For a standard Brownian Motion the events $\{W_{1}>0\}$ and $\{W_{2}>1\}$ are not independent

Question

I'm trying to prove that the events $\{W_{1}>0\}$ and $\{W_{2}>1\}$ are not independent, where $\{W_{t}\}_{t\geq 0}$ is a standard Brownian Motion. So,I'd like to prove that $$P(W_{1}>0,W_{2}>1)\neq P(W_{1}>0)P(W_{2}>1)$$ but I have some troubles computing $P(W_{1}>0,W_{2}>1).$

Here comes my attempt: \begin{eqnarray*} P(W_{1}>0,W_{2}>1)&=&P(W_{1}>0,W_{2}-W_{1}>1-W_{1})=E(1_{\{W_{1}>0\}}1_{\{W_{2}-W_{1}>1-W_{1}\}})\\\\ &=&E(E(1_{\{W_{1}>0\}}1_{\{W_{2}-W_{1}>1-W_{1}\}}|W_{1}))=E(1_{\{W_{1}>0\}}E(1_{\{W_{2}-W_{1}>1-W_{1}\}}|W_{1})) \end{eqnarray*}

I'm stuck in the last part; I can't see a way to compute $E(1_{\{W_{2}-W_{1}>1-W_{1}\}}|W_{1})$

Is there a easier way to compute this? Or there is an alternative way to prove such events are not independent?

Any kind of help is thanked in advanced.

Answer 1

From the definition of Brownian motion, you know that $Z_1=W_1$ and $Z_2 = W_2-W_1$ are independent $N(0,1).$

So $$ P(W_2>1\mid W_1>0) = P(Z_2>1-Z_1\mid Z_1>0).$$ It should be clear that this is larger than $P(W_2>1)=P(Z_2>1-Z_1)$ since guaranteeing $Z_1>0$ makes $Z_2>1-Z_1$ easier to satisfy (this is water-tight since they are independent).

If we really need to clinch it, we can write $$ P(Z_2>1-Z_1) = \int_{-\infty}^\infty \phi(z_1) (1-\Phi(1-z_1))dz_1 < 2\int_0^\infty \phi(z_1)(1-\Phi(1-z_1))dz_1$$ where $\phi$ and $\Phi$ are the normal PDF and CDF. The inequality holds since $\Phi(1-|z_1|)\le \Phi(1-z_1).$ But on the other hand if $Z_1'$ is a random variable independent of $Z_2$ that has the distribution of $Z_1$ conditional on $Z_1>0,$ we have $$ P(Z_2>1-Z_1\mid Z_1>0) =P(Z_2>1-Z_1') =2\int_0^\infty \phi(z_1)(1-\Phi(1-z_1))dz_1.$$ But this is really no more than a glorified rephrasing of the last paragraph.

I'm not sure if there's a good way to compute these exactly... we can easily do $P(W_1>0)=1/2$ and $P(W_2>1) = 1-\Phi(1/\sqrt{2}),$ but the computation of $P(W_2>1,W_1>0) = P(Z_2>1-Z_1,Z_1>0)$ seems an awkward region to integrate over. Rotating by 45 degrees seems the way to go.

EDIT

Indeed, rotating by 45 degrees to the independent standard normals $Z_1' = \frac{Z_1+Z_2}{\sqrt{2}}$ and $Z_2' = \frac{Z_2-Z_1}{\sqrt{2}}$ (sorry, not the same $Z_1'$ I defined before) gives an integral $$ P(W_2>1,W_1>0) = P(Z_1'>1/\sqrt{2}, Z_2' < Z_1') \\= \int_{1/\sqrt{2}}^\infty \phi(z_1') \int_{-\infty}^{z_1'} \phi(z_2')dz_2'dz_1' \\= \int_{1/\sqrt{2}}^\infty \Phi(x)\phi(x)dx \\ = \int_{\Phi(1/\sqrt{2})}^1 u du \\= \frac{1}{2}(1-\Phi(1/\sqrt{2})^2)\\= P(W_1>0)P(W_2>1)(1+\Phi(1/\sqrt(2)).$$

Answer 2

Let $\Phi$ be the CDF of the standard normal. Let $\varphi=\Phi'$ be the PDF of the standard normal (so it is also the PDF of $W_1$).

