Density Function of Random Variable Related to Brownian Motion 
Above is my question. I've done the first two parts, that's no problem. I'm stuck on finding the density of the rv $R = W_1 / M$. I have got as far as
$$g(x,y) = \frac{\partial^2}{\partial x \partial y} \Bbb P(M \le x, W_1 \le y) = \frac{2(2x-y)}{\sqrt{2\pi}}\exp \left(-\frac{1}{2}(2x - y)^2 \right)$$
for $x \ge y$ and $g(x,y) = 0$ for $x \le y$. I also have
$$ \Bbb P(R \ge r) = \Bbb P (W_1 \ge rM),$$
and so I'm now interested in evaluating a probability of the form "$\Bbb P(X \ge Y)$" for rvs $X = W_1$ and $Y = rM$. My attempt is then
$$\Bbb P(W_1 \le rM) = \int_{y < x} \frac{1}{r} g\left(\frac{x}{r},y\right) d(x,y),$$
where the $1/r$ factors come from the fact that $Y = rM$, not just $M$.
My issue is that this then gives $1$, for each $r$. So my density function is $1$ everywhere... which is of course not right (eg doesn't integrate to $1$). Of course, I only want to consider $r < 1$ as the probability is $0$ for $r > 1$, by definition of the two rvs. Also, I am concerned with the $1/r$ factors... when $r$ could be $0$.
Any advice would be most appreciated. Thanks.
PS - This is a probability question - not a finance question - so please don't suggest migrating it to quant.SE - thanks! :)
 A: I don't see anything wrong in your calulations.
Integration yields
$$\begin{align*} \mathbb{P}(W_1 \leq r M) &= \frac{1}{r} \int_{\mathbb{R}} \int_{y<x} \frac{2(2x/r-y)}{\sqrt{2\pi}} \exp \left(- \frac{1}{2} (2x/r-y)^2 \right) \, dy \, dx \\ &= \frac{1}{r} \frac{2}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left(- \frac{(2x/r-x)^2}{2} \right) \, dx \\ &= \frac{1}{r} \frac{2}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left( -\frac{x^2 (2/r-1)^2}{2} \right) \, dx\end{align*}$$
Using the identity $$\int_{[0,\infty)} \exp \left(- \frac{x^2}{2\sigma^2} \right) \, dx = \frac{1}{2} \int \exp \left(- \frac{x^2}{2\sigma^2} \right) \, dx = \frac{1}{2} \sqrt{2\pi \sigma^2},$$ we get
$$\begin{align*} \mathbb{P}(W_1 \leq r M) = \frac{1}{r} \frac{1}{|2/r-1|} = \frac{1}{2-r} \tag{1} \end{align*}$$
for any $r \leq 1$, $r \neq 0$. Since
$$\mathbb{P} \left( \frac{W_1}{M} \leq 0 \right) = \mathbb{P}(W_1 \leq 0) = \frac{1}{2},$$
$(1)$ holds also for $r=0$. Differentiating $(1)$ with respect to $r$ gives the density of $W_1/M$.
