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for a practice problem sheet I was asked to compute $\mathbb{P}(B_T>0, B_{2T} < 0)$ for $T>0$ and a Brownian Motion $B$.

I initially thought that since $B_T$ is independent of $B_{2T}$, and each $B_T$ is distributed $ \mathcal{N}(0, t)$, the computation would be:

$\mathbb{P}(B_T>0, B_{2T} < 0) = \mathbb{P}(B_T > 0)\mathbb{P}(B_{2T}<0) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$.

However, I was told that the answer is $\frac{1}{8}$.

What am I missing? What is another way of going about this problem?

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    $\begingroup$ $B_T$ is not independent of $B_{2T}$, though $B_T$ is independent of $B_{2T}-B_T$ $\endgroup$
    – Henry
    Commented Feb 7, 2020 at 18:01

1 Answer 1

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The two events are not independent. Here's one way to think about it: Let $X$ and $Y$ be independent normals with mean $0$ and variance $T$. Then the pair $(B_T,B_{2T})$ is distributed like $(X,X+Y)$. Then you are looking for $\mathbb{P}(X > 0 , X+ Y < 0)$. Try plotting this event in the $xy$-plane and see what its probability is in this case.

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  • $\begingroup$ How do I go about computing this though? And why are these events not independent? Isn't $B_s$ independent of $B_t$ for $s<t$? $\endgroup$
    – arnavlohe15
    Commented Feb 7, 2020 at 18:14
  • $\begingroup$ That is not correct. $B_s$ and $B_{t - s}$ are independent for $s < t$. To see how to compute this event, try my hint: draw this event in the $xy$-plane. $\endgroup$
    – Marcus M
    Commented Feb 7, 2020 at 18:16
  • $\begingroup$ Got it. Thank you so much. $\endgroup$
    – arnavlohe15
    Commented Feb 7, 2020 at 18:22

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