Probability of Brownian motion being positive at time 1, given it's positive at time 1/2

Question

As title said, what is $\mathbb{P}(W_1>0 | W_\frac{1}{2} > 0)$ where $W$ is a standard Brownian Motion. By shading regions I arrived to 3/4, but I am not sure.

Answer 1

$$ P(W_1 > 0 | W_{\frac 12} > 0) = P((W_\frac 12) + (W_1 - W_{\frac 12}) > 0 | W_\frac 12 > 0) \\ = P(N_1 + N_2 > 0 | N_1 > 0) $$

where $N_1,N_2$ are independent $N(0,\frac 12)$ random variables. Can you do it now? See if you can use symmetry at this point as well, noting that $-N_1$ would also be identically distributed to $N_1$ and independent of $N_2$.

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Here's a complete answer, to make up for a fairly poor hint here.

Of course we first write $$ P(N_1+N_2>0|N_1>0) = \frac{P(N_1+N_2>0, N_1>0)}{P(N_1>0)} = 2P(N_1+N_2>0, N_1>0) $$

as $N_1$ is symmetric, therefore $P(N_1>0) = P(N_1<0) = \frac 12$.

For the quantity $P(N_1+N_2>0, N_1>0)$, we use the following decomposition. Either both $N_1,N_2$ are positive, or $N_1$ is positive and $N_2$ is not negative enough ,so that $N_1+N_2$ is also positive. $$ P(N_1+N_2>0, N_1>0) = P(N_1>0,N_2 > 0) + P(N_1>0 , 0>N_2 > -N_1) $$

The former quantity is $\frac 14$ using independence. For the latter quantity, use the fact that $N_2$ is symmetric to get $$ P(N_1>0 , 0>N_2 > -N_1) = P(N_1>0 , 0>-N_2 > -N_1) \\= P(N_1>0, 0<N_2<N_1) $$

Now use the fact that $N_1$ is symmetric to get $$ P(N_1>0, 0<N_2<N_1) = P(-N_1>0, 0<N_2<-N_1) = P(N_1<0, 0<N_2<-N_1) $$ followed by symmetry of $N_2$ yet again to get $$ P(N_1<0, 0<N_2<-N_1) = P(N_1<0, 0<-N_2<-N_1) = P(N_1<0, N_1<N_2<0) $$

However, observe that $$ P(N_1<0, N_1<N_2<0) + P(N_1<0, 0<N_2<-N_1)\\ + P(N_1>0 , 0>N_2 > -N_1) + P(N_1>0 , 0<N_2 < N_1) \\= P(|N_2| \leq |N_1|) $$

and $P(|N_2| \leq |N_1|) = \frac 12$ because $N_1,N_2$ are identically distributed, so that $P(|N_2| \leq |N_1|) = P(|N_2| \geq |N_1|)$. Therefore, by everything we've written so far, it follows that $$ 4 P(N_1>0 , 0>N_2 > -N_1) = \frac 12 \implies P(N_1>0 , 0>N_2 > -N_1) = \frac 18 $$

Therefore, we get that $$ P(N_1+N_2>0, N_1>0) = \frac 14+ \frac 18 = \frac 38 $$

Followed by $P(N_1+N_2>0 | N_1>0) = \frac 34$. Note that symmetry arguments are supremely helpful in answering questions related to Brownian motion increments of identical distributions, like is the case here. One may use "shading methods" to informally figure out the answer, and then implement the symmetry arguments to make things rigorous.