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 sigle 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.

  • $\begingroup$ Is E the true expectation or is it just the sample mean? Either way this seems badly posed: sometimes the median is less than the sample mean, sometimes more. $\endgroup$
    – Ian
    May 31 '17 at 17:38
  • $\begingroup$ @Ian: $y$ is the sample mean as defined in the question and therefore a random variable. $\mathbf E[y]$ is the expectation of the random variable $y$ the sample mean. Why is it badly posed? These are mathematically rigorous statements. Each is just a definitive scalar. What time is median less and what time is it more than the expectation of the sample mean? Could you please prove your claims? $\endgroup$
    – Hans
    May 31 '17 at 17:48
  • $\begingroup$ Oh, sorry, I missed some notation at the start that clarified what was going on; Med is not the median of the sample, it's the median of the distribution of the sample mean of the sample. Now it makes sense. $\endgroup$
    – Ian
    May 31 '17 at 18:29
  • $\begingroup$ In your edit do you mean to say "median" rather than "mean"? $\endgroup$
    – Ian
    Jun 1 '17 at 19:44
  • $\begingroup$ @Ian: I mean to say "mean". (Double meaning. Ha) Median is below a certain point is equivalent to the probability of the variable below that point is greater than 0.5 or "concentrating below" that point. $\endgroup$
    – Hans
    Jun 1 '17 at 19:56

Yes, it seems the result holds.

Case: $n = 1$

Here, $Y = X_1$. First note that,

$$Med(X_i) = e^\mu$$ $$E(X_i) = e^\mu e^{\sigma^2/2}$$

Since $\sigma^2 > 0$, we know that $e^{\sigma^2/2} > 1$ which implies that $E(Y) > Med(Y)$.

Case: $n = \infty$

In the limit, it is clear by the central limit theorem that $$y \stackrel{d}{\rightarrow} N(\mu^*, \sigma^*/\sqrt{n})$$

Where $\mu^*$ and $\sigma^*$ are the mean and standard deviation of the $X_i$ respectively. Since this is converging to a normal distribution, in the limit we have $E(Y) = Med(Y)$.

Case: $1 < n < \infty$

This is not at all rigorous, but when $n=1$ we have a positive skew distribution and when $n\rightarrow\infty$ we converge to a symmetric (normal) distribution for $Y$. In between, the sampling distribution of $Y$ will remain positively skewed and hence $E(Y)$ will be greater than $Med(Y)$ for finite $n$.

This plot, which was constructed via simulation illustrates this point informally. The vertical lines represent the median (green) and mean (blue).

enter image description here

A (Partial) More Rigorous Approach

Define $D_n = E(Y_n) - Med(Y_n)$, we wish to show that $\forall \ n, \ \mu, \ \sigma$, it is the case that $D_n \geq 0$. By the extreme cases above, we have that: $$D_1 > 0$$ $$\lim_{n\rightarrow\infty}D_n = 0$$

Thus, if we can show that $D_{n} > D_{n+1}$ the by MCT we can argue that for any finite $n$ it must be the case that $D_n \geq 0$.

In attempt to show that $D_n - D_{n+1} > 0$, we can write:

$$D_n - D_{n+1} = \left[E(Y_n) - E(Y_{n+1}\right] + \left[Med(Y_{n+1}) - Med(Y_n)\right]$$

Since $Y$ is the sample mean, the expected value will be the same as $X_1$, regardless of $n$ (as seen in the simulation above). Hence we need only show that:

$$Med(Y_{n+1}) > Med(Y_n)$$

One way we can accomplish this, is by showing that

$$P(Y_{n+1} < M_n) < \frac{1}{2}$$

where $M_n = Med(Y_n)$. We can write this as:

$$\begin{align*} P(Y_{n+1} < M_n) &= P\left(\frac{n}{n+1}Y_n + \frac{x_{n+1}}{n+1} < M_n\right) \\ &= P\left(Y_n + \frac{X_{n+1} - Y_n}{n+1} < M_n\right) \end{align*}$$

Informally, it seems that if $\frac{X_{n+1} - Y_n}{n+1}$ is positive more often than it is negative, the above probability should indeed be less than 1/2.

Perhaps somebody smarter than me can figure out where to go from here. (:

  • $\begingroup$ Your implication "distribution of $Y$ will remain positively skewed and hence $E(Y)$ will be greater than $Med(Y)$ for finite $n$" does not hold. Also, have you tried different $\sigma$'s? Is there a scaling argument relating all $\sigma$'s? $\endgroup$
    – Hans
    May 31 '17 at 19:46
  • $\begingroup$ Hmm I agree with you that it is a leap. However, I suspect the implication holds under certain conditions (Uni-modal, etc). The base case holds for any $\sigma$ since $\sigma^2 > 0$. It would be great to see a more rigorous approach, but I remain convinced that the difference should be montone as a function of $n$ between the $n=1$ and $n=\infty$ case. $\endgroup$
    – knrumsey
    May 31 '17 at 21:49
  • $\begingroup$ By the way, would you mind changing the phrase "this should hold" to "this seems to hold"? "Should" means "indeed does" while "seem" indicates there is evidence but great uncertainty. You have not proved the conjecture. It is more appropriate to use "seem" than "should". $\endgroup$
    – Hans
    Jun 2 '17 at 3:25

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