# Unbiased estimator for $\mu_1/\mu_2$

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?

• Aug 12 at 6:21
• en.wikipedia.org/wiki/… Aug 12 at 6:22
• @user2974951 Those references might lead readers to confuse the random variables $X_i/Y_i$ with the ratio of means $\mu_1/\mu_2.$ What the question seeks is an estimator $t_n(X_1,\ldots, Y_n)$ for which $E[t_n(X_1,\ldots,Y_n)]=\rho.$ What is relevant about your points is that it would be hopeless to construct $t_n$ out of ratios of the variables or of their sample means, suggesting an unbiased estimator does not exist. When we couple that observation with the idea of using a minimal sufficient statistic, which obviously is $(\bar X,\bar Y),$ we're well on the way to a (negative) solution.
– whuber
Aug 12 at 13:31
• @whuber I was trying to build a known distribution from the samples, like a Gamma or something like that but so far I haven't achieved anything...
– ATXR
Aug 12 at 14:11
• @whuber the references might not be so bad. For many practical cases the distribution of $X/Y$ might not need to be like a pathological distribution without moments (this happens when $Y$ is only approximately normal distributed and does not approach zero unlike the normal distribution by which it is approximated). In such practical cases an approximation with a normal distribution does not need to be so bad. Aug 13 at 8:31

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.

• wow. amazing and thanks because I was wondering how one could obtain the correct estimator but had no clue. Aug 14 at 17:38
• This is a nice answer, but at the points where it states that an unbiased estimator must be a function of the sufficient statistic it goes too fast for me. It seems intuitive, but is there also some theorem or question/proof that states that an unbiased estimator does not exist if we can not construct an unbiased estimator based on the sufficient statistic? Aug 14 at 18:42
• The change with the integral domain from $\int_\mathbb{R}$ to $\int_0^\infty$ is not so clear. Aug 14 at 18:52
• @Sextus (1) Rao-Blackwellization, (2) Split into positive and negative parts and change the variable in the negative part.
– whuber
Aug 15 at 0:50
• $$\int_\mathbb{R} t(z) e^{-(z-\mu)^2/2}\,\mathrm{d}z = \int_0^\infty \left(t(z) e^{-(z-\mu)^2/2}+t(-z) e^{-(-z-\mu)^2/2}\right)\,\mathrm{d}z = \int_0^\infty \left( t(z) e^{-(z^2+\mu^2-2\mu z)/2}+t(-z) e^{-(z^2+\mu^2+2\mu z)^2/2}\right)\,\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) \,\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) \,\mathrm{d}z = \int_0^\infty \left( g(z) e^{\mu z}+g(-z) e^{-\mu z} \right) \,\mathrm{d}z + \int_0^\infty \left( g(z) + g(-z) \right) e^{-\mu^2/2} \,\mathrm{d}z$$ Aug 15 at 1:08