# Sampling Distribution of Reciprocal of Sample Mean

Given $$X_1, X_2,..., X_n$$ i.i.d. random variables.

$$E[X_i] = \mu_1 \in \mathbb{R}$$ $$\&$$ $$V[X_i] = \sigma_1^2 \in \mathbb{R}^+$$ $$\forall i \in \{1,2,3,...,n\}$$.

The statistics $$\bar{X} = \frac{X_1 + X_2 + ... + X_n}{n}$$ converges in distribution to $$Z_1 \sim N(\mu_1, \frac{\sigma_1^2}{n}$$) according to the Central Limit Theorem.

Does $$\frac{1}{\bar{X}}$$ converge in distribution to $$\frac{1}{Z_1}$$?

I guess the answer is NO, but I decided to ask the question here for more insights. A simple yes/no answer without any explanation would be appreciated too.

The reason I ask this question is relevant to ratio estimator in survey sampling as I read in some books that we could assume the ratio estimator $$\hat{R} = \frac{\bar{X}}{\bar{Y}}$$ to be normally distributed.

If I am not mistaken, if $$X_n$$ converges in distribution to $$X$$, and $$Y_n$$ converges in distribution to $$Y$$, then $$X_nY_n$$ converges in distribution to $$XY$$

Given $$Y_1, Y_2,..., Y_n$$ i.i.d. random variables.

$$E[Y_i] = \mu_2 \in \mathbb{R}$$ $$\&$$ $$V[Y_i] = \sigma_2^2 \in \mathbb{R}^+$$ $$\forall i \in \{1,2,3,...,n\}$$.

The statistics $$\bar{Y}$$ converges in distribution to $$Z_2 \sim N(\mu_2, \frac{\sigma_2^2}{n}$$) according to the Central Limit Theorem.

If $$\frac{1}{\bar{Y}}$$ converges in distribution to $$\frac{1}{Z_2}$$, $$\frac{\bar{X}}{\bar{Y}}$$ converges in distribution to $$\frac{Z_1}{Z_2}$$, which is a fat-tailed "Cauchy" distribution with not well-defined expected value and variance.

Therefore, it is incorrect to assume that the ratio estimator $$\hat{R} = \frac{\bar{X}}{\bar{Y}}$$ is normally distributed.

• en.wikipedia.org/wiki/Inverse_distribution Commented Jun 7, 2022 at 10:36
• Commented Jun 7, 2022 at 10:49
• Commented Jun 7, 2022 at 10:49
• $\bar X$ actually converges to $\mu_1$ (the law of large numbers) and so long as this is not $0$ you will have $\frac1{\bar X}$ converging to $\frac{1}{\mu_1}$. But $\overline {\frac1{ X}}$ will behave differently Commented Jun 7, 2022 at 19:01

Does $$\frac{1}{\bar{X}}$$ converge in distribution to $$\frac{1}{N(\mu_1, \frac{\sigma_1^2}{n})}$$?

No, because $$\frac{1}{N(\mu_1, \frac{\sigma_1^2}{n})}$$ is not a distribution/variable, and so it is not something that a sequence of distributions/variables can converge to.

You could alternatively describe the following two convergence behaviours

### When $$\mu_1 = 0$$

You might be thinking of the reciprocal normal distribution. When $$\mu_1 = 0$$ then you can have some expression where the reciprocal normal distribution is what the sequence converges to.

Let $$Y \sim N(0,1)$$ then $$\frac{\sqrt{n}}{\sigma} \bar{X}_n \xrightarrow{P} Y$$ so we could have $$\frac{\sigma}{\sqrt{n}} \frac{1}{ \bar{X}_n} \xrightarrow{D} \frac{1}{Y}$$

without this additional factor $$\frac{\sigma}{\sqrt{n}}$$ the sequence does not converge to anything because you get a sequence of distributions with an increasing variance. Eventually the distribution function converges to zero in every point (but zero in the entire range is not a distribution).

