The Brownian Motion Reflection Principle gives: For $X_t$ BM starting at $a$ and $b>0$
$\displaystyle P(X_s \ge b, 0 \le s \le t) = 2P(X_t \ge b | X_0 = a) = 2\int_b^\infty \frac{1}{\sqrt{2\pi t \sigma^2}}\exp(\frac{-(x-a)^2}{2t\sigma^2})dx$
Q: Calculate $P(X_2 > 2)$
There is a conflict in my answer though, using the above I will have
$X_2 \sim N(0,2)$ so $P(X_2 > 2) = 2\int_2^\infty \frac{1}{\sqrt{2\pi \cdot 2 \cdot 2^2}}\exp(\frac{-(x)^2}{2\cdot2 \cdot 2^2})dx=\int_b^\infty \frac{1}{2\sqrt{\pi}}\exp(\frac{-(x)^2}{16}dx$
Whereas the correct answer is
$\int_b^\infty \frac{1}{2\sqrt{\pi}}\exp(\frac{-(x)^2}{4})dx$
What did I miss to get the factor of $\frac{1}{4}$ rather than $\frac{1}{16}$ in the exponential.
Best Answer