Suppose $W_1(t)$ and $W_2(t)$ are two independent standard Brownian motions. What is the probability that both processes are larger than 0.667 at t=1.0?


$W_i(t) \sim N(0,t)$, at time $t=1$, $W_i(1) \sim N(0,1)$; and $\Phi(\cdot)$ is the cdf of a standard normal random variable. We compute:

$$\begin{align} \mathbb{P}(W_1(1)>0.667,W_2(1)>0.667)&=\mathbb{P}(W_1(1)>0.667)\times\mathbb{P}(W_2(1)>0.667)\\ &= (\mathbb{P}(W_1(1)>0.667))^2\\ &= \Phi(-0.667)^2\\ &\approx 0.0637 \end{align}$$

Is that it?

  • 1
    $\begingroup$ Yes, that's it!. $\endgroup$ – Kavi Rama Murthy Apr 10 at 5:38

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