If $X_1, X_2, \ldots X_n, \ldots$ are iid integrable real random variables, and $S_n = X_1 + \cdots + X_n$, $n \geq 1$, the standard strong law of large numbers expresses that $\frac 1n S_n \to E(X_1)$ almost surely. Let, for each $n\geq 1$, $X_n^* = X_k$ if $|X_k|$ is the maximum of $|X_1|, \ldots, |X_n|$ (the one with the smallest index e.g. in case of equality). What would be the almost sure asymptotic behavior (limit?) of $\frac 1n (S_n - X_n^*)$, and under which, possibly necessary and sufficient, condition on the law of $X_1$?
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1$\begingroup$ The SLLN actually tells you that single terms can't have a big effect on the sample mean. Suppose $S_n/n$ converges. Then it follows that $X_n/n$ converges to $0$ (otherwise the process $S_n/n$ would keep having significant jumps). That in turn implies that $X_n^*/n$ converges to $0$ (since if there are infinitely many $n$ with $|X_n^*|/n>\epsilon$, then there are infinitely many $k$ with $|X_k|/k>\epsilon$). So as soon as you have the SLLN, you also have almost sure convergence of $(S_n-X_n^*)/n$ to the mean. $\endgroup$– James MartinCommented Jan 11, 2023 at 11:57
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$\begingroup$ Yes. Given $\epsilon>0$, Integrability and equidistribution yields the convergence of the series $\sum_n P[|X_n|>\epsilon n]$. Then, independence and Borel Cantelli lemma shows that almost surely $|X_n| \le n \epsilon$ eventually. Therefore, $X_n/n \to 0$ almost surely, and $X^*_n/n \to 0$ almost surely. $\endgroup$– Christophe LeuridanCommented Jan 11, 2023 at 12:59
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$\begingroup$ Thank you. But suppose the SLLN does not hold, i.e.$ E(|X_1|) = \infty$. Could it still be that $\frac 1n (S_n - X_n^*)$ is stabilized almost surely, and under which minimal assumption on the law of $X_1$? $\endgroup$– Ed PanzaCommented Jan 11, 2023 at 19:22
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$\begingroup$ Consider your process $S_n-X_n^*$. Every time that we see a new record $X_n$, your process has a jump equal to the size of the previous record. Now either: (a) the records are $o(n)$ as $n\to\infty$. Then subtracting $X_n^*$ makes no difference to the limit behaviour once we divide by $n$. Or (b) the process $S_n-X_n^*$ keeps having jumps which are not negligible compared to $n$, and you won't get convergence of $(S_n-X_n^*)/n$. Hence you can't get convergence for your process unless the SLLN holds. Various other approaches also possible, e.g. as outlined by Christophe. $\endgroup$– James MartinCommented Jan 11, 2023 at 21:51
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$\begingroup$ @JamesMartin : I am not sure if I can follow the logic in your latter comment: Whatever size the record jumps are, they are immediately eliminated by the subtraction. What seems to be important, then, is the size of the second-largest jump. In any case, it would be good if you could present your comments, especially the latter one, as a formal answer. $\endgroup$– Iosif PinelisCommented Jan 12, 2023 at 2:04
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