Show that $P^0( a<W_\theta \leq b \vert W_t=y)\leq P^0( a<W_\theta \leq b \vert W_t=\frac{a+b}{2})$ is true for all $y \in \mathbb{R}$ where $0 \leq t <\theta<T$ and $W$ is a $1$-dimensional Brownian motion starting at $z \in \mathbb{R}$ with respect the probability measure $P^z$.
My first attempt and the problems My first idea was to use the formula of conditional probability $P(A \vert B)=\frac{P(A \cap B)}{P(B)}$ but in this case $P^0(W_t=y)=0$ and $P^0(W_t=\frac{a+b}{2})=0$ and therefore I don't know how to actually proceed since the denominator is zero
My Second attempt and the problem The other idea was to use the markov property of the Brownian motion to write
$P^0( a<W_\theta \leq b \vert W_t=y)=P^y( a<W_{\theta-t} \leq b )$
and
$P^0( a<W_\theta \leq b \vert W_t=\frac{a+b}{2})= P^{\frac{a+b}{2}}( a<W_{\theta-t} \leq b)$
Now If I can show that $P^y( a<W_{\theta-t} \leq b ) \leq P^{\frac{a+b}{2}}( a<W_{\theta-t} \leq b)$ then the proof is complete. I am stuck here since I can't show rigorously why this is true.
Any hints would be highly appreciated.
Using the fact that for Brownian motion $B$ w.rt $P^x$ ; $B_t$ is $N(x,t)$ distributed we have that $P^y(a<W_{\theta-t}\leq b)=\frac{1}{\sqrt{2\pi (\theta-t)}} \int_{a}^b \exp(\frac{(z-y)^2}{2(\theta-t)})dz= \frac{1}{\sqrt{2\pi }} \int_{\frac{a-y}{\sqrt{(\theta-t)}}}^{\frac{b-y}{\sqrt{(\theta-t)}}} \exp{\frac{-k^2}{2}}dk$ where the last inequality is a consequence of the subtitution $k=\frac{z-y}{\sqrt{\theta-t}}$
Similarly we get $P^{\frac{a+b}{2}}(a<W_{\theta-t}\leq b)=\frac{1}{\sqrt{2\pi }} \int_{\frac{a-b}{2\sqrt{(\theta-t)}}}^{\frac{b-a}{2\sqrt{(\theta-t)}}} \exp{\frac{-k^2}{2}}dk$
It is obvious that the length of the interval over which the two integral are computed is the same but in the latter integral, the interval is centered around zero(where the gaussian density has the highest weight) and therefore the claim follows!