What's the distribution of $\bar{X}^{-1}$?

Question

What's the distribution of $\bar{X}^{-1}$ with X being a continuous iid random variable that is uniformly distributed? Can I use the CLT here?

Answer 1

I'm going to assume you mean "uniformly continuous" and by your reference to the CLT I'm going to assume that you are asking for the asymptotic distribution of $(\bar X)^{-1}$ and also that $\mbox{Var}(X_1) =\sigma^2 < \infty$ with $X_1, X_2, ...$ being an iid sequence from some distribution.

If all of that was what you meant to suppose then you can use the delta method to get the asymptotic distribution of $(\bar X)^{-1}$. Let $g(\mu) = \frac 1 \mu$ where $E(X_1) = \mu \ne 0$. The delta method implies that $\sqrt n (g(\bar X) - g(\mu)) \to \mathcal N(0, g'(\mu)^2 \sigma^2)$ in distribution. So $\sqrt n (\bar X^{-1} - \mu^{-1})$ has an asymptotic $\mathcal N\left(0, \frac{\sigma^2}{\mu^4}\right)$ distribution.

Edit: Since I've received skepticism that this applies when $X$ is allowed to get close to $0$ (the delta method supposedly not applying) here is code that shows that it does, in fact, work when $X \sim \mathcal U(-2, 1)$. Obviously it won't work if $\mu = 0$ since then $g(\mu)$ is undefined. The delta method makes no moment assumptions with regard to $g(X)$; the only assumptions are that $g(\mu)$ exists and has nonvanishing derivative at $\mu$ and that $\sqrt n (\bar X - \mu)$ obeys a CLT. Maybe it is weird that $g(X)$ can have an asymptotic variance given that it doesn't have an existing variance, but such is life.

Z <- numeric(1000)
for(i in 1:1000) {
  sigma.sq <- 3^2 / 12
  mu <- -0.5
  n <- 100000
  X <- runif(n, -2, 1)
  Y <- 1 / mean(X)
  Z[i] <- sqrt(n) * (Y - 1 / mu) / sqrt(sigma.sq / mu^4)
}
hist(Z)
ks.test(Z, pnorm)

Answer 2

In the absence of a response on the questions, I'll make some mention of both possibilities for the order of the mean and reciprocal, and discuss why the limits on the domain of the Uniform matter.

Let $X$ have a continuous uniform distribution on $[a,b]\,, b>a>0$

Let $Y = 1/X$.

Then $f_Y(y) = \frac{1}{(b-a)y^2}$.

$\text{E}(Y) = \frac{1}{(b-a)} \int_a^b y^{-1} dy = \frac{\ln(b)-\ln(a)}{(b-a)}$

The mean doesn't exist if $a$ is not bounded above zero, given $b$ is positive. More generally, you need both limits on the same side of zero and both bounded away from it for the mean to exist.

If the mean doesn't exist, the CLT doesn't apply.

If the mean and variance exist ($b$ and $a$ on the same side of 0 and both bounded away from it), then the CLT should apply to $Y$, and $\sqrt{n}(\overline{Y}-\mu_Y)$ should be asymptotically normal.

But what about $\overline{X}^{-1}$? Note that - again, if $b$ and $a$ on the same side of 0 and both bounded away from it, then $\overline{X}$ will also have limits between $b$ and $a$ on the same side of 0 and bounded away from it, and so its reciprocal will have a mean and variance. While the CLT will apply to $X$ (so $\sqrt{n}(\overline{X}-\mu_X)$ would be asymptotically normal), here you take its reciprocal. At sufficiently large sample sizes, the reciprocal should also be approximately normal (see the Delta method).

However, if $b>0$ and $a = 0$ then the CLT applies to $X$ but the reciprocal has no mean or variance at any finite sample size.

http://en.wikipedia.org/wiki/Reciprocal_distribution

Answer 3

Let U~Uniform[0,1]. Let Z = 1/U.
Then P[Z < z] = P[1/U < z] = P[U > 1/z] = 1 - P[U < 1/z] = 1 - 1/z.
Note that z > 1.

On second thought I see this doesn't answer your question. But perhaps some of it can be useful along with your idea of using the CLT.