= Solution
A standard <Brownian motion> starts at zero, has continuous sample paths, independent stationary increments, and $W_t-W_s\sim\mathcal N(0,t-s)$ for $t>s$. The <Brownian reflection principle> reflects the increments after the first hitting time of a positive level and preserves Brownian law, by symmetry and the strong Markov property.
Reflect after the first hit of $x$. On the event that this hit occurs by time one, $W_1$ changes to $2x-W_1$. This is a measure-preserving involution taking the event $W_1\leq x-y$ to $W_1\geq x+y$, proving the specified identity for $x>0,y\geq0$; the $x=0$ case follows by symmetry. With $y=0$, the maximum event splits, up to zero-probability equality, into reflected equal halves. Thus
$$
\mathbb P(M\geq x)=2\mathbb P(W_1\geq x)=\mathbb P(|W_1|\geq x),\qquad x\geq0.
$$
Hence \b[$M$ and $|W_1|$ have the same distribution], and $M>0$ almost surely.
For the ratio we need the joint law, not just these marginals. For $w<m$, the reflected identity gives $\mathbb P(M\geq m,W_1\leq w)=\mathbb P(W_1\geq2m-w)$. Differentiating yields
$$
f_{M,W_1}(m,w)=2(2m-w)\varphi(2m-w),\qquad m>0,\quad w<m,
$$
where $\varphi$ is the standard normal density. Put $w=rm$; the change of variables has Jacobian $m$. Therefore, for $r<1$,
$$
f_R(r)=\int_0^\infty2(2-r)m^2\varphi((2-r)m)\,dm
=\frac{2}{(2-r)^2}\int_0^\infty v^2\varphi(v)\,dv
=\boxed{\frac1{(2-r)^2}}.
$$
The density is zero for $r>1$ and has no atom at one. Its integral on $(-\infty,1)$ is one; negative ratios have no lower bound.
This gives the complete <Brownian terminal-to-maximum ratio> law.
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