Does there exist an explicit formula for the moment generating function $\psi(u, v) = E e^{u W_T + v M_T}$ of the pair $(W_T, M_T)$ where $M_T = \max_{0\leq t\leq T} W_t$?
Using the well-known pdf of the pair, the mgf can be expressed as
$$\psi(u,v) = \int_0^\infty\int_{-\infty}^y \exp\left(u x + v y\right) \frac{2(2y-x)}{\sqrt{2\pi T^3}}\exp\left(-\frac{(2y-x)^2}{2T}\right) dx dy$$
Answer:
Using the expression of $\psi$ above and the change of variable $z = 2y-x$, I managed to get the following expression from Wolfram Alpha:
\begin{align} \psi(u,v) &=2\Phi\left(\sqrt{T}(u+v)\right) e^{\frac{1}{2}T(u+v)^2}\frac{u+v}{2u+v}\\ &\quad+\Phi\left(\frac{\sqrt{T}v}{2}\right) e^{\frac{1}{8}Tv^2}\left(\frac{2u e^{\frac{1}{4}Tu(2u+v)}}{2u+v}-1\right) \end{align} where $\Phi$ is the standard normal distribution function.
To make this work, I had to first calculate the inner integral (after changing variables) using Wolfram Alpha, and then calculate the outer integral. Finally, the result was expressed in terms of the error function instead of $\Phi$, so I had to rewrite that myself.
Note that Wolfram Alpha was not able to calculate the integrals directly. Instead, I had to first calculate the indefinite integral, and then apply the limits myself. Also, I calculated some of the integrals one summand at a time, though I don't know if that was necessary.
Update: The "answer" I stated previously was wrong. The new answer is double-checked and correct. It is much nicer to look at as well.:)