# proof of hitting times are stopping times for right-continuous processes

Let ${(\mathcal{F})}_{t\in \mathbb{T}}$ be a filtration^{} (http://planetmath.org/FiltrationOfSigmaAlgebras) on the measurable space^{} $(\mathrm{\Omega},\mathcal{F})$, It is assumed that $\mathbb{T}$ is a closed subset of $\mathbb{R}$ and that ${\mathcal{F}}_{t}$ is universally complete for each $t\in \mathbb{T}$.

Let $X$ be a right-continuous and adapted process taking values in a metric space $E$ and $S\subseteq E$ closed. We show that

$$\tau =inf\{t\in \mathbb{T}:{X}_{t}\in S\}$$ |

is a stopping time. Assuming $S$ is nonempty and defining the continuous function^{} ${d}_{S}(x)\equiv inf\{d(x,y):y\in S\}$, then $\tau $ is the first time at which the right-continuous process ${Y}_{t}={d}_{S}({X}_{t})$ hits $0$.

If $\mathbb{P}$ is a probability measure^{} on $(\mathrm{\Omega},\mathcal{F})$ and ${\mathcal{F}}_{t}^{\mathbb{P}}$ represents the completion^{} (http://planetmath.org/CompleteMeasure) of the $\sigma $-algebra ${\mathcal{F}}_{t}$ with respect to $\mathbb{P}$, then it is enough to show that $\tau $ is an $({\mathcal{F}}_{t}^{\mathbb{P}})$-stopping time. By the universal^{} completeness of ${\mathcal{F}}_{t}$ it would then follow that

$$\{\tau \le t\}\in \bigcap _{\mathbb{P}}{\mathcal{F}}_{t}^{\mathbb{P}}={\mathcal{F}}_{t}$$ |

for every $t\in \mathbb{T}$ and, therefore, that $\tau $ is a stopping time.
So, by replacing ${\mathcal{F}}_{t}$ by ${\mathcal{F}}_{t}^{\mathbb{P}}$ if necessary, we may assume without loss of generality that ${\mathcal{F}}_{t}$ is complete^{} with respect to the probability measure $\mathbb{P}$ for each $t$.

Let $\mathcal{T}$ consist of the set of measurable times $\sigma :\mathrm{\Omega}\to \mathbb{T}\cup \{\mathrm{\infty}\}$ such that $$ for every $t$ and that $\sigma \le \tau $. Then let ${\sigma}^{*}$ be the essential supremum^{} of $\mathcal{T}$.
That is, ${\sigma}^{*}$ is the smallest (up to sets of zero probability) random variable^{} taking values in $\mathbb{R}\cup \{\pm \mathrm{\infty}\}$ such that ${\sigma}^{*}\ge \sigma $ (almost surely) for all $\sigma \in \mathcal{T}$.

Then, by the properties of the essential supremum, there is a countable^{} sequence ${\sigma}_{n}\in \mathcal{T}$ such that ${\sigma}^{*}={sup}_{n}{\sigma}_{n}$. It follows that ${\sigma}^{*}\in \mathcal{T}$.

For any $n=1,2,\mathrm{\dots}$ set

$$ |

Clearly, ${\sigma}_{1}\le \tau $ and, choosing any countable dense subset $A$ of $\mathbb{T}$, the right-continuity of $Y$ gives

$$ |

So, ${\sigma}_{1}\in \mathcal{T}$, which implies that ${\sigma}_{1}\le {\sigma}^{*}$ with probability one. However, by the right-continuity of $Y$, ${\sigma}_{1}>{\sigma}^{*}$ whenever ${\sigma}^{*}$ is finite and ${Y}_{{\sigma}^{*}}>1/n$, so

$$ |

This shows that ${Y}_{{\sigma}^{*}}=0$ and therefore ${\sigma}^{*}\ge \tau $ whenever $$. So, ${\sigma}^{*}=\tau $ almost surely and $\tau \in \mathcal{T}$ giving,

$$ |

So, $\tau $ is a stopping time.

Finally, suppose that $\mathbb{T}$ does not have a minimum element. Choosing a sequence ${t}_{n}\to -\mathrm{\infty}$ in $\mathbb{T}$ then the above argument^{} shows that

$${\tau}_{n}=inf\{t\in \mathbb{T}:t\ge {t}_{n},{Y}_{t}=0\}$$ |

are stopping times so,

$$\{\tau \le t\}=\bigcup _{n}\{{\tau}_{n}\le t\}\in {\mathcal{F}}_{t}$$ |

as required.

Title | proof of hitting times are stopping times for right-continuous processes |
---|---|

Canonical name | ProofOfHittingTimesAreStoppingTimesForRightcontinuousProcesses |

Date of creation | 2013-03-22 18:39:12 |

Last modified on | 2013-03-22 18:39:12 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 60G05 |

Classification | msc 60G40 |