Solved – Distribution of ratio of sample means from two independent normal variables

The Question

We have a sample of size $N$ with mean $\bar{x}$ and SD $\bar{\sigma_x}$ from a random variable $X \sim \mathcal{N} (\mu, \sigma^2)$

We have a sample of size $M$ with mean $\bar{y}$ and SD $\bar{\sigma_y}$ from a random variable $Y \sim \mathcal{N} (c\mu, c^2\sigma^2)$

We wish to find estimates of $\mu$ and $c$, along with the distributions of those estimates

$\bar{x}$ is one obvious estimate of $\mu$, and we know it has a t-distribution, but using only this statistic ignores the information about $\mu$ contained in $\bar{y}$
$\frac{\bar{y}}{\bar{x}}$ would give us an estimate of $c$, but what distribution would it have? $\frac{Y}{X}$ has a Cauchy distribution, but what is the analogous distribution when using the ratio of sample means? Said another way:

Normal : t-distribution :: Cauchy : ???

Once we have an estimate of $c$, we could divide $\bar{y}$ by that estimate to get another estimate of $\mu$, and thereby extract the additional info about $\mu$ contained in $\bar{y}$. But what is the distribution of that estimate, and how to combine it with our $\bar{x}$ estimate? Things seem to be getting confusing since now we'd have a t-distribution random variable divided by our sample-equivalent-of-a-Cauchy random variable…. Is there a more straightforward way to do this?