$(a)$ Calculate $P(X_5>0|W_3=1)$ given that $\{W_t\}$ is an Simple Brownian Motion and $\{X_t\}$ a stochastic process

Question

Let $\{W_t\}$ be an SBM and define the process $\{X_t\}$ by

$X_t= {W_t}^{2}-t ,\ \ t\geq0.$

$(a)$ Calculate $P(X_5>0|W_3=1)$

MY WORKING

$a)$ I know that since $\{W_t\} $ is simple brownian motion that means $W_t$~$N(0,t)$ for $t\geq0$ which also means that $X_t$ is also normal random variable since it depends upton $W_t$. I also know the required result in part $a)$ is that of conditional probability but I can't solve it since it involves long expression of probability distributions of normal random variables $W_t$ and $X_t$ OR May be I am making a mistake. I don't know. I get stuck and confused at the beginning step. Any guide/help will be appreciated. Thanks

Answer 1

Hint

\begin{align*} \mathbb P\{X_5>0\mid W_3=1\}&=\mathbb P\{W_5^2>5\mid W_3=1\}\\ &=\mathbb P\{(W_5-W_3)^2+2(W_5-W_3)W_3+W_3^2>5\mid W_3=1\}\\ &=\mathbb P\{(W_5-W_3)^2+2(W_5-W_3)+1>5\}, \end{align*} where in the last equality we used the fact that $W_5-W_3$ and $W_3$ are independents. I let you conclude.