### When $$\mu_1 \neq 0$$

In this case the we can make a Taylor series approximation in the point $$x = \mu_1$$

$$\frac{1}{x} \approx \frac{1}{\mu} - \frac{1}{\mu^2} (x-\mu)$$

and we can approximate the variable $$1/\bar{X}_n$$ with this

$$\frac{\sqrt{n}}{\sigma}\left[\mu_1^2\left( {\frac{1}{\bar{X}_n}-\frac{1}{\mu_1}}\right) \right]\approx \frac{\sqrt{n}}{\sigma}\left[\bar{X}_n - \mu_1 \right] \xrightarrow{D} N(0,1)$$

or re-arranging the terms a bit

$$\frac{\mu_1^2\sqrt{n}}{\sigma} \frac{1}{\bar{X}_n} - \frac{\mu_1\sqrt{n}}{\sigma} \xrightarrow{D} N(0,1)$$

### Histograms of simulations

The simulations below show that there is a reasonable agreement between the histogram for sample sizes of 1000 and the limiting distributions.

The case when $$\mu_1 = 0$$ is actually interesting. When $$\mu_1 \approx 0$$ (ie. $$|\mu_1| << \sigma$$), then the convergence is very slow and for small samples you get that the reciprocal normal distribution can also be used as an approximation.

simulate1 = function(n1, n2, mu, sig, xlab = "") {

X = matrix(rnorm(n1*n2,mu,sig), nrow = n2)
X_m = rowMeans(X)
Y = 1/X_m
Z = mu^2*sqrt(n1)/sig * Y - mu*sqrt(n1)/sig

hist(Z, breaks = c(-10^9,seq(-5,5,0.2),10^9), xlim = c(-5,5), ylim = c(0,0.45), freq = 0, main = "", xlab = xlab,
cex.lab = 0.6, cex.axis = 0.8)
Z
}

set.seed(1)
Z = simulate1(10^3, 10^4, 4, 2, xlab = expression(Z == over(mu^2*sqrt(n),sigma) %.% over(1,X) - over(mu* sqrt(n),sigma)))
t = seq(-5,5,0.01)
lines(t,dnorm(t))
title("sample of size 1000 \n mu = 4, sigma = 2", cex.main = 1, line = -1)

simulate2 = function(n1, n2, mu, sig, xlab = "") {

X = matrix(rnorm(n1*n2,mu,sig), nrow = n2)
X_m = rowMeans(X)
Y = 1/X_m
Z = sig/sqrt(n1) * Y

hist(Z, breaks = c(-10^9,seq(-5,5,0.2),10^9), xlim = c(-5,5), ylim = c(0,0.45), freq = 0, main = "", xlab = xlab,
cex.lab = 0.6, cex.axis = 0.8)
Z
}

set.seed(1)
Z = simulate2(10^3, 10^4, 0, 2, xlab = expression(Z == over(sigma,sqrt(n)) %.% over(1,X)))
t = seq(-5,5,0.01)
lines(t,1/t^2/sqrt(2*pi) * exp(-1/2/t^2) )
title("sample of size 1000 \n mu = 0, sigma = 2", cex.main = 1, line = -1)

• Thanks for your detailed answer! I really appreciate that. I managed to understand some parts of your answer. Anyway, i just edited my question by using more "accurate" notations. Can you take a look at it again? Thank you so much!
– user340483
Commented Jun 7, 2022 at 11:47
• @mathworkswizard your new expressions are still difficult. You ask whether $1/\bar{X}$ converges to $1/Z$, but $1/Z$ is not a fixed distribution. That makes it ambiguous what is meant with convergence. That's why in my answer I use two parameters and look at the convergence of $$a_n \frac{1}{\bar{X}_n} + b_n$$ where one needs a suitable choice of $a_n$ and $b_n$ to get a converging sequence of distributions. Commented Jun 7, 2022 at 12:14
• The problem is that I can make any distribution $X$ converge to a distribution $Z$ by letting the $X_1, X_2, \dots, X_n$ and $Z_1, Z_2, \dots, Z_n$ be both a sequence of variables that converge towards a single point. For instance imagine, a normal distribution with standard deviation $\sigma = 1/n$ and a Cauchy distribution with scale parameter $\gamma = 1/n$. These distributions of these variables will converge to zero everywhere except in the point of the location parameter $\mu$. So then you could say that the distributions converge to each other? Commented Jun 7, 2022 at 12:14
• Thank you so much! I will read it when I get home. Your answer is great anyway!
– user340483
Commented Jun 7, 2022 at 12:51