Let $B=(B_t)_{t \ge 0}$ bethe one dimensional Brownian motion.
Let futheremore $s, t \in [0,1]$.
I want to know how to deduce following formulas:
1) $Var(B_t \vert B_1) = t - t^2$
2) $\mathbb{E}[B_s,B_t \vert B_1] = s/t\mathbb{E}[B_t^2 \vert B_1] $
My attempts:
I know that $\mathbb{E}[B_t \vert B_1]= tB_1$ and ${\displaystyle \operatorname {Var} (B_t|B_1)=\operatorname {E} ((B_t-\operatorname {E} (B_t\mid B_1))^{2}\mid B_t)}$ already holds,
futhermore, if $B_t$ is BM, then also $W_t := tB_{1/t}$.
But I don't see how to combine this facts to get 1) and 2).
Recall that the density of $B_t$ is $$f_t(x) = \frac1{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} $$ and so the conditional density of $B_t$ given $B_u=b$, where $t<u$, is $$ f_{t\mid u}(x\mid b) = \frac{f_t(x)f_{u-t}(b-x)}{f_u(b)}\propto \exp\left(-\frac{u(x-bt/u)^2}{2t(u-t)} \right), $$ which is a normal distribution with mean $bt/u$ and variance $t(u-t)/u$ It follows that the conditional expectation and variance of $B_t$ given $B_u$ (where here $u=1$) are \begin{align} \mathbb E[B_t\mid B_1] &= tB_1\\ \mathsf{Var}(B_t\mid B_1) &= t(1-t). \end{align} The third expression seems like it should follow from $$ \mathbb E[B_sB_t\mid B_1] = \mathbb E[(B_s/B_t)B_t^2\mid B_1], $$ but I'm not certain about the computation.