# Central limit theorem for independent random variables, with a Gumbel limit

Consider independent random variables $Y_i$, $i>0$, such that $\mathbb{E}(Y_i)\approx \frac{1}{i}$ and $\text{Var}(Y_i)\approx \frac{1}{i^2}$, where $\approx$ means asymptotically equivalent up to a multiplicative constant. Define the partial sum $S_n=Y_1+\cdots+Y_n$. I am looking for a limit distribution of $S_n - \mathbb{E}(S_n)$. Note that Kolmogorov's three-series theorem applied to the centered random variables $Y_i - \mathbb{E}(Y_i)$ shows that an almost sure limit does exist.

A few directions that have revealed unsuccessful include:

• the traditional central limit theorem, since the $Y_i$'s are independent but not identically distributed
• triangular arrays CLT with i.d. variables, for the same reason
• Lindeberg-Feller type CLT, since the limit of $\text{Var}(S_n)$ is finite
• a direct characterization of the limit distribution by the characteristic function

An example in Durrett's Probability book (Coupon collector's problem, Example 2.2.3. p 57) has the same kind of first two moments, and provides a Gumbel limit distribution. I tested this hypothesis numerically, and the adequation of the distribution of $S_n$ to a Gumbel distribution, for $n$ large, seems reasonably good. See the plot here, which represents an histogram for an independent sample of $S_n$ of size 50 000, with $n=5 000$, along with an adjusted Gumbel distribution:

I suspect some kind of Poisson limit behind this, as in Example 3.6.6. in Durrett's book, or as in Chapter 26 of Gnedenko and Kolmogorov's book (Limit distributions for sum of independent random variables). But I am not able to adapt the proofs to my setting.

Does anyone know if the Gumbel can occur as a limit distribution for such a sum? (for sure in extreme values theory, it represents the limit distribution of the maximum of an iid $n$-sample, but this is not the case here.) EDIT: The Gumbel is an Infinitely Divisible Distribution, and the way to prove it is to show that it is the limit distribution of $X_1+\ldots+X_n-\log(n)$, where $X_i\sim\text{Exp}(i)$.

Just to specify, the random variables are defined by $Y_j=-\log(1-V_j)$, where $V_j\sim\mbox{beta}(1-\sigma,\theta+j\sigma)$, $\sigma\in(0,1)$ and $\theta>-\sigma$. The reason for studying the sum $S_n$ is that it corresponds to $-\log$ of the tail sum, or rest, in a stick-breaking representation where the weights are defined by $\pi_i=V_i\prod_{j=1}^{i-1}(1-V_j)$.

• Have you looked into the theory of Infinitely Divisible Distributions? It seems to be what you are looking for. – guest Apr 25 '14 at 0:54
• Do you have more info on your random variables? Gumbel distributions tend to come up a lot in extreme value theory where you're looking at the maximum over a set of iid random variables. Is there maybe a way to see your sum as a maximization over another set of random variables? – Alex R. Apr 25 '14 at 4:22
• The problem is still confusing as stated. You assert that you are looking at a sum of independents. However in the standard stick breaking, and in your formulation in terms of V values, the Y values are not independent since they depend on shared Vs. Could you specify whether you intend standard non-independent stick breaking or a different model? – guest Apr 28 '14 at 9:25
• As pointed out by Anthony in his answer, you could in principle obtain pretty much any limiting distribution since each of the $Y_i$'s gives an order $1$ contribution to the limit. Since your problem is very specific, it might be possible to explicitly compute it, but there's no reason to believe that it would be one of the "universal" distributions. And no, it has also no reason to be infinitely divisible (and Gumbel isn't). – Martin Hairer Apr 29 '14 at 12:13
• @MartinHairer: Good point. In fact it is misleading to think of this problem as being related to the central limit theorem, in which the whole point is that, thanks to the rescaling, no one term has any effect on the limiting distribution. In this question there is no rescaling so the situation is completely different. – Nate Eldredge Apr 29 '14 at 16:14

There's no hope of obtaining a limit distribution based on this information. As you say, you might as well centre all the $Y_i$'s before you start. Then $S_n$ is a martingale, uniformly integrable (because $\mathbb ES_n^2$ is bounded) and hence $S_n(\omega)$ is convergent for almost every $\omega$.
Now imagine replacing $Y_1$ with a random variable $Y_1'$ with a different distribution, but with the same variance. Let $\hat S_n=Y_2+\ldots+Y_n$, so that $S_n=\hat S_n+Y_1$ and $S_n'=\hat S_n+Y_1'$. The random variable $\hat S_n$ has a limiting distribution, and so $S_n$ and $S_n'$ have distinct limiting distributions (but both families of rvs $(Y_1,Y_2,\ldots)$ and $(Y_1',Y_2,\ldots)$ satisfy your conditions).