I am self studying some stochastic calculus material and come across this question to show that the distribution of $P(W(s)\in dy|W(t)=x)$ with $W(0)=0$ is normal with mean $\frac{s}{t}W(t)$ and variance $s(t-s)$.
So I use: $P(W(s)\in dy|W(t)=x)=\frac{P(W(s)\in dy)P(W(t)\in dx|W(s)=y)}{P(W(t)\in dx)}$
In addition we know that: $P(W(t)\in dx|W(s)=y)=\frac{1}{\sqrt{2\pi(t-s)}}\exp\{-\frac{(x-y)^{2}}{2(t-s)}\}dx$
Hence, \begin{align*} P(W(s)\in dy|W(t)=s) & =\frac{1}{\sqrt{2\pi s}}\exp\{-\frac{y^{2}}{2s}\}\frac{1}{\sqrt{2\pi(t-s)}}\exp\{-\frac{(x-y)^{2}}{2(t-s)}\}\times\sqrt{2\pi t}\exp\{\frac{x^{2}}{2t}\}\\ & =\frac{t}{\sqrt{2\pi ts(t-s)}}\exp\{-\frac{(yt-sx)^{2}}{2ts(t-s)}\} \end{align*}
Now I can see that the mean can be shifted to be $\frac{s}{t}W(t)$ but how do I see that the variance is $s(t-s)$ from this expression ?
Thanks