How to solve this equation $\mathbb{P}\{ W_1 > 2 | W_4 = 4\}$?

Question

Assume that $W_t$ is a standard Wiener (Brownian motion) process. Calculate:

$$\mathbb{P} \{ W_1 > 2 | W_4 = 4 \}$$

This my way of solving this problem but I am not sure is it right: ^src)

\begin{align*} &\mathbb{P}\{ W_1 > 2 | W_4 = 4\} \\ &= \mathbb{P}\{ W_0 > 0 | W_3 = 2\} = \mathbb{P}\{ W_3 = 2 \} \\ &= \mathbb{P}\{ N(0, 3) > 2 \} = \mathbb{P}\{ \sqrt{3} \cdot N(0, 1) > 2 \} = \mathbb{P}\left\{ N(0, 1) > \tfrac{2}{\sqrt{3}} \right\} \\ &= 1 - \mathbb{P}\left\{ N(0, 1) \leq \tfrac{2}{\sqrt{3}} \right\} = 1 - \Phi\left(\tfrac{2}{\sqrt{3}}\right) \end{align*}

Answer 1

You can use this formula: W(s)|W(t) = a ~ (s/t * a; s (1-s/t))

P{W1>2|W4=4} =

P(N(1/4 * 4, 1 (1 - 1/4)) =

P(N(1, 3/4) > 2) =

1 - P(N(1, 3/4) <=

1 - P(N(0, 3/4) <= 1) =

1 - P(sqrt(3/4) N(0, 1) <= 1) =

1 - P(N(0, 1) <= 2/sqrt3) =

1 - Φ(2/sqrt3)

Answer 2

I'm getting the same answer with a slightly different approach. There are also some notable issues in your calculation; note that $\mathbb{P}(W_3=2)=0$.

We have $W_4|W_1\sim \mathcal{N}(W_1,3)$ while $W_1\sim \mathcal{N}(0,1)$. Using Baye's theorem we have $$f_{W_1|W_4}(x|y)=\frac{f_{W_4|W_1}(y|x)f_{W_1}(x)}{\int_{\mathbb{R}}f_{W_4|W_1}(y|x)f_{W_1}(x)\mathrm{d}x}=\sqrt{\frac{2}{3\pi}} \exp \Bigg\{-\frac{2}{3}\left(x-\frac{y}{4}\right)^{2}\Bigg\}$$ Evidently $W_1|W_4\sim \mathcal{N}\left(\frac{W_4}{4},\frac{3}{4}\right)$ so that $$\mathbb{P}\left(W_1>2|W_4=4\right)=\mathbb{P}\left(Z>\frac{2-1}{\sqrt{3}/2}\right)=1-\Phi\left(\frac{2}{\sqrt{3}}\right)$$