In practice, we often assume that the process we are examining has a mean, and so statistics involving averages are defined.

As pure conjecture, I was wondering if one can construct a hypothesis test for the existence of a mean. This seemed a bit too broad, so how about the more limited case where the data are iid observations of a symmetric distribution?

Here's what I've tried so far, and where I got stuck:

Assume: $E[X_i]=\mu \in (-\infty,\infty)\;\textrm{and} \;X_i \sim F(x_i;\mu)$


$$\bar{X}_n := \frac{\sum_{i=1}^n X_i}{n} \xrightarrow{wp1} \mu\; \textrm{by SLLN}$$

We also know that the distribution of $\bar{X}_n$ will converge to a stable distribution $f(x; \alpha,\beta,c,\mu)$, where $\beta=0,\alpha>1$ (to ensure it is both symmetric and has a mean), thus:

$$\bar{X}_n \xrightarrow{d} f(x;\alpha,0,c,\mu)$$

I didn't find this very helpful, because there are still too many free parameters. Therefore, I tried another adjustment:

Let $Y_n=X_{n}-\bar{X}_{n-1},\; n>1$ then:

$$\lim_{n\to \infty} E[Y_n] = E[X_n] - E[\bar{X}_{n-1}] = \mu - \lim_{n\to \infty} E[\bar{X}_{n-1}] = \mu-\mu = 0 $$

Where the third expression is justified by observing that $E[\bar{X}_k]$ is a constant function in $k$:

$$E[\bar{X}_k] = \frac{1}{k}\sum_{i=1}^k E[X_i] = \mu \;\forall k\implies \lim_{k\to\infty} E[\bar{X}_k] = \mu$$

So, we can now say that:

$$Y_n \xrightarrow{d} f(y;\alpha,0,c,0)$$


What I really want it so find a series of transformations of $Y_n$ that allow me to get a sequence of random variables $G_n$ such that:

$$G_n \xrightarrow{d} f(g;\alpha,0,1,0)$$

Then, I want to construct an asymptotically correct hypothesis test from an observation of $G_n$:

$$H_0: \alpha > 1;\; H_a: \alpha\leq 1$$

So we have the test:

$$T(a;R\subset \mathbb{R})=1 \implies a \notin R,$$

$$T= 0\;o/w$$

Where $a=G_n$, and we define $R:=R_n$ such that $E_{\alpha>1}[T(G_n;R_n)]\leq P(\textrm{Type I error})$, where we have some pre-specified Type I error rate.

Thoughts so far

I don't know how to find such a sequence, but it appears that some work has been done on direct inference on $\alpha$. See here.

I think that a reasonable test would be as follows:

  1. Partition the sample into $K$ groups.
  2. Let $Y_{i,k}:= X_{i,k} - \bar{X}$ be the mean corrected values using the grand mean.
  3. For each $k \in K$ calculate the partition sample mean $\bar{Y}_k$
  4. Assume that $\bar{Y}_{k}$ are observations from an $\alpha-$stable distribution with $\beta=\mu=0$
  5. Estimate $\alpha$ using one of the methods listed here and approx. (1-p)%-confidence interval $I_p$.
  6. If $\alpha<1 \in I_p$, reject at level $p$ and conclude a mean does not exist at level $p$.

Not sure if there area any issues with this though.

  • 1
    $\begingroup$ I would attack the higher moments first. If mean doesn't exist, then the variance shouldn't too (me thinks). It could be easier to show that the variance is growing with a sample size. $\endgroup$
    – Aksakal
    Commented Sep 16, 2015 at 18:47
  • 1
    $\begingroup$ @Aksakal this is true and an interesting idea (+1). I'd imagine the test is just as hard, since the $n>1^{th}$ moment of $X$ is the $1^{st}$ moment (expected value) of some other RV $Z$. $\endgroup$
    – user75138
    Commented Sep 16, 2015 at 20:36
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    $\begingroup$ I thought that higher moments would be easier to handle because they should blow up faster. $\endgroup$
    – Aksakal
    Commented Sep 16, 2015 at 20:49
  • $\begingroup$ @Aksakal perhaps, but I'm having a hard time coming up with a test that doesn't reduce to testing a mean again. Note, if $Z=X^2$ and $E[X]=0$ then $Var[X]=E[Z]$. I'm open to a concrete example or link. $\endgroup$
    – user75138
    Commented Sep 16, 2015 at 21:44

1 Answer 1


Not really. See here for example: Test for finite variance?

  • 1
    $\begingroup$ (+1) Thanks for the very relevant link! I think there's an issue in your test tho (not mathematically, but logically)....as Aksakal mentioned, if the $p^{th}$ moment exists, then so do the lower moments, including the one I am after. $\endgroup$
    – user75138
    Commented Sep 16, 2015 at 21:53

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