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$(X_k)_k$ is a sequence of independent r.v uniformly bounded by $c.$ If $\sum_{k}X_k$ converges a.s then $\sum_{k}E[X_k]$ converges.

The above is proved using the following inequality ($X_k$ should be centered): $$P\left(\max_{p \leq k}|Y_p|>\epsilon\right) \geq 1-\frac{(\epsilon+c)^2}{E[Y^2_k]} \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ where $Y_k=\sum_{p=1}^kX_k.$

If $X_k$ are not centered then they consider independent copy of $X_k.$

The objective is to find a way to solve the problem without the use of symmetric random variable, after searching, the following inequality (it seems it's a generalization of $(1)$) seems to solve the problem: $$P\left(\max_{1 \leq p \leq k}|Y_p| >\epsilon\right) \geq 1-4\frac{(\epsilon+\max_{1 \leq p \leq k } |X_p-E[X_p]|)^2}{\operatorname{Var}(Y_k)} \ \ \ \ \ \ \ \ (2)$$

Once $(2)$ is proved we can see that $\max_{1 \leq p \leq k}|X_p-E[X_p]| \leq 2c$

How to prove $(2)$ ? (Of course the natural way to prove it is to consider $E_p:=\{|Y_1| \leq \epsilon,...,|Y_{p-1}| \leq \epsilon,|Y_p|>\epsilon\}$).

Any other inequalities to solve the problem are welcomed!

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  • $\begingroup$ What is objectionable about symmetrization? $\endgroup$ Apr 28, 2021 at 0:15
  • $\begingroup$ It seems you asked essentially the same question at the same time on math.stackexchange.com. Please don’t do this. It leads to a duplication of effort. This question is probably more suitable for this site, so you should probably delete the other one. $\endgroup$ Apr 28, 2021 at 1:18
  • $\begingroup$ @IosifPinelis no objection! (We can also deduce the above theorem from the CLT...). So, it's just an interesting question, especially the inequality $(2)$! $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 2:02
  • $\begingroup$ Inequality (2) looks strange, with its right-hand side being a random variable. $\endgroup$ Apr 28, 2021 at 3:36
  • $\begingroup$ After a long search, it turns out $(2)$ is taken from : link.springer.com/content/pdf/10.1007/BF01459098.pdf, unfortunately not in english, possibly in german (according to google translate) $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 13:19

1 Answer 1

1
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The statement

If $(X_k)$ is a sequence of independent r.v.'s uniformly bounded in absolute value by some real $c$ such that $\sum_{k}X_k$ converges a.s., then $\sum_{k}EX_k$ converges

follows almost immediately from the Hoffmann–Jørgensen inequality $$EM^2\le2.4^2c^2+32t_0^2$$ (see e.g. Proposition 6.8), where $M:=\sup_k|S_k|$, $S_k:=\sum_{j=1}^k X_j$, and $t_0>0$ is such that $P(M>t_0)<1/32$. Indeed, since $S_\infty$ is finite a.s., we have $M<\infty$ a.s. and hence we can assume that $t_0<\infty$, so that $Var\, S_n\le ES_n^2\le EM^2\le2.4^2c^2+32t_0^2<\infty$ for all natural $n$.

Now Kolmogorov's two-series theorem implies that $\sum_k(X_k-EX_k)$ converges a.s. Since $\sum_k X_k$ converges a.s., we conclude that $\sum_k EX_k$ converges, as desired.

Extended comment: As the OP noted in a comment, inequality (2) does appear, as Satz III, in Kolmogorov's paper (of 1928, in German). However, apparently there is a mistake in the proof of this inequality in that paper. Namely, with standard modern notations: $E$ in place of $\mathfrak D$ (the expectation symbol in that paper), $A_k$ in place of $\mathfrak f_k$ (a certain event), and $EX1_{A_k}$ in place of $\mathfrak D_{\mathfrak f_k}(X)$, the equality in line 10 on p. 312 of Kolmogorov's paper becomes $$|t_k-Et_k1_{A_k}|=|s_k-Es_k1_{A_k}|,$$ which was apparently thought of as the consequence of the equality $$t_k-Et_k1_{A_k}=s_k-Es_k1_{A_k};\tag{$*$}$$ here, according to p. 310 of Kolmogorov's paper, $t_k=s_k-Es_k$. However, actually $$t_k-Et_k1_{A_k}=s_k-Es_k1_{A_k}-Es_k(1-P(A_k)).$$ So, ($*$) holds only if $Es_k=0$ or $P(A_k)=1$, which latter is not true in general. Thus, the proof seems to only work if $Es_k=0$ for all $k$. This is the first time in my life where I seem to see a mistake made by Kolmogorov!

It appears that this mistake was (tacitly?) corrected in later publications by symmetrization (which obviously provides for the zero-mean condition). If that correction was made indeed tacitly, it appears to be a disservice to people interested in this matter, as the question on this webpage shows.

Anyhow, the Hoffmann–Jørgensen inequality seems to be a more effective and more transparent way to achieve the desired result, as shown above in this answer.

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  • $\begingroup$ After running a search on the above publication (google books), it appears that Kai Lai Chung (A course in probability theory, third edition) use inequality $(2)$ (the correct version) and he provided a proof (it's theorem 5.3.2).. $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 16:07
  • $\begingroup$ The funny thing is that inequality $(1)$ and the symmetric argument are asked in the exercises right after this section $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 16:14
  • $\begingroup$ @Kurt.W.X : I see "Finally, and this is mentioned here only in response to a query by Doob, I chose to present the brutal Theorem 5.3.2 in the original form given by Kolmogorov because I want to expose the student to hardships in mathematics." on p.434 at play.google.com/books/… . Can you give me a link to the Google book by Chung? I cannot find it right away. $\endgroup$ Apr 28, 2021 at 16:18
  • $\begingroup$ I can post a picture - pdf file for the page from Chung books (because in Google books - amazon... it only shows a preview for the book and it happens that the page from where theorem 5.3.2 is taken is not omitted) $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 16:26
  • $\begingroup$ Here's the link for google books: books.google.com.lb/… $\endgroup$
    – Kurt.W.X
    Apr 28, 2021 at 16:29

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