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

• One could show that the third moment skewness is positive easily enough, but that doesn't automatically imply that the mean will exceed the median (I certainly believe the mean will exceed the median here but we'd need to prove that it does). What's this for? May 31, 2017 at 0:38
• @Glen_b: 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.
– Hans
May 31, 2017 at 0:56
• Incidentally, somewhat related (at least in the limit): Peter Hall, (1980), "On the Limiting Behaviour of the Mode and Median of a Sum of Independent Random Variables", Ann. Probab. Volume 8, Number 3, 419-430. Open Access at Project Euclid May 31, 2017 at 1:06
• I mean "no mgf" in the sense expressed here. By "other distributions with the same moment sequence exist" I mean there are distributions with the same sequence of moments as the lognormal / the lognormal is an example of a distribution not uniquely defined by its moments: if $f$ is a standard lognormal density ($\mu=0$, $\sigma=1$) and $g(x)=f(x)\cdot (1+\epsilon\sin[2\pi k\log(x)])$ then the contribution of $g-f$ to the $n$th moment is $0$ for each $n=1,2,...$. This extends to the general lognormal (indeed to the 3 parameter case) May 31, 2017 at 2:53
• Oh, this may have been it; it's more a generalization of the above case to a broader class: ... stats.stackexchange.com/questions/25010/… May 31, 2017 at 8:43

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 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 $$P[\sqrt{X_1 X_2}, and is approximately $$\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.

• There is an extra $\sigma$ on the denominator of Equation $(1)$ compared to the right hand side of the expression in the lemma. Does that matter?
– Hans
Oct 6 at 1:53
• Good catch -- I have now simplified the proof and avoided that expression entirely. Oct 6 at 14:26
• What about $n>2$? Oct 6 at 15:53
• Good question -- it took a couple years of thinking about this on and off to find a nice solution for $n=2$; I have some ideas for extending it but no results yet. Oct 6 at 16:55
• @PontusHultkrantz, I updated this for $2\le n\le 4$. And what about $n>4$? I have no idea, since I wouldn’t trust numerical integrals over five variables, and I don’t see how to extend the proof to all $n$. Oct 17 at 17:45