Given a time t between (0, 1), the standard brownian path is at $w$ at $t=1, b(1)=w$. Show: $b(t) \text{ is } N(tw, t(1-t) )$.
I let $y = b(t)$ and $x=b(1)=w$
For the conditional distribution of $y$ given $x=w$, i got: $$\frac{exp(\frac{-y^2}{2t} + \frac{-(1-y)^2}{2(1-t)}+ \frac{w^2}{2})}{\sqrt{2\pi t(1-t)}}$$
after simplification, for the exp part I got: $$\frac{-(y-t)^2 + (1-w^2)(t^2-t)}{2t(1-t)}$$ but i don't see how to get into the form $$\frac{-(y-tw)^2}{2t(1-t)}$$
i'm not sure where i went wrong?
Let $Y = B(t), X = B(1)$ where $t \in (0, 1)$. Then $$ f_{Y|X=w}(y|w) = \frac {f_{Y,X}(y, w)} {f_X(w)} = \frac {f_Y(y)f_{X-Y}(w-y)} {f_X(w)} \propto \exp\left\{-\frac {y^2} {2t} - \frac {(w-y)^2} {2(1-t)} + \frac {w^2} {2} \right\} $$
and the remaining part is left for you to simplify the kernel and fill up the remaining constants.