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More options … # Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# One sided strong laws for random variables with infinite mean

• Corresponding author
• Department of Applied Mathematics, Illinois Institute of Technology, Chicago, Illinois, 60616, USA
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Published Online: 2017-06-22 | DOI: https://doi.org/10.1515/math-2017-0070

## Abstract

This paper establishes conditions that secure the almost sure upper and lower bounds for a particular normalized weighted sum of independent nonnegative random variables. These random variables do not possess a finite first moment so these results are not typical. These mild conditions allow us to show that the almost sure upper limit is infinity while the almost sure lower bound is one.

MSC 2010: 60F15

This paper extends work done in  and . We use those Weak Laws to obtain our almost sure upper and lower bounds. We present conditions that allow us to achieve these bounds for our normalized weighted sums. As in  the random variables at the heart of these theorems are independent without a first moment where P{Xj = 0} = 1 − aj and $P\left\{{X}_{j}>x\right\}=\frac{1}{x+1/{a}_{j}},$ for x > 0. We set ${A}_{n}={\sum }_{j=1}^{n}{a}_{j},$ which must go to infinity in order to establish our theorems.

In terms of notation we use lg x = ln x, but whenever we have ln 1 in a denominator we will just set that equal to one, so we won’t be dividing by zero. Hence lg x = ln x, except lg 1 = 1. We set lg1 x = lg x, and lgk+1 x = lg(lgk x), so the i in lgi x is not the base, it is the iteration of the logarithm. Many times in the proofs and the examples we will pull the logarithm and any other slowly varying function out from an integral and the infinite summations since they are slowly varying, see  pages 279-284. Finally, note that the constant C will be used as a generic bound that is not necessarily the same in each appearance. From  we have a Weak Law that helps us produce the almost sure lower bound.

#### Theorem 1

If An → ∞, then $∑j=1najXjAnlg⁡An→P⁡1.$(1)

We next establish the almost sure upper bound for these same weighted sums. The condition that ensures that the upper bound is infinite is $∑n=1∞anAnlg⁡An=∞.$(2)

#### Theorem 2

If (2) holds, then $lim supn→∞∑j=1najXjAnlg⁡An=∞almostsurely.$

#### Proof

For all M > 0 $∑n=1∞P{anXnAnlg⁡An>M}=∑n=1∞P{Xn>MAnlg⁡Anan}=∑n=1∞1MAnlg⁡Anan+1an=∑n=1∞anMAnlg⁡An+1=∞.$

Thus $lim supn→∞anXnAnlg⁡An=∞almostsurely$ hence $lim supn→∞∑j=inajXjAnlg⁡An≥lim supn→∞anXnAnlg⁡An=∞almostsurely$ which completes this proof. ☐

In order to get the lower limit we assume there is a sequence {dn, n ≥ 1} where $limn→∞∑j=1najlg⁡(ajdj)Anlg⁡An=1$(3) and $∑n=1∞an2dnAn2lg2⁡An<∞.$(4)

Note that we need both An and an dn to go to infinity to achieve our results. But these are mild conditions.

What is fascinating is that even though we can split (3) into two pieces, we need both parts. When aj = jα, where α > −1, we select dj = j which gives us $∑j=1najlg⁡ajAnlg⁡An→αα+1$ and $∑j=1najlg⁡djAnlg⁡An→1α+1.$

This shows that both terms are necessary. Combining that with An ~ nα+1/(α + 1), we see that (3) and (4) hold, see Example 4.

