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For $X_t$ a brownian motion defined for $t\in[0,T]$,
How to show the equality of following events: $$ \{ \displaystyle \max_{[\tau,T]}(X_t)\ge 2u-d, \tau\leq T \}=\{\displaystyle \max_{[0,T]}(X_t)\ge 2u-d\} $$ where,

$\tau=min\{t:X_t=u\} $ and $0<d <u $

I Thank You in advance.

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  • $\begingroup$ which part you have difficulty? $\endgroup$ – Gordon Oct 12 '16 at 20:57
  • $\begingroup$ With equality, it is true? or not ? $\endgroup$ – M. A. Kacef Oct 12 '16 at 21:06
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Let $\max_{[0, T]} \left( X_t \right) = M_T$. We have $2u - d > u$ and thus $M_T \geq 2u - d$ implies that $M_T > u$. Note that $X$ is a continuous process with initial value $X_0 = 0 < u$. Then, in order for a path of $X$ to start at $X_0 = 0$ and reach a point $M_T > 2u - d$ it has to cross the level $u$ first.

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  • $\begingroup$ Thank you to answer me @LocalVolatility, I found a simple approach but I'd like your opinion. Thanks again $\endgroup$ – M. A. Kacef Oct 20 '16 at 10:42
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My idea to answer is:

Since $2u-d>u$, then

$$\{ \displaystyle\max_{[\tau,T]} (X_t)\ge 2u-d \}=\{\displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}$$ because the random variable $\tau$ is the time of the first crossing of the barrier $u$ by the process $X$.

In addition, we have the following equality: $$\{\tau \le T\}=\{\displaystyle\max_{[0,T]} (X_t)\ge u \} $$ Finally, since $$\{ \displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}\subset \{\displaystyle\max_{[0,T]} (X_t)\ge u \} $$ Therefore

$$\{ \displaystyle\max_{[\tau,T]} (X_t)\ge 2u-d, \tau \le T \}=\{\displaystyle\max_{[0,T]} (X_t)\ge 2u-d \}$$

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