Infinite Second Moment implies Growth condition on Expected Maximum

Question

Let $X, X_i$, $1 \leq i \leq n$ be IID random variables such that:

$$\mathbb{E}X^2 = +\infty$$ I am trying to show that this implies a growth condition on $$\mathbb{E}\left[\max_{1\leq i \leq n}{|X_i|}\right].$$ Specifically, I want to show for any $\epsilon > 0$ there exists some $c = c(\epsilon)$ such that

$$\mathbb{E}\left[\max_{1 \leq i \leq n}|X_i|\right] \geq c n^{1/(2+\epsilon)}.$$

Our instructor suggested that we prove the statement using contraposition, e.g. we show that: $$ \liminf \frac{\mathbb{E}[\max_{1 \leq i \leq n}|X_i|]}{n^{1/(2+\epsilon)}} = 0 \implies \mathbb{E}[X^2] < +\infty $$ My attempt: The above condition is equivalent to saying that there exists a subsequence $n_k$ that: $$ \lim_{k \to \infty} \frac{\mathbb{E}[\max_{1 \leq i \leq n_k}|X_i|]}{n_k^{\frac{1}{2+\epsilon}}} = 0 $$ Let $a_n$ be any monotone increasing sequence of positive real numbers such that $\lim a_n = +\infty$. By the monotone convergence theorem and Fubini's theorem: $$ \mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(X^2 > t) dt = \sum_{n=1}^\infty \int_{a_n}^{a_{n+1}}\mathbb{P}(X^2 > t) dt \leq \sum_{n=1}^\infty \underbrace{(a_{n+1} - a_n)}_{:= \Delta a_n} \mathbb{P}(X^2 > a_n) $$ Thus, it suffices to show that the sum on the right hand side is finite for $a_n$ appropriately chosen.

To use information about the maximum, we have for $t> 0$ and $M_n = \max_{1\leq i \leq n} |X_i|$ the running maximum that: $$ \mathbb{P}(M_n >t) = 1 - (\mathbb{P}(X\leq t))^n = 1 - (1 - \mathbb{P}(X > t))^n $$ Rearranging gives: $$ \mathbb{P}(X > t) = 1 - (1 - \underbrace{\mathbb{P}(M_n > t)}_{x})^{1/n} $$ The right hand side is monotone as a function of the argument $x$. Hence, applying Markov's inequality to $M_n$ gives: $$ \mathbb{P}(X > t) \leq 1 - \left(1 - \frac{\mathbb{E}[M_n]}{t}\right)^{1/n} $$ for any integer $n$ and nonnegative number $t$. Replacing $X$ with $X^2$ gives:

$$ \mathbb{P}(X^2 > t) = \mathbb{P}(X > t^{1/2}) \leq 1 - \left(1 - \frac{\mathbb{E}[M_n]}{t^{1/2}}\right)^{1/n} $$ Hence, for $a_k$ as above, we have: $$ \mathbb{E}[X^2] \leq \sum_{k=1}^{\infty} \left(1 - \left(1 - \frac{\mathbb{E}[M_n]}{a_k^{1/2}}\right)^{1/n}\right)\Delta a_k $$

My question now is - what is the right choice of sequence $a_n$? I was considering using the subsequence along which the quotient:

$$ \mathbb{E}[M_{n_k}]/{n_k}^{1/(2+\epsilon)} \to 0 $$

But this scares me a bit because I can't control the size of the increments $\delta n_k$ (the subsequence could be very sparse). One way I was thinking of dealing with that was through summation by parts, but I'm not sure if that would do anything.

Any hints or ideas would be appreciated!

Answer 1

Here are some thoughts. First, if $(Y_i)$ is a collection of identically distributed and non-negative random variables, then $\mathbb E[\max_{1\leqslant i\leqslant n}Y_i]\leqslant \lVert \max_{1\leqslant i\leqslant n}Y_i\rVert_p\leqslant n^{1/p}\lVert Y_1\rVert_p$. Doing a truncation argument, one can see that if $X_1\in\mathbb L^p$, then $n^{-1/p}\mathbb E\left[\max_{1\leqslant i\leqslant n}\lvert X_i\rvert\right]\to 0$. Therefore, if we want the rates that are mentioned in the opening post, that is, that for each $\alpha\in (0,1/2)$, there exists $c_\alpha$ such that $c\geqslant c_\alpha n^{1/2-\alpha}$, then necessarily, we cannot have $X_1\in\mathbb L^p$ for some $p>2$.

An other observation is that if $X_1$ has density $t^{-3}\mathbb{1}_{t>1}$ (for which the square integrability of $X_1$ is barely violated), then $\max_{1\leqslant i\leqslant n}X_i$ has density $nt^{-3}\mathbb{1}_{t>1} \left(1-t^{-2}\right)$. As a consequence, after doing the substitution $u=t^{-2}$, we find that $$ \mathbb E\left[\max_{1\leqslant i\leqslant n}\lvert X_i\rvert\right]=n\int_0^1u^{-1/2}(1-u)^{n-1}du=\operatorname{Beta}(1/2,n)=n\frac{\Gamma(1/2)\Gamma(n)}{\Gamma(n+1/2)}, $$ which behaves as $n^{1/2}$.

A last observation is that we can reduce to the case where $X_1$ take values that are dyadic numbers: indeed, we first reduce to the case where the $X_i$ are non-negative and we define $Y_i:=\sum_{k=1}^\infty 2^k\mathbf{1}_{2^k\leqslant X_i<2^{k+1}}$. Then $Y_i\leqslant X_i\leqslant 2Y_i$. Letting $p_k:=\mathbb P(Y_1=2^k)$, one can compute the law of $\max_{1\leqslant i\leqslant n}Y_i$ in the following way: $\mathbb P(\max_{1\leqslant i\leqslant n}Y_i=2^\ell)=\mathbb P(\max_{1\leqslant i\leqslant n}Y_i\leqslant 2^\ell)-\mathbb P(\max_{1\leqslant i\leqslant n}Y_i\leqslant 2^{\ell-1})=\left(\sum_{k=1}^\ell p_k\right)^n-\left(\sum_{k=1}^{\ell-1} p_k\right)^n$ hence $$ \mathbb E\left[\max_{1\leqslant i\leqslant n}Y_i\right]=\sum_{\ell=1}^\infty 2^\ell \left(\left(\sum_{k=1}^\ell p_k\right)^n-\left(\sum_{k=1}^{\ell-1} p_k\right)^n\right). $$ Letting $s_\ell:=\sum_{k=1}^\ell p_k$, this can be rewritten as $$ \mathbb E\left[\max_{1\leqslant i\leqslant n}Y_i\right]=\sum_{\ell=1}^\infty 2^\ell \int_{s_{\ell-1}}^{s_\ell}nt^{n-1}dt. $$ If $t\in [s_{\ell-1},s_\ell]$, then $2^{\ell}$ behaves as $Q_{Y_1}(1-t)$, where $Q_{Y_1}(u)=\inf\{ s, \mathbb P(Y_1>s)\leqslant u)$. Therefore
$$ \mathbb E\left[\max_{1\leqslant i\leqslant n}Y_i\right]\geqslant 2^{-1} \int_{0}^{1}nQ_{Y_1}(1-t)t^{n-1}dt. $$ The assumption we have at our disposal is that $\int_0^1Q_{Y_1}^2(u)du=+\infty$. Maybe some reversed Hölder's inequality can help.