#### Theorem 3

If there exist a sequence {dn, n ≥ 1} satisfying (3) and (4) then $lim infn→∞∑j=1najXjAnlg⁡An=1almostsurely.$

#### Proof

From Theorem 1 we have $lim infn→∞∑j=1najXjAnlg⁡An≤1almostsurely.$

We will show that conditions (3) and (4) establishes $lim infn→∞∑j=1najXjAnlg⁡An≥1almostsurely.$(5)

Using our sequence {dn, n ≥ 1} we have $∑j=1najXjAnlg⁡An≥∑j=1najXjI(Xj≤dj)Anlg⁡An=∑j=1naj[XjI(Xj≤dj)−EXjI(Xj≤dj)]Anlg⁡An+∑j=1najEXjI(Xj≤dj)Anlg⁡An.$

By the Khintchine-Kolmogorov Convergence Theorem, see  page 113, and the Kronecker lemma, the first term converges to zero almost surely since $∑n=1∞an2EXn2I(Xn≤dn)An2lg2⁡An=∑n=1∞an2An2lg2⁡An∫0dnx2dx(x+1/an)2≤∑n=1∞an2dnAn2lg2⁡An<∞$ by (4). We now examine our expectation $∑j=1najEXjI(Xj≤dj)Anlg⁡An=1Anlg⁡An∑j=1naj∫0djxdx(x+1/aj)2=1Anlg⁡An∑j=1naj∫1/ajdj+1/aj(u−1/aj)duu2=1Anlg⁡An∑j=1naj∫1/ajdj+1/ajduu−1Anlg⁡An∑j=1n∫1/ajdj+1/ajduu2.$

The first term converges to one by (3) since $1Anlg⁡An∑j=1naj∫1/ajdj+1/ajduu=1Anlg⁡An∑j=1naj[lg⁡(dj+1/aj)−lg⁡(1/aj)]=1Anlg⁡An∑j=1najlg⁡(ajdj+1)→1$ while the second term goes to zero since $−1Anlg⁡An∑j=1n∫1/ajdj+1/ajduu2≤1Anlg⁡An∑j=1n∫1/aj∞duu2=1Anlg⁡An∑j=1naj=AnAnlg⁡An=1lg⁡An→0.$

Hence (5) does hold, concluding the proof. ☐

We conclude with three examples, each showing how to select {dn, n ≥ 1} so that (3) and (4) hold. What is quite remarkable is that if we select {dn, n ≥ 1} correctly we see that all three conditions hold even though (2) and (4) seem contradictory.

#### Example 4

Let L(x) be any slowly varying function and α > −1, then$lim infn→∞∑j=1njαL(j)Xjnα+1L(n)lg⁡n=1almostsurely$and$lim supn→∞⁡∑j=1njαL(j)Xjnα+1L(n)lg⁡n=∞almostsurely.$

#### Proof

Since aj = jαL(j) we have An ~ nα+1L(n)/(α + 1) and lg An ~ (α + 1) and lg n, thus $∑n=1∞anAnlg⁡An≥C∑n=1∞nαL(n)nα+1L(n)lg⁡n=C∑n=1∞1nlg⁡n=∞.$

In this case we set dj = j. Thus ajdj = jα+1L(j) and lg(ajdj) ~ (α + 1) lg j, hence $∑j=1najlg⁡(ajdj)Anlg⁡An∼(α+1)∑j=1njαL(j)lg⁡jnα+1L(n)lg⁡n→1$ and $∑n=1∞an2dnAn2lg2⁡An≤C∑n=1∞n2α+1L2(n)n2α+2L2(n)lg2⁡n=C∑n=1∞1nlg2⁡n<∞.$

Next we look at a borderline case, where once again An → ∞ is necessary.

#### Example 5

If α > −1, then for all δ$liminfn→∞∑j=1n(lg⁡j)α(lg2⁡j)δjXj(lg⁡n)α+1(lg2⁡n)δ+1=1almostsurely$and$limsupn→∞∑j=1n(lg⁡j)α(lg2⁡j)δjXj(lg⁡n)α+1(lg2⁡n)δ+1=∞almostsurely.$

#### Proof

Since aj = (lgj)α(lg2 j)δ/j we have An ~ (lgn)α+1(lg2n)δ /(α + 1) and lg An ~ (α + 1) lg2 n, thus $∑n=1∞anAnlg⁡An≥C∑n=1∞(lg⁡n)α(lg2⁡n)δ/n(lg⁡n)α+1(lg2⁡n)δ+1=C∑n=1∞1nlg⁡nlg2⁡n=∞.$

