# Probability Density Function of a Wiener Process Minimum

Let $$W_t$$ be a standard Wiener process. Find the probability density function of $$m_T = min_{t\in [0,T ]}W_t$$.

I know that it is based of the concept of the reflection principle, but I wasn't too sure on how to compute the Probability density function for this.

Firstly, $$m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T$$. So, you can either consider the running maximum or minimum.
Let $$\tau$$ be a stopping time and $$(B_t)$$ a Brownian motion. Then, \begin{align*} W_t =\begin{cases} B_t & t\leq \tau, \\ 2B_\tau - B_t & t\geq \tau, \end{cases} \end{align*} is again a standard Brownian motion (This is the reflection principle).
For $$a\geq 0$$ and $$t>0$$, the reflection principle implies that \begin{align*} \mathbb{P}[\{M_T\geq a\}] &= 2\mathbb{P}[\{B_t\geq a\}] \\ \implies \mathbb{P}[\{M_T\leq a\}] &= 2\mathbb{P}[\{B_t\leq a\}]-1 \\ &= 2\Phi\left(\frac{a}{\sqrt{t}}\right)-1. \end{align*}
Thus, the probability density function of $$(M_t)$$ is given by \begin{align*} f_{M_t}(x) &= \frac{\partial }{\partial x} \left(2\Phi\left(\frac{x}{\sqrt{t}}\right)-1\right) \\ &= \frac{2}{\sqrt{t}}\varphi\left(\frac{x}{\sqrt{t}}\right) \\ &= \sqrt{\frac{2}{t\pi}}e^{-\frac{1}{2t}x^2} \end{align*} for $$x\geq0$$ and $$f_{M_t}(x)=0$$ for $$x<0$$. The function is clearly non-negative and you can easily see that it integrates to one.
Here, $$\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$$ is the pdf of a standard normally distributed random variable and $$\Phi$$ the corresponding cdf.