mu(w),\Gamma),\,\quad \mu(w)=\begin{bmatrix}
wt/T\\
ws/T\\
\end{bmatrix},\,\quad \Gamma=\begin{bmatrix}
t&s\\
s&s
\end{bmatrix}-\frac{1}{T}\begin{bmatrix}
t^2&ts\\
ts&s^2
\end{bmatrix}$$
Now recall
$$\textrm{Cov}[W_t,W_s|W_T]=E[W_tW_s|W_T]-E[W_t|W_T]E[W_s|W_T]$$
So
$$E[W_tW_s|W_T=w]=s-\frac{st}{T}+\frac{st w^2}{T^2}$$