# Median and mean of the sample mean of i.i.d. log-normal

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 $$\forall (n,\mu,\sigma)$$? $$\text{Med}[y]<\mathbf E[y]$$

Motivation: I am computing the sample mean of the lognormal random variables via Monte Carlo. The sample mean seems tend to concentrate below the mean for large $$\sigma$$. I am wondering whether this is true for all cases. It is true for a single sample. However, there is no explicit formula for the distribution of the mean of finite number of --- not even two -- samples i.i.d. lognormal variables. I have no idea how to prove it.

• A discrete parallel to consider: if the x's are Bernoulli with probability 40%, then Med[x1] < E[x1], Med[x2] < E[x2], but Med[x1+x2] > E[x1+x2].
– user44143
Commented Jun 6, 2017 at 9:59
• @MattF. is there a general name for such comparison problem between mean and median, I thought there is such a name but cannot remember... Commented Jun 6, 2017 at 13:21
• @HenryL, I know of no good name for this. If you say "positive Pearson's second skewness", some people might understand you; that skewness is defined as 3(mean-median) / (standard deviation). But I don't recommend that, since I haven't seen any good citation of a text of Pearson where he proposed this particular measure, and I prefer quartile skewness in any case. mathworld.wolfram.com/PearsonsSkewnessCoefficients.html
– user44143
Commented Jun 9, 2017 at 14:45
• "a single sample". $\qquad$ Here I am tempted to say that you are using the word "sample" incorrectly. But instead I will say only that it may be of use for you to know that a different convention concerning the meaning of that word is widespread, and universal or nearly so among statisticians. The whole set $\{\,x_i\,\}_{i\,=\,1}^n$ is one sample. Each $x_i$ is one observation within that sample. That is why one calls the number $n$ the size of the sample, and it is why one speaks of a two-sample t-test. Commented Jun 15 at 23:32
• Clearly the proposed inequality can fail for only finitely many values of $n.$ Commented Jun 15 at 23:34

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

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 $$(x_1+x_2)/2 for some positive $$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]

• this is no more than a simulation result.... Commented Jun 25, 2017 at 14:38
• Here is my derivation for your formula. You can see if you would like to incorporate it into your answer. \begin{align}P[y_2 < m] &= \int_{y_2=\frac{x_1+x_2}2<m}dF_2(x_1,x_2) \quad\text{for arbitrary joint CDF }F_2 \\ &= \int\int_{\frac{x_1+x_2}2<m}dF(x_1)dF(x_2)\quad\text{for arbitrary i.i.d. with CDF }F \\ &= \int_{x_1} F(2m-x_1)\, dF(x_1)\quad\text{integrate over }x_2 \\ &= \int_{0}^{F_\sigma(2m)} F_\sigma\big(2m - F_\sigma^{-1}(p)\big)dp, \end{align} where $p=F(x_1)$ and $F = F_\sigma$ thus $x_1\in[0,2m]$ for a CDF $F_\sigma$ on a positive random variable $x_1$.
– Hans
Commented Jun 27, 2017 at 8:21

I would like to provide an asymptotic solution, as the sample size $$n\rightarrow\infty$$.

Using the Fenton-Wilkson empirical asymptotic result in [1], we know that the sample sum of i.i.d. lognormals $$X_i\sim \text{LN}(\mu,\sigma^2)$$ can be approximated by lognormal with different mean and variance $$Y=\sum_{i=1}^nX_i$$ is distributed as $$Y\sim \text{LN}(\mu_n,\sigma^2_n)$$ where $$\sigma_{n}^{2}=\log\left(\frac{e^{\sigma^{2}}-1}{n}+1\right)$$ and $$\mu_{n}=\log\left(n\cdot e^{\mu}\right)+\frac{1}{2}\left(\sigma_{n}^{2}-\sigma^{2}\right)$$. We know from properties of lognormal distributions that $$\text{Median}(Y)=e^{\mu_n}$$ and $$\text{Mean}(Y)=\exp\big(\mu_n+\frac{\sigma^2_n}{2}\big)$$and therefore $$\text{Median}(Y)\le \text{Mean}(Y)$$ is asymptotically proved since $$\sigma^2_n\rightarrow 0$$. However this argument holds only when the Fenton-Wilkson approximation works within the order of accuracy of $$O\Big(e^{\frac{\sigma^2_n}{2}}\Big)$$, which is not always the case. If we know the exact distribution of $$Y$$, this argument can be modified to see if such an argument holds in general.

[1] Cobb, Barry R., R. Rumi, and Antonio Salmerón. "Approximating the distribution of a sum of log-normal random variables." Statistics and Computing 16.3 (2012): 293-308. http://leo.ugr.es/pgm2012/submissions/pgm2012_submission_6.pdf (Wayback Machine)

• Is the error Fenton-Wilkson approximation proved or a numerical empirical result tested for some large number of examples? Would you mind specifying the definition of the "accuracy" that you say the Fenton-Wilkson works within? Thank you for your answer, Henry L.
– Hans
Commented Jun 6, 2017 at 18:22
• @Hans AFAIK, it is a numerical empirical result tested for a wide range of applications, but it may fail, you should look for expertise on the topic. For the accuracy I think they match the first two moments, so it depends on how good a second order approximation is around the point you are to estimate. Commented Jun 6, 2017 at 20:52
• For comparing mean and median via moment-matching, I would want an approximation that matched the first three moments. After all, if we matched the first two moments with a normal distribution, we would get the wrong answer to this problem.
– user44143
Commented Jun 6, 2017 at 21:11
• @MattF. Sorry but why first 3 moments? Not sensical to me. Commented Jun 6, 2017 at 22:25
• Henry.L: I think @MattF. was suggesting that the third moment could point towards skewness in a way the first two might not. Stating that a log-normal approximation to the sum of i.i.d. log-normals is better than a CLT-type normal approximation to the sum for finite $n$ needs some justification which could come from this third moment. Otherwise, as $n$ increases without limit the log-normal approximation will converge in distribution to the normal approximation (in a standardised sense). Commented Jun 21, 2017 at 17:53