Let $W(t)$ a standard Brownian motion. In the book "Introduction to stochastic differential equation" of Evans, page 41 they try to find $$\mathbb P\{a_1\leq W(t_1)\leq a_2, a_2\leq W(t_2)\leq b_2\}.$$
In a formal way they tried as follow :
$$\mathbb P\{a_1\leq W(t_1)\leq a_2\}=\int_{a_1}^{a_2}\frac{e^{\frac{-x_1^2}{2t_1}}}{\sqrt{2\pi t_1}}dx_1.$$ and, given $W(t_1)=x_1$, $a_1\leq x_1\leq b_1$, then presumably the process is $N(x_1,t_2-t_1)$ on the interval $[t_1,t_2]$.
Q1) How can they get that ?
Thus the probability that $\mathbb P\{a_1\leq W(t_2)\leq b_1\mid W(t_1)=x_1\}$ should be $$\int_{a_2}^{b_2}\frac{1}{\sqrt{2\pi(t_2-t_1)}}e^{-\frac{|x_2-x_1|}{1(t_2-t_1)}}dx_2.$$
Q2) I'm a bit in trouble since $\mathbb P\{a_2\leq W(t_2)\leq b_2\mid W(t_1)=x_1\}=\frac{\mathbb P\{a_2\leq W(t_2)\leq b_2, W(t_1)=x_1\}}{\mathbb P\{W(t_1)=x_1\}}$, and since $\mathbb P\{W(t_1)=x_1\}=0$, it looks that we have a problem. So, how does it work here ?
I think it's more an "unformal" way to deduce the joint PDF rathan than a proof. If we give that $W(t_1)=x_1$ (so it's not random : we say that $W(t_1)=x_1$), then $$W(t_2)=W(t_2)-W(t_1)+W(t_1)=W(t_2)-W(t_1)+x_1.$$ Since $W(t_2)-W(t_1)\sim \mathcal N(0,t_2-t_1)$, then $$W(t_2)=W(t_2)-W(t_1)+x_1\sim \mathcal N(x_1,t_2-t_1).$$
At the end, $$\mathbb P\{a_1\leq W(t_1)\leq b_1, a_1\leq W(t_2)\leq b_1\}=\int_{a_1}^{b_1}\mathbb P\{a_2\leq W(t_2)\leq b_2\mid W(t_1)=x_1\}\frac{1}{\sqrt{2\pi t_1}}e^{-\frac{x_1^2}{2t_1}}\,\mathrm d x_1=...$$
Notice that $\mathbb P(A\mid B)=\frac{\mathbb P(A\cap B) }{\mathbb P(B)}$ if $P(B)>0$. Otherwise, it's not !