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! :)
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 $W_1/M$.