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If $B$ is the standard brownian motion and $a,b >0$ I want to show, using the reflection principle

$$\mathbb{P}\left(B_t\geq a-b | \inf_{s\leq t} B_s \geq -b\right) = \frac{\mathbb P(|B_t+x|\leq b)}{\mathbb P(|B_t|\leq b)}$$

and the reflection principle that I know is for $x>0$ and $y<x$:

$$\mathbb P\left(\sup_{s\leq t} B_s \geq x, B_t < y\right) = \mathbb P\left(B_t > 2x-y\right)$$

Now I wrote the conditional probability

$$\mathbb{P}\left(B_t\geq a-b | \inf_{s\leq t} B_s \geq -b\right) = \frac{\mathbb{P}(B_t\geq a-b , \inf_{s\leq t} B_s \geq -b)}{\mathbb{P}(\inf_{s\leq t} B_s \geq -b)}$$

and I rewrote the numerator, trying to get something like the reflection principle

$$\mathbb P(B_t\geq a-b , \inf_{s\leq t} B_s \geq -b)=\mathbb P\left(-\sup_{s\leq t} (-B_s) \geq -b, B_t \geq a-b\right)=\\ = \mathbb P\left(\sup_{s\leq t}(-B_s)\leq b, B_t \geq a-b\right)$$

but I don't know how to procede, it does not look like what I need...

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