# Clarification: Brownian Motion Reflection Principle

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 $$\sigma^2 = 2$$.