Preliminaries : My question follow the proof of the Reflection principle that you can consult on wikipedia (I do not copy/paste for the sake of the global understanding of my question). Every notations are taken from the wikipedia link, except this one $X(t)$ that I type $X_t$.
I want to show : $$ \mathbb{P}\left(\tau_a \leq t, X_{t-\tau_a} < 0\right) \overset{?}= \mathbb{P}\left(\tau_a \leq t, X_{t-\tau_a} > 0\right) \ \ \ (*) $$
I do not understand why my brief proof could not work ? I detail.
My proof : For proving that I have simply said : $(X_t)_{t\geq 0}$ is a brownian motion so it is equal by law to $(-X_t)_{t\geq 0}$. Moreover $\tau_a$ and $X$ are independant so : $\mathbb P_{(\tau_a,X)}=\mathbb P_{(\tau_a,-X)}$
Versus
Common proof : Every proofs I have seen about, prove (*) like that :
Knowing that $X$ and $\tau_a$ are independant, it exists $F:\mathbb R_+\times C^0(\mathbb R_+,\mathbb R)\rightarrow \mathbb R$, bounded, borelian : \begin{align} \mathbb{P}\left(\tau_a \leq t, X(t-\tau_a) < 0\right) &= \mathbb E \left(F(\tau_a,X)\right)\\ &=\phi(\tau_a) \end{align}
such that : $$ \forall s\in\mathbb R_+,\ \phi(s)=\mathbb P(s\leq t, X_{t-s}<0) $$
Since $-X$ is a brownian motion we can conclude : $$ \forall s\in\mathbb R_+,\ \phi(s)=\mathbb P(s\leq t, X_{t-s}>0) $$ That achieve the common proof.
Question:
I understand the common proof but my question is why we cannot do, as simply as it is, mine ?
Why we do the common proof rather than my basic idea ?
- Does my brief and basic idea work actually ?
- Is it something flawed, missed in my basic proof ?