Let $X_t$ and $Y_t$ be two independent standard Brownian motions, and for each time $t$ let $F_t$ be the $σ$−algebra consisting of all events determined by the paths $(X_s)_{s≤t}$ and $(Y_s)_{s≤t}$
(a) What is $P \{X_3 > X_2 + X_1 + 1\}$?
(b) What is $P \{X_t < 2Y_t + 1 \text{ for all } t ≤ 1\}$?
I am confused with part (a). How should i proceed with it , one solution = $P\{X_3 - X_2 - X_1 > 1\} = P\{\sqrt{3}\cdot X_1-\sqrt{2}\cdot X_1-X_1 >1\}$. But not sure if this approach is correct. Part(b) i have no clue?
Hint For b): consider $Z_t=\frac {X_t-2Y_t} {\sqrt 5}$. Then $\{Z_t\}$ is a standard Brownian motion. So the required probability is $P\{Z_t < \frac 1 { \sqrt 5} \,\text {for}\, t \leq 1\}$.Do you know what this probability is?
Your answer to a) is wrong because $X_3-X_2-X_1$ does not have the same distribution as $\sqrt 3 X_1-\sqrt 2 X_1-X_1$. In fact the former is normal with mean $0$ and variance $2$ whereas the latter is normal with mean $0$ and variance $(\sqrt 3 -\sqrt 2 -1)^{2}$. [$X_3-X_2$ and $X_1$ are independent normal variables with mean $0$ and variance $1$ each so their difference is normal with mean $0$ and variance $2$. This also tells you how to answer part a)].