Probability of Wiener process infimum

Question

$(W_t)_{t\in[0,1]}$ a standard Wiener process. I would like to calculate $\mathbb{P}\left(\inf_{t\in[\frac{1}{2},1]}W_t>0\right)$, but I'm stuck :( Can anyone help, please?

Answer 1

We have \begin{align} \{\inf_{t\in[\frac{1}{2},1]}W_t>0 \} &= \{W_{\frac{1}{2}}+\inf_{t\in[\frac{1}{2},1]}(W_t-W_{\frac{1}{2}})>0 \} \\ &= \{W_{\frac{1}{2}}+\inf_{t\in[0,\frac{1}{2}]}(Z_t)>0 \} \\ &= \{W_{\frac{1}{2}}+\sup_{t\in[0,\frac{1}{2}]}(-Z_t)>0 \} \\ &= \{W_{\frac{1}{2}}-\sup_{t\in[0,\frac{1}{2}]}(B_t)>0 \} \tag{1} \end{align}

where $Z_t$ and $B_t$ are two brownian motions independant to $W_t$ and $B_t = -Z_t$.

We know also the law of supremum of BM, then $$\sup_{t\in[0,\frac{1}{2}]}(B_t) \stackrel{d}=|B_{\frac{1}{2}}|$$ Then, from $(1)$ we have $$\{\inf_{t\in[\frac{1}{2},1]}W_t>0 \} = \{W_{\frac{1}{2}}-|B_{\frac{1}{2}}|>0 \} = \{\frac{1}{\sqrt{2}}\left(X-|Y|\right)>0 \}=\{X-|Y|>0 \} $$ with $X,Y$ are independent and follow the standard normal distribution.

So, \begin{align} \mathbb{P}\left(\inf_{t\in[\frac{1}{2},1]}W_t>0\right) &= \mathbb{P}\left(X-|Y|>0\right) \\ &= \mathbb{P}\left(\{ X>Y\}\cap \{Y>0 \}\right) +\mathbb{P}\left(\{ X>-Y\}\cap \{Y<0 \}\right) \\ &= \mathbb{P}\left(X>Y>0\right) +\mathbb{P}\left(-X<Y<0\right) \\ \end{align} As $X,Y \stackrel{d}= \mathcal{N}(0,1)$ then the two probabilities are both $\frac{1}{8}$ the plan $Oxy$. Hence

$$\mathbb{P}\left(\inf_{t\in[\frac{1}{2},1]}W_t>0\right) = \frac{1}{8}+\frac{1}{8} = \frac{1}{4} $$