Let $(W_t)_{t\ge0}$ be a Standard Brownian Motion. I would like to compute the conditional probability
$$P\left(\inf_{t\in [1,2]} W_t < 0 ~ \middle| ~ W_1 >0,~ W_2 >0\right).$$
By using the reflection principle twice, I obtain the result $1\over 3$. Can anyone verify my answer?
My computation is the following:
$$P\left(\inf_{t\in [1,2]} W_t < 0 ~ \middle| ~ W_1 >0,~ W_2 >0\right)~=~{P\left(\inf_{t\in [1,2]} W_t < 0 , W_1 >0, W_2 >0 \right)\over P\left(W_1>0, W_2>0\right)}.$$
Frist compute the denominator:
$$P\left(W_1>0, W_2>0\right)~=~\int_0^\infty {1\over \sqrt{2\pi}}e^{-{x^2\over 2}} \int_{-x}^\infty {1\over \sqrt{2\pi}} e^{-{y^2\over 2}}\, dy\, dx~=~{3\over 8}.$$
Now consider the numerator:
$$P\left(\inf_{t\in [1,2]} W_t < 0 , W_1 >0, W_2 >0 \right)$$
$$~=~\int_0^\infty P\left(\inf_{t\in [1,2]} W_t < 0 , W_2 >0 \middle | W_1 = x\right){1\over \sqrt{2\pi}}e^{-{x^2\over 2}} \, dx$$
For the conditional density, from reflection principle,
$$P\left(\inf_{t\in [1,2]} W_t < 0 , W_2 >0 \middle | W_1 = x\right) ~=~{1\over 2} P\left( \inf_{t\in [1,2]} W_t < 0 ~|~ W_1 = x \right) $$
Use reflection principle again,
$$P\left( \inf_{t\in [1,2]} W_t < 0 ~|~ W_1 = x \right) ~=~2 P\left( W_2<0~ \middle | ~ W_1 = x \right) $$
$$ = ~ 2 \int_{-\infty}^{-x} {1\over \sqrt{2\pi}}e^{-{y^2\over 2}}\, dy$$.
Combine all equation, I got the answer $1\over 3$.
Best Answer
By reflection principle, $$ \mathbb{P}(\min_{t\in [1,2]} W_t < 0, W_1 >0 , W_2>0) = \mathbb{P}(W_1 > 0, W_2 <0). $$ Now, since $\mathbb{P}(W_1 > 0, W_2 <0) + \mathbb{P}(W_1 > 0, W_2 >0) = \frac{1}{2}$, we are done if we know the value of $\mathbb{P}(W_1 > 0, W_2 >0)$. Can you compute this probability?