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.

Answer

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.