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Assume that all processes to be considered are regular (say cadlag). Assume $X^1$ and $X^2$ are stochastic processes such that $X^1_t = X^2_t$, that $Y^1$ and $Y^2$ are processes such that $Y^1_t=Y^2_t$ and that $S$ is a processes independent of both $Y$'s (not the X's). Is it immediately true that if $$X^1_t = S_t Y^1_t$$ then $$X_t^2 = S_t Y^2_t$$ or are there counterexamples?

htd
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  • when you write $X^1_t=X^2_t$ do you mean for every $\omega$, almost surely, in law ? Best regards. – TheBridge May 20 '16 at 11:22
  • @TheBridge I would be interested in both in law and a.s., but initially I meant in law. – htd May 20 '16 at 11:37

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In the case of equality in law then the answer is clear from the independence hypothesis you have formally : $P(X^1_t=S_t.Y_t^1<a)=P(Y_t^1<a/s).P(S_t<s)=P(Y_t^2<a/s).P(S_t<s)=P(S_t.Y_t^2<a)$ and as $P(X^1_t<a)=P(X_t^2<a)$ this leads to the conclusion, $P(X_t^2<a)=P(S_t.Y_t^2<a)$.

For the case of almost sure equality (or perfect equality) you have almost surely, for a set $A$ of probability $1$ where both $X^1_t=X^2_t$ and $Y^1_t=Y^2_t$ , for any $\omega \in A$ : $X_t^1(\omega)=Y_t^1(\omega).S_t(\omega)=Y_t^2(\omega).S_t(\omega)$ but you also have : $X_t^1(\omega)=X_t^2(\omega)$ so $X_t^2(\omega)=Y_t^2(\omega).S_t(\omega)$

best regards

TheBridge
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