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.

  • Variance of $X_2$ is $2$, i.e. $\sigma^2 t = 2$, with $\sigma^2 = 1$, rather than $\sigma^2 = 2$.
  • $\begingroup$ haha, oh geez. And I of course know that. $\endgroup$ – all.over Apr 17 at 17:34

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