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.


1 Answer 1


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.


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