For an iid sample of $n$ realizations of random variable $X$ with $\mathbb{E}\left[X\right] = \mu > 0$, let $\bar{X}_n$ denote its sample mean. Is there a distribution for $X$ such that $\mathbb{E}\left[\frac{1}{\bar{X}_n}\right] < \frac{1}{\mathbb{E}\left[\bar{X}_n\right]}$?
Jensen's inequality shows that if the support of $X$ only contains positive numbers, this cannot be the case, but is there a good counterexample if we don't impose this restriction?
Thanks in advance!
At least when $n = 1$, the counterexample is easy to construct. For example, let $X$ be a triple-point distribution: \begin{align} P[X = -2] = 0.2, P[X = -1] = 0.25, P[X = 2] = 0.55. \end{align} Then \begin{align} & E[X] = -0.4 - 0.25 + 1.1 = 0.45 > 0, \\ & E[1/X] = -0.1 - 0.25 + 0.275 = -0.075 < 0. \end{align} Clearly, $E[1/X] < 1/E[X]$.
For general $n$, the following fact (Asymptotic Statistics by A. W. van der Vaart, Exercise 5.8) is also interesting: although it is not an example of $E[\bar{X}_n^{-1}] < 1/E[\bar{X}_n]$, it can be used as to show that $E[\bar{X}_n^{-1}] \geq 1/E[\bar{X}_n]$ does not always hold (in fact, it shows that for a large family of distributions, the expectation of $\bar{X}_n^{-1}$ either does not exist or is infinite):
Let $X_1, X_2, \ldots, X_n$ be i.i.d with expectation $1$ and finite variance. If the random variables are sampled from a density $f$ that is bounded and strictly positive in a neighborhood of zero, then $E[|\bar{X}_n^{-1}|] = \infty$ for every $n$.
Proof of the fact. It can be shown by induction that the density of $\bar{X}_n$ is bounded away from zero in a neighborhood of zero for every $n$.
When $n = 1$, this is the condition.
Suppose the density of $\bar{X}_k$ is bounded away from zero in a neighborhood of zero, i.e., there exists $\delta > 0$ such that $f_{\bar{X}_k}(x) \geq \epsilon > 0$ for all $x \in (-\delta, \delta)$. By condition, there exists $\delta' > 0$ such that $f_{X_{k + 1}}(x) = f(x) \geq \epsilon' > 0$ for all $x \in (-\delta', \delta')$. Let $\delta_0 = \min(\delta, \delta')$, $\epsilon_0 = \min(\epsilon, \epsilon')$, then $\min(f_{\bar{X}_k}(x), f(x)) \geq \epsilon_0 > 0$ for all $x \in (-\delta_0, \delta_0)$.
Since $\bar{X}_{k + 1} = \frac{k}{k + 1}\bar{X}_k + \frac{1}{k + 1}X_{k + 1}$ and the two summands are independent, the density of $\bar{X}_{k + 1}$ can be obtained by the convolution formula (and densities of the scale family): \begin{align} & f_{\bar{X}_{k + 1}}(y) = \int_{-\infty}^\infty f_{(k + 1)^{-1}X_{k + 1}}(y - x)f_{(k + 1)^{-1}k\bar{X}_k}(x)dx \\ =& \int_{-\infty}^\infty (k + 1)f((k + 1)(y - x))k^{-1}(k + 1)f_{\bar{X}_k}(k^{-1}(k + 1)x)dx \\ =& k^{-1}(k + 1)^2\int_{-\infty}^\infty f((k + 1)(y - x))f_{\bar{X}_k}(k^{-1}(k + 1)x)dx \\ \geq & k^{-1}(k + 1)^2\int_{y - (k + 1)^{-1}\delta_0}^{y + (k + 1)^{-1}\delta_0} f((k + 1)(y - x))f_{\bar{X}_k}(k^{-1}(k + 1)x)dx. \tag{1} \end{align}
If $|y| < \frac{1}{2}\delta_0$, then when $x \in (y - (k + 1)^{-1}\delta_0, y + (k + 1)^{-1}\delta_0)$, we have $(k + 1)(y - x) \in (-\delta_0, \delta_0)$ and $k^{-1}(k + 1)x \in \left(-\frac{k + 3}{2k}\delta_0, \frac{k + 3}{2k}\delta_0\right) \subset (-\delta_0, \delta_0)$ (when $k \geq 3$), hence the last integral in $(1)$ is at least $2k^{-1}(k + 1)\delta_0\epsilon_0^2 > 0$. This completes the inductive step. Therefore the density of $\bar{X}_n$ is bounded away from zero in a neighborhood of zero for every $n$.
For fixed $n$, suppose $f_{\bar{X}_n}(x) \geq \epsilon^*$ for all $x \in (-\delta^*, \delta^*)$ for some $\epsilon^* > 0$ and $\delta^* > 0$. For any $a \in (0, \delta^*)$:
\begin{align}
& E[|\bar{X}_n^{-1}|] = \int_\mathbb{R} \frac{1}{|x|}f_{\bar{X}_n}(x)dx \\
\geq & \int_a^{\delta^*}\frac{1}{x}f_{\bar{X}_n}(x)dx \\
\geq & \epsilon^*\int_a^{\delta^*}\frac{1}{x} dx \\
=& \epsilon^*(\log\delta^* - \log a),
\end{align}
which diverges to $+\infty$ as $a \downarrow 0$. Therefore, $E[|\bar{X}_n^{-1}|] = \infty$. This completes the proof.
