# probability question about brownian motion

Assume $W_{t}$ is a standard Brownian Motion, calculate the the probability that $W_{t}*W_{2t}$ is negative, i.e., $P(W_{t}*W_{2t}<0)$. I find it tricky to calculate the probability.Thank you.

• Looks like homework. What have you tried? You probably want to use the fact that $W$ has independant increments. – AFK May 4 '15 at 12:01
• This is conceptually simple, but you may need some tedious computations. – Gordon May 4 '15 at 12:53
• I have tried to solve like: P(W_{t}*W_{2t}<0)=P(W_{t}<0,W_{2t}>0)+P(W_{t}>0,W_{2t}<0)=P(W_{2t}>0|W_{t}<0)P(W_{t}<0)+P(W_{2t}<0|W_{t}>0)P(W_{t}>0)=P(W_{2t}-W_{t}+W_{t}>0|W_{t}<0)P(W_{t}<0)+P(W_{2t}-W_{t}+W_{t}<0|W_{t}>0)P(W_{t}>0). – cmd1991 May 4 '15 at 13:51 • I have no idea how to continue, Is there any hint? Or are there any other solutions? – cmd1991 May 4 '15 at 13:53 ## 2 Answers Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= P(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=E\Big(\mathbb{1}_{\{W_t/\sqrt{t} < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=\int_{-\infty}^0 \phi(x) \Phi(x) dx\\ &=\frac{1}{2}\Phi(x)^2\mid_{-\infty}^0\\ &=\frac{1}{8}, \end{align*} where \begin{align*} \phi(x)=\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \end{align*} is the density of a standard normal random variable, and $$\Phi(x)$$ is the cumulative distribution function. • In the first line P(W_t < 0, W_{2t} >0) = E(W_t < 0, W_{2t}-W_t > -W_t), why is probability changed to expectation? Which result are you using? – Idonknow Oct 30 '19 at 13:40 • @Idonknow: Note that P(A) = E(\pmb{1}_A). – Gordon Oct 30 '19 at 13:53 • Yes, I am aware of this equality. But there is no indicator function in E[W_t<0, W_{2t} - W_t > - W_t]. – Idonknow Oct 30 '19 at 13:55 • It is a typo. See the revision. – Gordon Oct 30 '19 at 13:56 • I see. Thanks for clarifying. – Idonknow Oct 30 '19 at 13:57 Since W_{2t}-W_{t} is independent of W_t and has the same law as W_{2t-t}=W_t we only have to computeP(X(X+Y)<0)$$where (X,Y) follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either X<0 and X+Y>0 or the opposite. The two events have the same probability since (-X,-Y)\sim (X,Y). You are left with the computation of$$ P(X<0,X+Y>0)$since the distribution of$(X,Y)$is invariant by rotation around the z-axis) this probability can be computed geometrically (think cutting a cake, the cake being the bivariate density) : it is equal to$1/8$. The final result is thus$1/4$. • Thank you very much for giving an intuitive answer. Just a bit confused that do you mean$(-X, -X-Y) \sim (X, X+Y)$instead of$(-X, -X-Y) \sim (X, Y)\$ – cmd1991 May 4 '15 at 15:53
• That was a typo sorry. – vanna May 4 '15 at 18:02