Two looks at one path · Problem 181

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. Split $W_{2}$ into $W_{1}$ plus an independent increment. Then $W_{1} + W_{2}$ is a sum of independent normals, and its variance is easy.

Step 1. $W_{1} + W_{2} = 2 W_{1} + (W_{2} - W_{1})$ . Here $W_{1} \sim N (0, 1)$ and $W_{2} - W_{1} \sim N (0, 1)$ are independent.

Step 2. So $W_{1} + W_{2} \sim N (0, 4 + 1) = N (0, 5)$ .

Step 3. $P (W_{1} + W_{2} > 2) = 1 - Φ (\frac{2}{5}) = 1 - Φ (0.8944) \approx 0.186.$

The trap. Adding the variances $1 + 2 = 3$ as if $W_{1}$ and $W_{2}$ were independent. They share the path up to time 1, so $Cov (W_{1}, W_{2}) = 1$ and the variance is $1 + 2 + 2 \cdot 1 = 5$ .