For identically distributed real-valued random variables $X_1, X_2, \ldots$ on the same probability space having finite expectations, is it true that $\lim_{n\rightarrow\infty} \frac{1}{n}E(\max_{1\leq i\leq n}|X_i|)=0$?

If not, could you explain or give an example?


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Here is an answer in full generality. (In particular, we are not assuming that $X$ has density.) As in @Tom Chen's answer, assume WLOG that $X_i$ are non-negative. Denoting $M_n = \max_{1\leq i \leq n} X_i$, by Tonelli's theorem we have

$$ \frac{1}{n}\Bbb{E}[M_n] = \frac{1}{n}\Bbb{E}\bigg[ \int_{0}^{\infty} \mathbf{1}_{\{ x < M_n \}} \, dx \bigg] = \frac{1}{n}\int_{0}^{\infty} \Bbb{E}[\mathbf{1}_{\{ x < M_n \}}] \, dx = \int_{0}^{\infty} \frac{1 - F_X(x)^n}{n} \, dx, $$

where $F_X(x) = \Bbb{P}(X \leq x)$ is the CDF of the common distribution of $X_i$'s and we utilized the following usual trick

$$ \Bbb{P}(x < M_n) = 1 - \Bbb{P}(M_n \leq x) = 1 - F_X(x)^n. $$

Now since

$$ \frac{1 - F_X(x)^n}{n} = (1 - F_X(x)) \cdot \frac{1 + \cdots + F_X(x)^{n-1}}{n} \leq 1 - F_X(x)$$

and $\int_{0}^{\infty} (1 - F_X(x)) \, dx = \Bbb{E}[X] < \infty$, we can apply the dominated convergence theorem to conclude:

$$ \lim_{n\to\infty} \frac{1}{n}\Bbb{E}[M_n] = \int_{0}^{\infty} \bigg( \lim_{n\to\infty} \frac{1 - F_X(x)^n}{n} \bigg) \, dx = \int_{0}^{\infty} 0 \, dx = 0.$$


If all $X_i$ have finite expectation, then $\lim_{n \to \infty} \frac{1}{n}\ \mathbb{E}[\max_{i \in S}|X_i|]$ = $\lim_{n \to \infty} \frac{1}{n}\ \mathbb{E}_X[nF_X(x)^{n-1}f_X(x)]$. That is if all $X_i$ are i.i.d. random variables.

Then $$ \lim_{n \to \infty} \frac{1}{n}\ \mathbb{E}_X[nF_X(x)^{n-1}f_X(x)] =\\ \lim_{n \to \infty} \mathbb{E}_X[F_X(x)^{n-1}f_X(x)] = \\ \lim_{n \to \infty} \int_\mathbb{R} x \left( F_X(x)^{n-1}f_X(x) \right) \ \text{dx} $$

where $F_X(x)$ is the CDF and $f_X(x)$ is the pdf.

Hopefully I didn't make formality mistakes. If so, can you pick up from there?

  • $\begingroup$ This is incorrect. One cannot simply take the $\max$ out the the expectation. $\endgroup$ – Jack Oct 16 '16 at 1:04
  • $\begingroup$ @Jack You're right. I didn't know for sure the way $\max$ behaves in expectation. $\endgroup$ – Cehhiro Oct 16 '16 at 1:07
  • $\begingroup$ @Jack I've modified the answer to use the $\max$ into an order statistic. Is it well written now? (Thanks for the help. I'm also learning by answering.) $\endgroup$ – Cehhiro Oct 16 '16 at 1:28
  • $\begingroup$ I don't quite understand the notation $E_X$. $\endgroup$ – Jack Oct 16 '16 at 1:36
  • $\begingroup$ @Jack It specifies the expected value is taken wrt $X$, rather than just any $X_i$. $\endgroup$ – Cehhiro Oct 16 '16 at 1:38

Let me expand O. Von Seckendorff's point. Without loss of generality, suppose $X_i$'s are positive random variables. Define $Y_n = \max_{1\le i \le n} X_i$.

Case 1: Suppose $F(x) = 1$ for all $x \ge M$. This implies $X_i \overset{\text{a.s.}}{\le} M$, hence $Y_n \overset{\text{a.s.}}{\le} M$, so $\frac{1}{n}\mathbb{E} Y_n \le \frac{1}{n} M \rightarrow 0$ as $n \rightarrow \infty$.

Case 2: Suppose $F(x) < 1$ for all $x$. Let $$h_n(x) = x F(x)^{n-1}f(x)$$ Since $|h_n(x)| \le x f(x)$ and $x f(x)$ is integrable (since we assume $X_i's$ all have finite expectation), by the monotone convergence theorem, we can interchange limit and integral, giving us $$\lim_{n\rightarrow \infty}\frac{1}{n}\mathbb{E}\left[Y_n\right] = \int_{\mathbb{R}}x \left(\lim_{n\rightarrow \infty}F(x)^{n-1}\right)f(x) dx$$ Since $F(x) < 1$ for all $x$, the the limit will be 0, and hence we're integrating the 0 function, and hence the result follows.

  • $\begingroup$ Thanks for the help. This is much better than mine. $\endgroup$ – Cehhiro Oct 26 '16 at 18:52

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