Conditional expectation with brownian motion

Question

I want to find $E[e^{σB_t}|∫_0^1B_s ds]$.
I make the notation $∫_0^1B_s ds = z$, and I know that:
$E[B_t|Z]= 3t(1 - t/2)z $, $Var [B_t|Z] = t - 3t^2(1 - t/2)^2$ and $Var(∫_0^1B_s ds) = 1/3$. The answer has to be an integral.

Answer 1

For every centered normal $(X,Y)$ there exists $(a,b)$ such that $X=aY+bZ$ with $Z$ standard normal independent of $Y$. Thus, $$ E[\mathrm e^X\mid Y]=\mathrm e^{aY}E[\mathrm e^{bZ}]=\mathrm e^{aY+b^2/2}. $$ To find $(a,b)$, note that $E(X^2)=a^2E(Y^2)+b^2$ and $E(XY)=aE(Y^2)$.

Applying this to $X=\sigma B_t$ and $Y=\int\limits_0^1B_s\mathrm ds$, one sees that $E(X^2)=\sigma^2t$, $E(Y^2)=\int\limits_0^1\int\limits_0^1\min(u,v)\mathrm du\mathrm dv=\frac13$, and, if $t\leqslant1$, $E(XY)=\sigma\int\limits_0^1\min(t,s)\mathrm ds=\sigma t(1-\frac12t)$.

To write the result as an integral (!), consider $Y_t=\int\limits_0^tB_s\mathrm ds$, then $$ E[\mathrm e^{\sigma B_t}\mid Y]=1+a\mathrm e^{b^2/2}\int_0^t\mathrm e^{aY_s}B_s\mathrm ds. $$