# Probability of Brownian motion to have a zero in an interval

I have what should be a very simple questions for Brownian motion experts... Let $[a,b]$ be a given time interval. Let $f(x)$ be the probability that a linear Brownian motion with initial value $x$ at time $t=0$ has a zero in the interval $[a,b]$. I want to argue that $f(x)$ is maximal for $x=0$. This seems intuitively clear but I cannot figure out a simple proof of this (other than writing the exact expression for $f$ which is a double integral and analyzing its variations via long calculations). I would be interested, in order of preference, by such a simple proof or by a reference where this lemma is proven.

Thanks!

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There is an easy way to do it via calculation. Start by supposing that $x\geq 0$ Then we're asking for the probability that $x+W_t$ reaches $0$ in the interval $[a,b]$ where $W_t$ is a standard Brownian motion. This the same as $P(\min_{a\leq t\leq b}x+W_t<0)=P(\min_{a\leq t\leq b}W_t<-x)$ Then $$\begin{eqnarray} P(\min_{a\leq t\leq b}W_t<-x)&=& E[1\{\min_{a\leq t\leq b}W_t<-x \}| W_a]\\ &=& E[1\{\min_{a\leq t\leq b}W_{t-a}<-x-W_a \}| W_a]\\ &=& E[1\{\min_{0\leq t\leq b-a}W_{t}<-x-W_a \}| W_a]\\ &=& E[1\{-|W_{b-a}|<-x-W_a \}| W_a]\\ \end{eqnarray}$$ Where we used the fact that $W_t-W_a$ has the same law as $W_{t-a}$ and $\min_{0\leq t\leq b-a}W_{t}$ has the same law as $-|W_{b-a}|$ then we have
$$P(\min_{a\leq t\leq b}W_t<-x)=2\int_R \frac{\exp(-y^2/2a)}{\sqrt{2\pi a}}N(-y-x)dy$$
Where $N$ is the Gaussian cumulative. Now it is easy to see that the right hand side is decreasing as function of $x$. We treat then $x\leq 0$ in the same way by symmetry of the Brownian motion.