Given $X_1,...,X_n \stackrel{iid}{\sim} \text{Unif}(a,b)$, what is $$\mathbb{P}\left[\frac{\sum x_i}{\sum \frac{1}{x_i}} \geq \max(x_i) \cdot \min(x_i)\right].$$ I don't need a closed-form formula, just a description of the asymptotic behavior would be nice. Numerical experiments seem to show that the probability tends to 1.

Here $a,b$ are arbitrary. As wolfies pointed out in the comments, when $a=0$, the result is immediate since $\min(x_i)\to 0$. So we can assume $a>0$. The inequality is invariant by scaling, so we can assume w.l.o.g. that $a=1$ and $b$ is arbitrary.

A little background: I am studying an algorithm which has parameters $x_i \in \Bbb R_+, i =1,...,n$. The algorithm has provably good performances when $$\frac{\sum x_i}{\sum \frac{1}{x_i}} \geq \max(x_i) \cdot \min(x_i).$$ The left-hand side can be written as $AM(x_i) \cdot HM(x_i)$ where $AM$ and $HM$ are the arithmetic and harmonic means. Both quantities are between $\min(x_i)$ and $\max(x_i)$, so nothing can be said about their product in general.

However, numerical evidences seem to indicate that the event occurs with probability one as $n\to \infty$ if the $x_i$'s are sampled uniformly. But I am unable to come up with a proof of this fact.

  • $\begingroup$ What do you mean by $Unif(a,b)$? Uniform distributed on $(a,b)$ or on the image of $(a,b)$ under the transformation $f$. $\endgroup$ Apr 17, 2017 at 12:54
  • $\begingroup$ The uniform distribution on the interval $(a,b)$. $\endgroup$
    – Nitrogen
    Apr 17, 2017 at 12:55
  • $\begingroup$ I am not sure that $a$ and $b$ are arbitrary. If $a$ is 0 (as in the original posting of this question), then the sample minimum tends to 0 as $n$ becomes large, so the RHS tends to 0 too, so the result holds trivially (since the sample means are strictly positive). $\endgroup$
    – wolfies
    Apr 17, 2017 at 13:11
  • $\begingroup$ @wolfies As you pointed out, the case $a = 0$ is easy. So my question really refers to the other cases, i.e, when $a>0$. Then, numerically the probability still seems to tend to 1, but your argument for $a=0$ doesn't work anymore. $\endgroup$
    – Nitrogen
    Apr 17, 2017 at 13:16
  • $\begingroup$ BTW, in the $n = 2$ case, the two expressions (LHS and RHS) are identical. $\endgroup$
    – wolfies
    Apr 17, 2017 at 14:59

1 Answer 1


$X \sim \text{Uniform}(a,b)$ $\dots$ assume w.l.o.g. that $a=1$ and $b$ is arbitrary

Let $X \sim \text{Uniform}(1,b)$, and let ${X_1, \dots, X_n}$ denote a random sample of size $n$ on parent $X$.


As $n$ becomes large, the sample maximum tends to $b$, and the sample minimum to 1, so the product $$\max(x_i) \cdot \min(x_i) \quad \to \quad b$$

One can also formally derive the pdf of $\max(x_i) \cdot \min(x_i)$ as the product of the largest and smallest order statistics (functional form given at bottom) which may be helpful too, but is not necessary.

LHS $$\quad \large \frac{\frac{1}{n}\sum x_i}{\frac{1}{n}\sum \frac{1}{x_i}}$$

If we are interested in asymptotic (large sample) behaviour, we can apply Central Limit Theorem concepts.

  • The numerator is the sample mean of $X$, which has expected value: $\large \frac{b+1}{2}$

  • The denominator is the sample mean of $1/X$, which has expected value: $\large \frac{\log{b}}{b-1}$

The limit ratio of numerator to denominator is: $$\text{LHS} \large = \frac{b^2-1}{2 \log (b)}$$

The following diagram plots the LHS limit as a function of $b$. Note that since $b$ is the RHS limit, we are plotting the LHS limit as a function of the RHS limit:

enter image description here

[ The orange curve is the identity $b=b$ ]

... which completes the proof: the LHS limit exceeds the RHS limit, for all $b > 1$. The diagram also suggests that, the larger is $b$, the larger will be the probability that the ratio of samples means will exceed the product of the sample range.

Monte Carlo check

A Monte Carlo check is informative in checking the rate of convergence, and to confirm the limits above.

In the following:

  • $n = 500$
  • the blue curve is a Monte Carlo approximation of the pdf of the LHS ratio
  • the red curve is a Monte Carlo approximation of the pdf of the RHS $\max(x_i) \cdot \min(x_i)$.

enter image description here

As argued above:

  • the red curve is centred on $b$, where $b =2$ in the top diagram, and $b = 4$ in the bottom diagram.

  • the blue curve appears Normal (as per CLT) with mode at $\frac{b^2-1}{2 \log (b)}$ which is $\approx 2.164$ when $b = 2$ (top diagram), and $\approx 5.41$ when $b = 4$ (bottoms diagram).

Small Sample comparison

The following diagram compares the small sample pdf of the two products:

enter image description here

when $b = 2$. This time, the dashed red curve is the actual exact pdf of the product of the sample min and max.

Exact CDF of the product of the sample min and max

Given a Uniform parent, say with $b=2$:

f = 1;  domain[f] = {x, 1, 2}; 

... we can find the joint pdf of the sample min and sample max (I am using here the order statistic function from mathStatica to automate) as:

enter image description here

Let $Z = X_1 X_n$ denote product of the sample min and max. Then the cdf of $Z$, $P(Z<z)$ is:

enter image description here

The pdf can then be obtained by by differentiating wrt $z$. The latter was used to plot the exact dashed red curves above. I should perhaps note that I am one of the authors of the software used here.


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