$W(t)$ is a standard Brownian motion. Conditional on the event that $W(t)$ is positive at $t=1$, what is the probability that it is negative at $t=2$? I can't find the right numerical result for $\mathbb{P}[W(2)<0|W(1) > 0]$.
$$\mathbb{P}[W(2)<0|W(1) > 0] = \mathbb{P}[W(2)<0, W(1) > 0 ] / \mathbb{P}[W(1) > 0]$$
$W(1) \sim N(0,1)$ and
$$( W(1), W(2) ) \sim N ( (0,0) , ( [1,1],[1,2]))$$
 A: $$P(W_2<0|W_1>0)=\frac{P(W_2<0,W_1>0)}{P(W_1>0)}=2P(W_2<0,W_1>0)$$
$$\begin{aligned}P(W_2<0,W_1>0)&=E[\mathbf{1}_{\{W_2-W_1<-W_1\}\cap\{W_1>0\}}]=\\
&=E[E[\mathbf{1}_{\{W_2-W_1<-W_1\}}|W_1]\mathbf{1}_{\{W_1>0\}}]=\\
&=E[\Phi(-W_1)\mathbf{1}_{\{W_1>0\}}]=E[\Phi(-W_1)\mathbf{1}_{\{-W_1<0\}}]=\\
&=E[\Phi(-W_1)\mathbf{1}_{\{\Phi(-W_1)<1/2\}}]=\int_{(0,1/2)} x dx=\frac{1}{8}\end{aligned}$$
So $P(W_2<0|W_1>0)=1/4$
A: $W(1)$ and $W(2)-W(1)$ are independent and with the same distribution $p(x)=N(0,1)$.
Now therefore since $W(2)-W(1)$ has to be smaller than $-W(1)$ so that can change sign:
$P(W(2)<0,W(1)>0)=\int_0^{+\infty} p(x) \int_{-{\infty}}^{-x}p(x')dx'$
call $F(x)$ the c.d.f. of $p(x)$. Than this can be rewritten:
$\int_0^{+\infty} F'(x) F(-x)dx=\int_0^{+\infty} F'(x) (1-F(x))dx=
\int_0^{+\infty} F'(x)dx-\int_0^{+\infty} F'(x) F(x)dx=$
$=F(x)|_0^{+\infty}-1/2F^2(x)|_0^{+\infty}=1/2-3/8=1/8$
Instead:
$P(W(1)>0)=1/2$
, leading to:
$P(W(2)<0|W(1)>0)=1/4$.
A: $W(1)$ and $W(2)-W(1)$ are independent and with the same distribution, and moreover their absolute values are independent from their signs (normal distributions are symmetric with respect to 0)
If you know $A : W(1) > 0$, then for $W(2)$ to be negative you want
B: $W(2) - W(1) < 0$ and
C: $|W(2) - W(1)| > |W(1)|$
The events $A,B,C$ are mutually independent, so
$P(W(2) < 0 | W(1) = 0) = P(A \land B \land C)/P(A) = P(A)P(B)P(C)/P(A) = P(B)P(C)$
$P(B) = 1/2$, and $P(C) = 1/2$ because they have the same distribution.
and so $P(W(2) < 0 | W(1) >0) = 1/2 * 1/2 = 1/4$