For any $N>0$, let $0=x_0<x_1<x_2<\cdots<x_M=N$ be a partition of $\left[0,N\right]$ with equi-spaced points $x_j$'s. Note that $$ \mathbb{P}\left(N\ge W_1>0,W_2>1\right)=\sum_{j=0}^{M-1}\mathbb{P}\left(W_2>1|x_{j+1}\ge W_1>x_j\right)\mathbb{P}\left(x_{j+1}\ge W_1>x_j\right). $$ For a fixed $N$, as $M\to\infty$, the last equality tends to $$ \mathbb{P}\left(N\ge W_1>0,W_2>1\right)=\int_0^N\mathbb{P}(W_2>1|W_1=x)\varphi(x){\rm d}x. $$ Let $N\to\infty$ as well, and the above equality tends to $$ \mathbb{P}\left(W_1>0,W_2>1\right)=\int_0^{\infty}\mathbb{P}(W_2>1|W_1=x)\varphi(x){\rm d}x. $$

Therefore, \begin{align} \mathbb{P}\left(W_1>0,W_2>1\right)&=\int_0^{\infty}\mathbb{P}(W_2>1|W_1=x)\varphi(x){\rm d}x\\ &=\int_0^{\infty}\mathbb{P}(W_2-W_1>1-x|W_1=x)\varphi(x){\rm d}x\\ &=\int_0^{\infty}\mathbb{P}(W_2-W_1>1-x)\varphi(x){\rm d}x\\ &=\int_0^{\infty}\mathbb{P}(W_1>1-x)\varphi(x){\rm d}x\\ &=\int_0^{\infty}\left(1-\Phi(1-x)\right)\Phi'(x){\rm d}x\\ &=\int_0^{\infty}\Phi(x-1)\Phi'(x){\rm d}x\\ &=\int_0^{\infty}\Phi(x-1){\rm d}\Phi(x). \end{align}

Answer 3

Let us fix for a while an $\epsilon>0$.

Let $a$ run in the set countable set $A=\{2\epsilon.4\epsilon,6\epsilon,\dots\}$.

Then the event $W_1>0$ is $$\bigsqcup_{a\in A}\{a-\epsilon<W_1\le a+\epsilon\}\ . $$ We split correspondingly $W_2>1$, let us fix an $a\in A$, then conditioned by $\{a-\epsilon<W_1\le a+\epsilon\}$ we have $$ \{W_2-W_1 > 1-a+\epsilon\} \subseteq \{W_2 > 1\} \subseteq \{W_2 -W_1 > 1-a-\epsilon\} \ . $$ So $\Bbb P(\ W_1>0\ ,\ W_2>1\ )$ is between the bounds $$ \sum_{a\in A} \underbrace{\left(\int_{a-\epsilon}^{a+\epsilon}\rho_1(t)\; dt\right)} _{2\epsilon\rho_1(t(a))} \cdot \underbrace{\left(\int_{1-a\pm\epsilon}^\infty\rho_1(u)\; du\right)} _{\Phi(a\mp\epsilon-1)} \ . $$ Here $\rho_1$ is the standard gaussian density, and $\Phi$ its repartition. The point $t(a)$ is an intermediate point suitably chosen in the interval $(a-\epsilon,a+\epsilon)$.

We take $\epsilon$ of the shape $\frac 1n$ and $a=\frac {2k}n$ with a running $k\in\Bbb N$, so in the limit we get a Riemannian sum, so the value $$ \begin{aligned} \Bbb P(\ W_1>0\ ,\ W_2>1\ ) &= \lim_{n\to\infty} \frac 2n\sum_{k>0} \rho_1\left(\frac{2k}n\right)\cdot \Phi\left(\frac{2k}n-1\right) \\ &= \int_0^\infty \rho_1(x)\cdot \Phi(x-1)\; dx \\ &\approx 0.21101\ . \end{aligned} $$ (One may write directly the integral.)

This was the complicated side. The product $\Bbb P(\ W_1>0\ )\cdot \Bbb P(\ W_2>1\ )$ is simpler. The first factor is $\frac 12$. The second factor is approx. $0.23975$. (Late in the morning, hope it's ok.)