median and mean of the sample mean of i.i.d. log-normal

Question

Let $y:=\frac1n\sum_{i=1}^n x_i$, where $\{x_i\}_{i=1}^n$ is a set of i.i.d. random variables, and every $x_i$ has a lognormal distribution $x_i \sim\text{Lognormal}(\mu,\sigma^2)$. Let $\text{Med}[y]$ be the median of $y$. Is the following inequality true $\forall (n,\mu,\sigma)$? $$\text{Med}[y]<\mathbf E[y]$$

Answer 1

This is true for $n \le 4$, which we prove for $n=2$ in the following form, and then sketch for $n=3$ and $n=4$.

Main Result: $$P\Big[Y < Mean\Big] \ > \ \frac12$$

Proof: We change notation so that $X_i\sim LN (0, 2\sigma^2)$. Then $Mean=E[Y] = e^{\sigma^2}$.

Let \begin{align} U &= (\ln X_1 + \ln X_2)/2\\ V &= (\ln X_1 - \ln X_2)/2 \end{align} so that $U, V$ are iid $N (0, \sigma^2)$ with cdf $F$ and pdf $f$. Then: \begin{align} P\Big[Y<Mean\Big] &= P \left[\frac12(e^{U + V} + e^{U - V}) < e^{\sigma^2} \right] \\ &= P \left[e^U\cosh V < e^{\sigma^2} \right] \\ &= P \left[U < \sigma^{2^\phantom{2\!}} - \ln\cosh V \right] \\ &= E\! \left[F[\sigma^2 - \ln\cosh V]\right] \\ &> E\! \left[F[\sigma^2] - (\cosh V - 1)f(\sigma^2)\right]\\ &=F[\sigma^2]-\left(e^{\sigma^2/2}-1\right)f(\sigma^2)\\ &> 1/2 \end{align}

The first inequality here comes from the lemma below.

The term $F[\sigma^2]$ in both inequalities is also $P[\sqrt{X_1 X_2}<Mean]$, and is approximately $\frac12+\sigma/\sqrt{2\pi}$ for small $\sigma$.

The function of $\sigma$ in the second inequality is approximately $\frac12(1+\sigma/\sqrt{2\pi})$ for small $\sigma$ and $1-\sigma^{-1}/\sqrt{2\pi}$ for large $\sigma$.

Lemma: For $W>1$ (and in particular for $W=\cosh V$), $$F[\sigma^2 - \ln W] > F[\sigma^2] - (W-1)f(\sigma^2)$$ This is obviously true with equality when $W=1$. For $W>1$ it follows from the corresponding inequality for the derivatives of both sides, which can be written as: $$\frac{-f(\sigma^2 - \ln W)}{W} > -f(\sigma^2)$$ Cancelling many factors shows this equivalent to: $$\exp\left(\frac{-(\ln W)^2}{2\sigma^2}\right) < 1$$ which proves the lemma.

Sketch of Extension for $2\le n\le 4$:

A similar technique proves the same result for $n \le 4$, transforming $\ln(X)$ by an orthogonal matrix where all entries in the first row are equal. Then we replace $e^U$ by the geometric mean of the $X$'s, and we replace $\cosh V$ by a function with $n$ exponential terms in $n-1$ variables. The corresponding lemma is: $$F[k\sigma^2 - k\ln W]\, > $$ $$F[k\sigma^2]\ -kf(k\sigma^2)\Big(W-1+\frac{k^2-1}{2}(W-1)^2\Big)$$ where $k=\sqrt{n/2}$; this lemma would be false for higher $n$ (e.g. at $n=7$, $\sigma=1$, $W=1.05$), but Bernoulli's inequality with an exponent of $0\le k^2-1 \le 1$ shows that it is true for $2\le n\le 4$. When we take expectations of this lemma, we get a good bound for low $\sigma$, which we can combine with separate and easier bounds for high $\sigma$ to prove the result.