# Big-O bounds on the k-th largest element of iid Gaussians

I'm interested in the following problem. Let $$X_1, \dots, X_n$$ be iid samples with a $$N(0,1)$$ distribution. Let $$X_{[k]}$$ be the $$k$$-th largest element of $$\{X_1, \dots, X_n\}$$, so e.g. $$X_{[1]} = \max_i X_i$$.

Are there simple big-O bounds on $$X_{[k]}$$ for, say, all $$k < n/2$$? I know that $$X_{[1]} = O(\sqrt{\log{n}})$$ (see here), and more generally, $$X_{[c]} = O(\sqrt{\log{n}})$$ for constant $$c$$ (see here). But what about $$X_{[\log{n}]}$$ or $$X_{[n/4]}$$, for example?

My motivation is that I want to calculate $$\sum_{i=1}^k E[X_{[i]}]$$ for various values of $$k$$.

Thanks!

• It's addressed at stats.stackexchange.com/questions/9001/… Jun 28 '19 at 8:07
• @BjørnKjos-Hanssen These are not very nice-looking "simple big-O bounds" however (which I assume is the main point of the OP's question). Jun 28 '19 at 16:04
• @ClementC. is correct, I saw that answer but those are not big-O bounds Jun 28 '19 at 18:13

This is not a complete answer by any means, but just a quick estimate on $$\mathbb{E}[\sum_{i=1}^k X_{[i]}]$$ that is slightly better than the trivial bound of $$O(k\sqrt{\log n})$$. If this is your goal, I would think it is easier to go directly for it than consider any given $$X_{[k]}$$. Let $$X_S=\sum_{i\in S} X_i$$ for a subset $$S\subseteq [n]$$ and $$Y_k=\sum_{i=1}^k X_{[i]}$$. Using the same technique as in the first link, \begin{align} \exp(t\mathbb{E}[Y_k])&\leq \mathbb{E}[\exp(tY_k)]\\ &=\mathbb{E}\bigg[\max_{S\subseteq [n]:\vert S\vert=k}\exp(tX_S)\bigg]\\ &\leq \sum_{S\subseteq [n]:\vert S\vert=k}\mathbb{E}[\exp(tX_S)]\\ &=\sum_{S\subseteq [n]:\vert S\vert=k}\exp\bigg(\frac{kt^2}{2}\bigg)\\ &={n \choose k}\exp\bigg(\frac{kt^2}{2}\bigg). \end{align} Taking (natural) logarithms, we find $$\mathbb{E}[Y_k]\leq \frac{\log{n \choose k}}{t}+\frac{kt}{2}.$$ Optimizing gives $$t=\sqrt{\frac{2\log{n \choose k}}{k}}$$ so we deduce $$\mathbb{E}[Y_k]\leq \sqrt{2k\log{n \choose k}}.$$

For $$k=o(n)$$, it is known that $$\log{n \choose k}=(1+o(1))k\log(n/k)$$, so in this regime, we get $$\mathbb{E}[Y_k]\leq k\sqrt{2(1+o(1))\log(n/k)},$$ which for $$k=\log n$$, say, is a bit better than the naive bound.

When $$k=\alpha n$$ for some constant $$\alpha$$, one instead has $$\log{n \choose k}=(1+o(1))H(\alpha)n$$, where $$H(p)=-p\log p-(1-p)\log(1-p)$$ is the entropy function. So in this regime, that would give $$\mathbb{E}[Y_k]\leq n\sqrt{2(1+o(1))\alpha H(\alpha)}.$$

