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Let $(X_t)$ and $(B_t)$ two stochastic processes and $0\leq h\leq t$. Do we always have

$$p\left\{X_t\in A\mid \min_{0\leq u\leq t}B_u>a\right\}=p\left\{X_t\in A,\min_{0\leq u\leq h} B_u>a\mid \min_{h\leq u\leq t}B_u>a\right\}\ \ \ ?$$

I have that $$p\left\{X_t\in A,\min_{0\leq u\leq h} B_u>a\mid \min_{h\leq u\leq t}B_u>a\right\}=\frac{p\left\{X_t\in A, \min_{0\leq u\leq h} B_u>a, \min_{h\leq u\leq t} B_u>a\right\}}{p\left\{\min_{h\leq u\leq t} B_u>a\right\}}=\frac{p\left\{X_t\in A, \min_{0\leq u\leq t} B_u>a\right\}}{p\left\{\min_{h\leq u\leq t} B_u>a\right\}}$$

But I can't get $$p\left\{\min_{h\leq u\leq t} B_u>a\right\}=p\left\{\min_{0\leq u\leq t} B_u>a\right\},$$ therefore I have problem to conclude.

idm
  • 11,824

1 Answers1

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I'd say no. Your derivation looks correct to me and you arrive at $$p\left\{\min_{h\leq u\leq t} B_u>a\right\}=p\left\{\min_{0\leq u\leq t} B_u>a\right\}$$ This can't be true for all $0 \leq h \leq t$. Take e.g. $h = t$. $$p\left\{B_t>a\right\} = p\left\{\min_{t\leq u\leq t} B_u>a\right\}=p\left\{\min_{0\leq u\leq t} B_u>a\right\}$$ But for any $0 \leq t' \leq t$ it is true that $$p\left\{\min_{0\leq u\leq t} B_u>a\right\} \leq p\left\{B_{t'}>a, B_t>a\right\} < p\left\{B_t>a\right\}$$ That gives us a contradiction.

muaddib
  • 8,267