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Distribution of the Minimum of a Wiener Process

Article Quant Q&A · Author: Anon
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Summary

The document derives the probability density of the minimum of a standard Wiener process over a fixed time interval using the reflection principle. It first observes that the minimum has the same distribution as the negative of the process maximum, since negating a Wiener process preserves its law. The reflection principle then relates the running maximum’s cumulative distribution to the normal cumulative distribution, from which differentiation gives its density.

The resulting minimum distribution is the reflection of the maximum’s half-normal distribution: it is supported on nonpositive values, with a density that decays according to a Gaussian form and scales with the observation horizon. The response also states the corresponding maximum density and identifies the standard normal density and cumulative distribution used in the derivation. This is a mathematical result for standard Brownian motion with no drift over a fixed interval; it does not cover drifted processes, nonstandard volatility, or applications to financial price paths without further assumptions.

Key ideas

  • Negating a standard Wiener process turns its minimum into the negative of a maximum with the same law.
  • The reflection principle gives the cumulative distribution of the running maximum.
  • Differentiating that cumulative distribution yields a half-normal density for the maximum.
  • The minimum density is the reflected maximum density and is supported on nonpositive values.
  • The result assumes standard Brownian motion without drift over a fixed horizon.

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StatisticsVolatility
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# Probability Density Function of a Wiener Process Minimum


# 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.

## Answer by Kevin (score 7, accepted)

https://quant.stackexchange.com/a/49477

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.