Question

Let \( \{ B(t) \}_{t \geq 0} \) be a standard Brownian motion and let \( \Phi(\cdot) \) be the cumulative distribution function of the standard normal distribution. If \[ P\left( \left( B(2) + 2B(3) \right)>1 \right) = 1 - \Phi\left( \frac{1}{\sqrt{\alpha}} \right), \, \alpha>0, \] then the value of \( \alpha \) (in integer) is equal to ________

Hint:

To compute probabilities involving linear combinations of Brownian motions, calculate the mean and variance of the combination and use the cumulative distribution function of the normal distribution.

Answer 1

Correct Answer: 22

From the given, \( P\left( \left( B(2) + 2B(3) \right)>1 \right) = 1 - \Phi\left( \frac{1}{\sqrt{\alpha}} \right) \). To solve for \( \alpha \), we first compute the distribution of \( B(2) + 2B(3) \). Since \( B(t) \) is a standard Brownian motion, the random variable \( B(2) + 2B(3) \) is normally distributed. The mean of \( B(2) + 2B(3) \) is 0 (because Brownian motion has a mean of 0 at all times), and the variance is given by: \[ \text{Var}(B(2) + 2B(3)) = \text{Var}(B(2)) + 4\text{Var}(B(3)) + 4\text{Cov}(B(2), B(3)). \] Since \( B(2) \) and \( B(3) \) are correlated with covariance \( \text{Cov}(B(2), B(3)) = 2 \), we have: \[ \text{Var}(B(2) + 2B(3)) = 2 + 4 \times 3 + 4 \times 2 = 2 + 12 + 8 = 22. \] Thus, \( B(2) + 2B(3) \sim N(0, 22) \). The probability is given by: \[ P\left( B(2) + 2B(3)>1 \right) = 1 - \Phi\left( \frac{1}{\sqrt{22}} \right). \] From the equation, we match this to \( 1 - \Phi\left( \frac{1}{\sqrt{\alpha}} \right) \). Hence, we find \( \alpha = 22 \). Thus, the value of \( \alpha \) is \( \boxed{22} \).