Let $X_1,X_2,\ldots,X_n$ and $Y_1,Y_2,\ldots,Y_n$ be independent random samples from $N(\mu_1,1)$ and $N(\mu_2,1)$ populations respectively with $\mu_2\neq0$.
I need to find an unbiased estimator for $\rho=\frac{\mu_1}{\mu_2}$.
I've been trying to combine both distributions in different ways but haven't gotten anything interesting. Any idea how I can start?
Let's use basic statistical reasoning to simplify the problem, then solve it.
Because the $X_i$ are independent of the $Y_j$ and the former provide information only about $\mu_1$ and the latter only about $\mu_2,$ and there is an obvious unbiased estimator of $\mu_1$ (the mean of the $X_i$), it suffices to find an unbiased estimator of $1/\mu_2$ based on the $Y_j$ only. Such an estimator would be a function of a sufficient statistic, such as the mean $\bar Y$ of the $Y_j,$ which has a known variance of $1/n.$ Without loss of generality we may rescale $\bar Y$ to be a unit-variance Normal variable $Z$ with unknown mean $\mu = \mu_1 \sqrt{n}.$
According to the definition of bias, an unbiased estimator $t$ would be a function where for all $\mu\ne 0,$
$$\frac{\sqrt{2\pi}}{\mu} = \sqrt{2\pi}\,E[t(Z)] = \int_\mathbb{R} t(z) e^{-(z-\mu)^2/2}\,\mathrm{d}z.$$
The integral is a convolution: it represents a weighted average of the numbers close to $\mu.$ Consequently, as a function of $\mu$ it is asymptotically (for large $|\mu|$) close to $\mu.$ This allows us to re-express the integral in terms of the functions
$$g(z) = t(z)e^{-z^2/2};\ \check g(z) = g(-z)$$
as
$$\begin{aligned} e^{\mu^2/2}\int_\mathbb{R} t(z) e^{-(z-\mu)^2/2}\,\mathrm{d}z &= \int_0^\infty \left(g(z) e^{\mu z} + g(-z) e^{-\mu z}\right)\,\mathrm{d}z \\&=\mathscr{L}[g](-\mu) + \mathscr{L}[\check g](\mu) \end{aligned}\tag{*}$$
in terms of the Laplace transform $\mathscr{L}.$ This transform exists and is well-defined because $t,$ whence $g,$ increases slowly enough to assure convergence of the integral.
The resulting identity, which must hold for all $\mu\ne 0,$ is
$$-\frac{e^{\mu^2/2}\sqrt{2\pi}}{\mu} = \mathscr{L}[g](-\mu) + \mathscr{L}[\check g](\mu).$$
Now, this approach succeeds in finding unbiased estimates of $\mu^k$ for positive integral powers $k:$ the Laplace transform is well-defined and can be inverted. (You can check, for instance, that for $k=0,1,\ldots,4$ this method finds the functions $t(z) = $ $1,$ $z,$ $z^2-1,$ $z^3-3z,$ and $z^4 - 6z^2 + 3$ and that these are the usual unbiased estimators of $\mu^k.$)
For the present situation where $k=-1,$ the integrand in $(*)$ behaves like $1/z$ near $z\approx 0$ and therefore diverges, proving no unbiased estimator exists, QED.
