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Why is $f_{M_t,W_t}(m,w) = \frac{2 ( 2 m - w)}{t\sqrt{2 \pi t}} e^{-\frac{(2m-w)^2}{2t}}, m \ge 0, w \leq m$ ? I now know what running maximum is, but unsure why joint distribution goes as above formula.

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This is a consequence of the reflection principle: $$ \mathbb{P}\left(B_t \leqslant x, M_t \geqslant y\right) = \mathbb{P}\left(B_t \leqslant x, T_y \leqslant t \right) \stackrel{\text{refl.pr.}}{=} \mathbb{P}\left(\hat{B}_t \geqslant 2y-x, T_y \leqslant t \right) = \mathbb{P}\left(B_t \geqslant 2 y -x, T_y \leqslant y\right) $$ where the last equality holds because the hitting time $T_y$ is the same for $B_t$ and the reflected Brownian motion $\hat{B}_t$. Since $y>x$ and $\{B_t \geqslant 2y-x\} \subset \{T_t \leqslant t\}$ we have $$ \mathbb{P}\left(B_t \leqslant x, M_t \geqslant y\right) = \mathbb{P}\left(B_t \geqslant 2 y -x\right) $$ It now remains to differentiate to recover the probability density function.

Added: See F.C. Klebaner, "Introduction to Stochastic Calculus with Applications", 3rd ed., ch 3.6 for more details.

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