I came across the following problem: Given a standard Brownian Motion $B=(B_t)_{t\in R_0 ^+}$, the natural filtration $F=(F_t)_{t\in R_0 ^+}$, $F_t=\sigma(B_s : s \in R_0 ^+ ,s \leq t)$ and two real numbers 0<x<y, what ist $E[B_ye^{B_y}|F_x]$.
My Ansatz was:
$E[B_ye^{B_y}|F_x]$
$= E[(B_y -B_x +B_x)e^{B_y-B_x} e^{B_x}|F_x] $
$ = B_x e^{B_x} E[e^{B_y-B_x}|F_x] +e^{B_x}E[(E_y-B_x)e^{B_y-B_x}|F_x] $
$= B_x e^{B_x +\frac{y-x}{2}} + e^{B_x} E[(E_y-B_x)e^{B_y-B_x}|F_x] $
But here I am stuck. Does someone know how to continue?
$E[(B_y-B_x)e^{B_y-B_x}|F_x]=E[(B_y-B_x)e^{B_y-B_x}]$ (by independence)$ =EXe^{X}$ where $X \sim N(0,y-x)$. So $E[(B_y-B_x)e^{B_y-B_x}|F_x]=\int te^{t}\sqrt {\frac 2 {y-x}}\frac 1 {\sqrt {2\pi}} e^{-t^{2}/(y-x)}dt$. Hint for computing this integral: Write $t-t^{2}/(y-x)=-A (t-1/(2A))^{2}+\frac 1 {(2A)^{2}}$ where $A=1/(y-x)$.