The right-hand side of formula (3.1.5) reads, on the set \(y \geq 0, y-x \geq 0\),

\[\frac{\mathbb{P}\left(T_{y-x} \in d t\right)}{d t} d x d y=\frac{2 y-x}{t} p_{t}(2 y-x) d x d y\]

Check simply that this probability has total mass equal to 1!

Formula 3.1.5:

\[\mathbb{P}\left(W_{t} \in d x, M_{t} \in d y\right)=\mathbb{1}_{\{y \geq 0\}} \mathbb{1}_{\{x \leq y\}} \frac{2(2 y-x)}{\sqrt{2 \pi t^{3}}} \exp \left(-\frac{(2 y-x)^{2}}{2 t}\right) d x d y\]