Gaussian conditional distribution

Question

Let $Y_t$ be a gaussian process with $E[Y_t]=0$ and $Z=\frac{\int_0^1 Y_s ds}{\sqrt{V}}$ where $V=Var(\int_0^1 Y_s ds)$ (so Z has a standard normal distribution). I want to prove that conditionally on Z, $Y_t$ has a gaussian distribution. I'm thinking of proving that $(Y_t,Z)$ has a two-dimensional gaussian distribution, because then $Y_t|Z$ has a gaussian distribution. I would appreciate any idea. Thank you in advance.

Answer 1

Yes, one knows from general principles that, for each fixed $t$, the random couple $(Y_t,Z)$ is a linear functional of the gaussian process $(Y_s)_{0\leqslant s\leqslant1}$ hence $(Y_t,Z)$ is normal.

Let $C$ denote the covariance of the process $(Y_s)_{0\leqslant s\leqslant1}$, that is, $C(u,v)=E(Y_uY_v)$ for every $(u,v)$, then $(Y_t,Z)$ is centered, furthermore, $E(Y_t^2)=C(t,t)$, $E(Z^2)=A/V$ and $E(Y_tZ)=B_t/\sqrt{V}$ where $$A=\iint_{[0,1]^2}C(u,v)\mathrm du\mathrm dv,\qquad B_t=\int_0^1C(t,u)\mathrm du,$$ hence $$E(Y_t\mid Z)=(\sqrt{V}\,A^{-1}\,B_t)\cdot Z.$$