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Answered

2022-07-03

Setting We work on a filtered probability space with finite time horizon $T$. The filtration is assumed to be complete. Let $X$ be a stochastic process that satisfies a property (A) a.s. For example, if property (A) is being nonnegative at the time $T$, then $X$ satisfies $P[{X}_{T}\ge 0]=1$.

Question I would like to obtain that $X$ (up to indistinguishability) satisfies property (A) for each $\omega $.

My attempt Define a process $Y$ to be equal to 0 on the event $\{{X}_{T}<0\}$ and define it to be equal to $X$ on $\{{X}_{T}\ge 0\}$. Then ${Y}_{T}\ge 0$. Is it then correct that $X,Y$ are indistinguishable, i.e. $P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]=1$? Is this approach possible whenever the event where the stochastic process does not satisfy the desired propert has probability 0?

Question I would like to obtain that $X$ (up to indistinguishability) satisfies property (A) for each $\omega $.

My attempt Define a process $Y$ to be equal to 0 on the event $\{{X}_{T}<0\}$ and define it to be equal to $X$ on $\{{X}_{T}\ge 0\}$. Then ${Y}_{T}\ge 0$. Is it then correct that $X,Y$ are indistinguishable, i.e. $P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]=1$? Is this approach possible whenever the event where the stochastic process does not satisfy the desired propert has probability 0?

Answer & Explanation

Elias Flores

Expert

2022-07-04Added 24 answers

Well if you modify $X$ only at time $T$ over an event of 0 mass, yes all that happens strictly before $T$ is equal to $X$ for every $\omega $ and only for the path which are negative at time $T$ you have to modify your process so yes $X$ and $Y$ are indistinguishable. Formally:

$=\{{X}_{T}={Y}_{T}\}\cup \{{X}_{T}\ne {Y}_{T}\}$={XT=YT}∪{XT≠YT}

So as $P[{X}_{T}={Y}_{T}]=1=P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]$

$=\{{X}_{T}={Y}_{T}\}\cup \{{X}_{T}\ne {Y}_{T}\}$={XT=YT}∪{XT≠YT}

So as $P[{X}_{T}={Y}_{T}]=1=P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]$

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