Lower bounds added for completeness: for linear $$k$$, one can get decent linear lower bounds that degrade as $$\alpha\to 0$$ but is tight at $$\alpha=1/2$$. Let $$\alpha<1/2$$, then by the Chernoff bound, $$\Pr(X_{\alpha n+1}\leq 0)\leq \exp\bigg(\frac{-(1/2-\alpha)^2n}{2}\bigg).$$ It is known that for $$i, the conditional distribution of $$X_{[i]}$$ given $$X_{[j]}=x$$ is the same as the unconditional distribution of $$X_{[i]}$$ taken from a sample of size $$j-1$$ conditioned to be larger than $$x$$. As we are summing over all $$i\leq \alpha n$$ and taking expectations, we find that $$\mathbb{E}[\sum_{i=1}^{\alpha n} X_{[i]}\vert X_{\alpha n+1}\geq 0]\geq \mathbb{E}[\sum_{i=1}^{\alpha n} Y_i],$$ where the $$Y_i$$ are normal random variables conditioned to be larger than $$0$$, which are known to have expectation $$\sqrt{2/\pi}$$. We also find that $$\mathbb{E}[\sum_{i=1}^{\alpha n} X_{[i]}\vert X_{\alpha n+1}\leq 0]\geq 0,$$ as for each $$x<0$$, conditioning on $$X_{\alpha n}=x$$, by the same reasoning we are now taking expectations over a sample of size $$\alpha n$$ normal random variables conditioned to be at least $$x$$, which by symmetry of the normal distribution is at least $$0$$ (formally, we are using that the sum over the whole sample makes the labelling of the order statistics irrelevant). We conclude that $$\mathbb{E}[\sum_{i=1}^{\alpha n} X_{[i]}]\geq \bigg(1-\exp\bigg(\frac{-(1/2-\alpha)^2n}{2}\bigg)\bigg)\alpha n\sqrt{2/\pi}=(1-o(1))\alpha n \sqrt{2/\pi}.$$ This holds for all $$\alpha<1/2$$, and by monotonicity of the left hand side in $$\alpha$$ for $$\alpha<1/2$$ and taking limits, we can further conclude using the upper bound in the comments $$\mathbb{E}[\sum_{i=1}^{n/2} X_{[i]}]=(1-o(1))n\sqrt{1/(2\pi)}.$$

Obviously similar lower bounds hold for $$k=o(n)$$ but these aren't super useful as they are of vastly different order than the upper bound.

Numerical Simulations: Numerical simulations suggest that these bounds are surprisingly decent in the two regimes mentioned in the post. Note that for any $$k$$, the function $$f(x_1,\ldots,x_n)=\sum_{i=1}^k x_{[i]}$$ is Lipschitz, so by Gaussian concentration, one expects simulations to roughly approximate the expectation. For instance, for $$k=\log(n)$$, here is a plot of the upper bound with that of simulation, with the top curve being the naive bound of $$\sqrt{2}\log(n)^{3/2}$$ and the lower curve the given bound:

In the linear $$k$$ regime, for the few values I tried, the constant of the upper bounds seems to be off by a factor of at most $$3$$ (at $$\alpha=1/2$$, where we have computed the exact asymptotics). For instance, for $$k=n/4$$, the plot with the bound looks like:

The upper bounds seem pretty good for small constant $$\alpha$$, while the lower bounds degrade. For instance, here is $$\alpha=.001$$:

Here's one more plot (because I can't resist) with $$k=\sqrt{n}$$: Hope this helps!

• Wow, these bounds are actually perfect for my application!! I really appreciate the graphs too showing that they're tight haha. Your strategy of bounding the sum directly, instead of each of the individual terms, is much smarter. Thank you!! Jun 28 '19 at 17:06
• You're welcome! By the way, for the specific case of $k=n/2$, one has $\mathbb{E}[\sum_{i=1}^{n/2} X_{[i]}]=\frac{1}{2}\mathbb{E}[\sum_{i=1}^{n/2} X_{[i]}-\sum_{i=1}^{n/2} X_{[n-i]}]$, so by TI, $\mathbb{E}[\sum_{i=1}^{n/2} X_{[i]}]\leq \frac{1}{2}\mathbb{E}[\sum_{i=1}^n \vert X_{[i]}\vert]=n\sqrt{1/(2\pi)}$. Simulations suggests this is essentially exact; there's probably a way to get a lower bound for $k=\alpha n$, $\alpha<1/2$ of $(1-o(1))\alpha n\sqrt{2/\pi}$ (the intuition is the top $\alpha n$ will be positive w.h.p. and so should be at least the expected absolute value of a Gaussian).
– J.G
Jun 28 '19 at 21:37
• @UthsavChitra in case you're interested, I've added a proof of the linear lower bound that is tight at $\alpha=1/2$. Hope this is helpful!
– J.G
Jun 28 '19 at 23:49
• Note that using these bounds you can derive asymptotic bounds on $\mathbb{E} X_{[k]}$, since deterministically $kX_{[k]} \leq \sum_{i=1}^k X_{[k]}$. Thus $\mathbb{E} X_{[k]} \leq \tfrac{1}{k}\mathbb{E} Y_k \leq \sqrt{\tfrac{2}{k}\log\tfrac{n}{k}}$. For example, if $k = o(n)$ as above, this gives $\mathbb{E} X_{[k]} \leq \sqrt{(2+o(1))\log \tfrac{n}{k}}$. Jun 29 '19 at 17:15