$W(t)$ 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 $t=2$?

Question

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 $$( W(1), W(2) ) \sim N ( (0,0) , ( [1,1],[1,2]))$$

Answer 1

$$P(W_2<0|W_1>0)=\frac{P(W_2<0,W_1>0)}{P(W_1>0)}=2P(W_2<0,W_1>0)$$ $$\begin{aligned}P(W_2<0,W_1>0)&=E[\mathbf{1}_{\{W_2-W_1<-W_1\}\cap\{W_1>0\}}]=\\ &=E[E[\mathbf{1}_{\{W_2-W_1<-W_1\}}|W_1]\mathbf{1}_{\{W_1>0\}}]=\\ &=E[\Phi(-W_1)\mathbf{1}_{\{W_1>0\}}]=E[\Phi(-W_1)\mathbf{1}_{\{-W_1<0\}}]=\\ &=E[\Phi(-W_1)\mathbf{1}_{\{\Phi(-W_1)<1/2\}}]=\int_{(0,1/2)} x dx=\frac{1}{8}\end{aligned}$$ So $P(W_2<0|W_1>0)=1/4$

Answer 2

$W(1)$ 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:

$P(W(2)<0|W(1)>0)=1/4$.

Answer 3

$W(1)$ 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 $P(W(2) < 0 | W(1) >0) = 1/2 * 1/2 = 1/4$