Here is a positive result for the average of two iid lognormals. Recall that the average of a lognormal distributed as $\operatorname{LN}(\mu,\sigma)$ is $\exp(\mu+\sigma^2/2)$.

Let $y_2$ be the average of two iid lognormals each distributed as
$\operatorname{LN}(\mu,\sigma)$. Then:

For small $\sigma$, $\operatorname{Med}[y_2] \approx \exp(\mu+\sigma^2/4)$, which is less
than $E[y_2]$.

For large $\sigma$, $\operatorname{Med}[y_2] \approx \exp(\mu+.545\sigma)$, which is less
than $E[y_2]$.

From those we can establish that $\operatorname{Med}[y_2] < E[y_2]$ for all positive $\sigma$.

The approximations are valid to second order in $\sigma$, in the sense that

$$P\left[y_2 < \exp(\sigma^2/4)\right] = \frac{1}{2} + O(\sigma^3)$$

$$P\left[y_2 < \exp(0.545\sigma)\right] = \frac{1}{2} + O\left(\frac{1}{\sigma^3} \right)$$

where $.545$ is an abbreviation for $\ln(F_1^{-1}(\frac{1}{\sqrt{2}}))$, and $F_{\sigma}$ is the CDF of $\operatorname{LN}[0,\sigma]$, and we take $\mu=0$ for simplicity.

**Proof:**

We prove the first approximation from the general convolution formula

\begin{align}
P[y_2 < m]
&= \int\int_{\frac{x_1+x_2}2<m} \, dF(x_1) \, dF(x_2)\ \ \text{for arbitrary i.i.d. with CDF }F \\
&= \int_{x_1} F(2m-x_1)\, dF(x_1)\ \ \text{by integrating over }x_2 \\
&= \int_0^1 F\big(2m - F^{-1}(p)\big)\,dp\ \text{ where } p=F(x_1) \\
\end{align}

which in this particular case gives
$$P[y_2 < m] = \int_0^{F_\sigma(2m)} F_\sigma\big(2m - F_\sigma^{-1}(p) \big) \, dp.$$

$F_\sigma(2m)$ is the largest possible $p$ such that $(x_1+x_2)/2<m$ for some positive $x_2$.

The graph shows the line $1-p$ and the integrand for $\sigma=1/2, \ m=\exp(\sigma^2/4)$. The integrand approaches the line as $\sigma$ goes to $0.$ The median is $m$ iff the areas under the two curves are equal.

We use Mathematica to differentiate the formula for $P[y_2 < m]$ under the integral sign and at the limit of integration, and this computes that the coefficients in the power series are $1/2, 0, 0$.

```
F[u_] = Simplify[CDF[LogNormalDistribution[0, sigma], u], Assumptions -> u > 0]
x1 = u /. Solve[p == F[u], u][[1]]
x2 = u /. Solve[q == F[u], u][[1]]
qsol = q /. Solve[(x1 + x2)/2 == Exp[sigma^2/4], q][[1]]
plim = F[2 Exp[sigma^2/4]]
Limit[plim, sigma -> 0]
Limit[D[plim, sigma], sigma -> 0]
Limit[D[plim, {sigma, 2}], sigma -> 0]
Integrate[Limit[qsol, sigma -> 0], {p, 0, 1}]
Integrate[Limit[D[qsol, sigma], sigma -> 0], {p, 0, 1}]
Integrate[Limit[D[qsol, {sigma, 2}], sigma -> 0], {p, 0, 1}]
Plot[{qsol /. sigma -> 1/2, 1 - p, 1}, {p, 0, 1}, PlotRange -> {0, 1},
AxesLabel -> {"p=F[x1]", "q=F[x2]"}, PlotStyle -> {, , {Black, Thin}}]
```

We prove the second approximation from
$$\frac{1}{2} < P{\Large[}y_2 < e^{0.545\sigma}{\Large]} < F_\sigma(2 e^{0.545\sigma})^2\ \text{ and }
\lim_{\sigma\rightarrow\infty}F_\sigma(2 e^{0.545\sigma})^2=\frac{1}{2}$$

and the fact (not proved here) that $P[y_2 < e^{0.545\sigma}]$ has eventually monotonic derivatives.

The inequality comes from
\begin{align}
\frac{1}{2}
&= F_\sigma(e^{0.545\sigma})^2 \\
&= P[x_1 < e^{0.545\sigma}\ \& \ x_2 < e^{0.545\sigma}] \\
&< P[\ \ \ \ \ \ \ x_1 + x_2 < 2e^{0.545\sigma} \ \ \ \ \ \ \ \ \, ] \\
&< P[x_1 < 2e^{0.545\sigma}\ \& \ x_2 < 2e^{0.545\sigma}] \\
&= F_\sigma(2 e^{0.545\sigma})^2 \\
\end{align}

We use Mathematica to show that the limit of the right-hand side is 1/2.

```
x3 = u /. Solve[1/Sqrt[2] == F[u], u][[1]]
Limit[F[2 x3]^2, sigma -> Infinity] // FullSimplify
N[x3]
```