Let W be a Brownian motion. I have to evaluate $P\left(\int_{0}^{t} W(s)ds <y |W(t)=x\right)$.
I was able to find, without two much trouble, that $\int_{0}^{t} W(s)ds \sim N\left(0, \frac{t^3}{3}\right)$. Similarily, $W(t) \sim N(0,t)$. Now, $cov\left(\int_{0}^{t} W(s)ds, W(t)\right)=\frac{t^2}{2}$ and $\rho$, the coefficient correlation, is $\frac{\sqrt{3}}{2}$. To simplify, let $Y=\int_{0}^{t} W(s)ds$ and $X=W(t)$.
Now, since these are two normal distributions, they follow a bivariate normal distribution. I know the formula to calculate the conditional pdf in this case; I got that
$$f_{Y|X=x}(y) \sim N\left(\frac{x \rho\sigma_Y }{\sigma_X}, \sigma_X ^2(1-\rho^2)\right) \implies f_{Y|X=x}(y) \sim N\left(\frac{3x}{2t}, \frac{t}{4}\right)$$ Thus,
$$P\left(\int_{0}^{t} W(s)ds <y |W(t)=x\right)=\int_0^y \frac{1}{\sqrt{2\pi\frac{t}{4}}}e^{\frac{-2\left(z-\frac{3x}{2t}\right)^2}{t}}dz$$
which, unless I'm mistaken, doesn't really simplify... is there a better way to evaluate the conditional probability?