In this case we set dj = j lg j. Thus ajdj = (lg j)α+1(lg2 j)δ and lg(aj dj) ~ (α + 1) lg2 j, hence $∑j=1najlg⁡(ajdj)Anlg⁡An∼(α+1)∑j=1n(lg⁡j)α(lg2⁡j)δ+1/j(lg⁡n)α+1(lg2⁡n)δ+1→1$ and $∑n=1∞an2dnAn2lg2⁡An≤C∑n=1∞((lg⁡n)α(lg2⁡n)δn)2(nlg⁡n)((lg⁡n)α+1(lg2⁡n)δ+1)2=C∑n=1∞1nlg⁡nlg22⁡n<∞.$

#### Example 6

If α > −1, then for all δ$liminfn→∞∑j=1n(lg2⁡j)α(lg3⁡j)δjlg⁡jXj(lg2⁡n)α+1(lg3⁡n)δ+1=1almostsurely$and$limsupn→∞∑j=1n(lg2⁡j)α(lg3⁡j)δjlg⁡jXj(lg2⁡n)α+1(lg3⁡n)δ+1=∞almostsurely.$

#### Proof

Since aj = (lg2 j)α(lg3 j)δ/(j lg j) we have An ~ (lg2n)α+1(lg3 n)δ/(α + 1) and lg An ~ (α + 1) lg3 n, thus $∑n=1∞anAnlg⁡An≥C∑n=1∞(lg2⁡n)α(lg3⁡n)δ/(nlg⁡n)(lg2⁡n)α+1(lg3⁡n)δ+1=C∑n=1∞1nlg⁡nlg2⁡nlg3⁡n=∞.$

In this case we set dj = j lgj lg2 j. Thus ajdj = (lg2 j)α+1(lg3 j)δ and lg(ajdj) ~ (α + 1) lg3 j, hence $∑j=1najlg⁡(ajdj)Anlg⁡An∼(α+1)∑j=1n(lg2⁡j)α(lg3⁡j)δ+1/(jlg⁡j)(lg2⁡n)α+1(lg3⁡n)δ+1→1$ and $∑n=1∞an2dnAn2lg2⁡An≤C∑n=1∞((lg2⁡n)α(lg3⁡n)δnlg⁡n)2(nlg⁡nlg2⁡n)((lg2⁡n)α+1(lg3⁡n)δ+1)2=C∑n=1∞1nlg⁡nlg2⁡nlg32⁡n<∞.$

In each case we see that even though (2) and (4) are quite similar series and even though one must be finite and the other infinite, if one selects {dn, n ≥ 1} correctly, these two equations along with (3) hold every time. We can continue with $an=(lgd+1⁡n)α(lgd+2⁡n)δn∏i=1dlgi⁡n$ where α > −1, as long as we set ${d}_{n}=n{\prod }_{i=1}^{d+1}{\mathrm{lg}}_{i}n.$

#### Remark 7

It would be nice to replace (2) with the milder condition An → ∞, but they arent equivalent. If we have rapidly growing sequences we can see the difference. For example, let An = en2, then clearly An → ∞. However$an=An−An−1=en2−e(n−1)2and$∑n=1∞anAnlg⁡An<∑n=1∞en2en2lg⁡(en2)=∑n=1∞1n2<∞.$

## References

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Adler A., An exact weak law of large numbers, Bull. Inst. Math. Acad. Sinica, 2012, 7, 417-422 Google Scholar

• 

Nakata T., Weak law of large numbers for weighted independent random variables with infinite mean. Statist. Probab. Letters, 2016, 109, 124-129

• 

Feller W., An Introduction to Probability Theory and its Applications, Vol 2, 2nd edition, 1996, Wiley and Sons Google Scholar

• 

Chow Y.S., Teicher H., Probability Theory: Independence, Interchangeability, Martingales, 3rd edition, 1997, Springer-Verlag Google Scholar

Accepted: 2017-05-16

Published Online: 2017-06-22

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 828–832, ISSN (Online) 2391-5455,

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