Let $(X_i)_{i}$ be a sequence of iid positive variables of mean 1 and variance $\sigma^2$. Let $\bar{X}_n = \frac{\sum_{i=1}^n X_i}{n}$.

My question is: Can we can bound $\mathbb{E}(1/\bar{X}_n)$ as a function of $\sigma$ and $n$?

There seems to be some strategy that may work based on the taylor extension, but

  • I'm not sure about the hypothesis that need to be met;
  • if it works in this case; and
  • if we can say something definite on $\bar{X}_n$ or if we need to use the central limit theorem and can only say this for the normal approximation?

More details about the Taylor expansion. According to this wikipedia article, $$\mathbb{E}(f(X)) \approx f(\mu_X) +\frac{f''(\mu_X)}{2}\sigma_X^2$$

So in my case it would give something like: $$\mathbb{E}(1/\bar{X}_n) \approx 1 +\frac{\sigma^2}{4 n}$$ I'm trying to find maybe a formal proof of a similar result, or hypothesis so that it works. Maybe references? Thanks

EDIT: if needed, we can consider that the $(X_i)_i$ are discrete, there exists $v_1<\cdots<v_K$ such that $\mathbb{P}(X=v_k)=p_k$ and $\sum p_k = 1$. In this case we know that $\bar{X}_n \geq v_1$. Although I believe something can be said in the general case.

PS: this is almost a cross-post of this on Math.SE.

  • $\begingroup$ Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE $\endgroup$ – Ferdi Mar 25 at 7:58
  • $\begingroup$ It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both. $\endgroup$ – Gopi Mar 25 at 8:04
  • $\begingroup$ Perhaps Markov's inequality or Chebyshev's inequality are useful. $\endgroup$ – Ertxiem Mar 25 at 8:07
  • $\begingroup$ Without the discreteness criterion, the answer is no, because any finite bound based on $\sigma^2,n$ will be exceeded when the underlying distribution is a suitable mixture of a $\Gamma\left(\frac{1}{n+1}, (n+1)\right)$ distribution with some other distribution. The presence of this Gamma component assures the sum of $n$ iid values will have a Gamma component with shape less than $1,$ whose PDF diverges at $0,$ forcing the reciprocal sum to have infinite expectation. This demonstrates the hopelessness of using a Taylor expansion in the analysis. $\endgroup$ – whuber Mar 26 at 18:46
  • $\begingroup$ @whuber thanks, I'm not sure I've understood the details. This means that I've probably made a mistake in my answer but I can't see where? In addition, using the TCL $\bar{X}_n$ converges towards a gaussian, so if $X_i$ has finite moments, the gaussian approx of $\bar{X}_n$ should have finite moment? $\endgroup$ – Gopi Mar 27 at 1:42

You cannot bound that expectation in $\sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $\infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $\bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.

For a useful bound you will at least need to restrict the common distribution of $X_1, \dotsc, X_n$ much more.


After your new information, and with $v_1>0$, the expectation of $1/\bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $x\mapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $\DeclareMathOperator{\E}{\mathbb{E}}\E 1/\bar{X}_n \ge 1/\E \bar{X}_n$.

  • $\begingroup$ But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $\bar{X}_n$ is centered around 1 and has a variance of $\sigma^2 / n$. $\endgroup$ – Gopi Mar 25 at 9:18
  • 1
    $\begingroup$ But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $\infty$. Can you rule out that possibility? $\endgroup$ – kjetil b halvorsen Mar 25 at 9:26
  • $\begingroup$ In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement). $\endgroup$ – Gopi Mar 25 at 9:54
  • 1
    $\begingroup$ Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $\mu=1$ is really restrictive ... $\endgroup$ – kjetil b halvorsen Mar 25 at 9:58
  • 1
    $\begingroup$ Obviously the support is not positive integers ;). The support is the set of rational numbers. $\endgroup$ – Gopi Mar 25 at 10:07

I think I have the gist of it. Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:

There exists $\varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ \frac{f''(1)(x-1)^2}{2} + \frac{f'''(\varepsilon) (x-1)^2}{2}$.

In our case, if $X_i$ belongs in the domaine $[v_1;+\infty[$, then $\bar{X}_n$ has the same domain and we have $\varepsilon \geq v_1$.

Hence $\mathbb{E}(1/\bar{X}_n) = \mathbb{E}\left (1 - (\bar{X}_n-1) + \frac{(x-1)^2}{4}+ \frac{f'''(\varepsilon) (\bar{X}_n-1)^2}{2} \right)$, and \begin{align*} \mathbb{E}(1/\bar{X}_n) &= 1 + \frac{f'''(\varepsilon) \mathbb{E}\left ((\bar{X}_n-1)^2\right )}{2} = 1 +\frac{ V(\bar{X}_n)}{4} - \frac{ V(\bar{X}_n)}{12 \varepsilon^4}\\ \end{align*} and hence $$1 + \frac{ \sigma^2}{4 n}- \frac{\sigma^2}{12 v_1^4 n} \leq \mathbb{E}(1/\bar{X}_n) \leq 1 + \frac{ \sigma^2}{4 n}.$$

For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:

\begin{align*} \mathbb{E}(1/\bar{X}_n) &= \sum_{i=0}^{+\infty} \frac{f^{(i)}(1)}{i!}\mathbb{E}\left((\bar{X}_n-1)^i\right)\\ &= \sum_{i=0}^{+\infty} \frac{(-1)^i}{i!i!}\mathbb{E}\left((\bar{X}_n-1)^i\right) \end{align*}

Now if we can say something about the $k^{th}$ moment of $\tilde{X}_n$ being $O(1/n^{k/2})$ this validates that $\mathbb{E}(1/\bar{X}_n) \approx 1 + \frac{ \sigma^2}{4 n}$.

  • $\begingroup$ Turns out that $\bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf $\endgroup$ – Gopi Mar 25 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.