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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?

RiXaTorAgu
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  • Related Is the ratio distribution of two normally distributed variables ever normal? – user2974951 Aug 12 '22 at 06:21
  • https://en.wikipedia.org/wiki/Ratio_distribution#Normal_ratio_distributions – user2974951 Aug 12 '22 at 06:22
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    @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 '22 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... – RiXaTorAgu Aug 12 '22 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. – Sextus Empiricus Aug 13 '22 at 08:31

1 Answers1

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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.

whuber
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    wow. amazing and thanks because I was wondering how one could obtain the correct estimator but had no clue. – mlofton Aug 14 '22 at 17:38
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    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? – Sextus Empiricus Aug 14 '22 at 18:42
  • The change with the integral domain from $\int_\mathbb{R}$ to $\int_0^\infty$ is not so clear. – Sextus Empiricus Aug 14 '22 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 '22 at 00:50
  • $$ \int_\mathbb{R} t(z) e^{-(z-\mu)^2/2},\mathrm{d}z = \int_0^\infty \left(t(z) e^{-(z-\mu)^2/2}+t(-z) e^{-(-z-\mu)^2/2}\right),\mathrm{d}z = \int_0^\infty \left( t(z) e^{-(z^2+\mu^2-2\mu z)/2}+t(-z) e^{-(z^2+\mu^2+2\mu z)^2/2}\right),\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) ,\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) ,\mathrm{d}z = \int_0^\infty \left( g(z) e^{\mu z}+g(-z) e^{-\mu z} \right) ,\mathrm{d}z + \int_0^\infty \left( g(z) + g(-z) \right) e^{-\mu^2/2} ,\mathrm{d}z $$ – Sextus Empiricus Aug 15 '22 at 01:08
  • I have this additional term $\int_0^\infty \left( g(z) + g(-z) \right) e^{-\mu^2/2} ,\mathrm{d}z $ is this integral $\int_\mathbb{R} g(z) ,\mathrm{d}z$ equal to zero? – Sextus Empiricus Aug 15 '22 at 01:09
  • @Sextus I don't obtain that additional term. Please note that $e^{-\mu^2/2}$ is a factor of the original integral and therefore must be a factor of whatever you come up with. – whuber Aug 15 '22 at 01:11
  • I think I made a mistake in the comment and should use a minus $$ \int_\mathbb{R} t(z) e^{-(z-\mu)^2/2},\mathrm{d}z = \int_0^\infty \left(t(z) e^{-(z-\mu)^2/2}-t(-z) e^{-(-z-\mu)^2/2}\right),\mathrm{d}z$$ and end up with $$\int_0^\infty \left( g(z) e^{\mu z}-g(-z) e^{-\mu z} \right) ,\mathrm{d}z + \int_0^\infty \left( g(z) - g(-z) \right) e^{-\mu^2/2} ,\mathrm{d}z$$ but that is still not the same. – Sextus Empiricus Aug 15 '22 at 01:17