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Let $X, X1,X_2,\cdots$ be i.i.d random.variables. Check if the WLLN and the SLLN hold if the common d.f. is given by,

(1)$P(X=n)=P(X=-n)=\frac{C}{2n^2ln^2(n)}$, $n \geq 3$

(2)$P(X=n)=P(X=-n)=\frac{C}{2n^2ln(n)}$, $n \geq 3$

(3)At this time, $X_i$ are independent though not identical. Let $P(X_1=0)=P(X_2=0)=1$ and $P(X=n)=P(X=-n)=\frac{1}{2nln(n)}$,$P(X_n=0)=1-\frac{1}{nln(n)}$, $n \geq 3$ (Hint:This sequence obeys the WLLN but not the SLLN)

Solution:

According to the SLLN: Let $X_1,X_2,\cdots$ be pairwise independent with identically distributed random variables with $E|X_i|<\infty$. Let $E(X_1)=u$ and $S_n=X_1+X_2+\cdots+X_n$. Then, $S_n/n \rightarrow u$ as $n\rightarrow \infty$

According to the WLLN: Let $X_1,X_2,\cdots$ be i.i.d with $E|X_i|<\infty$. Let $E(X_1)=u$ and $S_n=X_1+X_2+\cdots+X_n$. Then, $S_n/n \rightarrow u$ as $n\rightarrow \infty$

As for the question (1):

$E(X_i)=\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}k+\sum_{k\leq -3}\frac{C}{2k^2ln^2(k)}k=0$

$E|X_i|=\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}|k|+\sum_{k\leq -3}\frac{C}{2k^2ln^2(k)}|k|=2\sum_{k\geq 3}\frac{C}{2k^2ln^2(k)}|k|=\sum_{k\geq 3}\frac{C}{2k ln^2(k)}=\infty$

It seems that (1) does not satisfy both WLLN and SLLN. Since $E|X_i|=\infty$\

As for the question (2):

$E(X_i)=\sum_{k\geq 3}\frac{C}{2k^2ln(k)}k+\sum_{k\leq -3}\frac{C}{2k^2ln(k)}k=0$

$E|X_i|=\sum_{k\geq 3}\frac{C}{2k^2ln(k)}|k|+\sum_{k\leq -3}\frac{C}{2k^2ln(k)}|k|=2\sum_{k\geq 3}\frac{C}{2k^2ln(k)}|k|=\sum_{k\geq 3}\frac{C}{2k ln(k)}=\infty$

It seems that (2) does not satisfy both WLLN and SLLN. Since $E|X_i|=\infty$\

What is the difference between the condition of WLLN and SLLN? Can the statement satisfy the WLLN, though not satisfy SLLN? Or can the statement satisfies the SLLN, though not satisfy WLLN? For me, the conditions for these two are the same.

Thanks a lot!

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    $\begingroup$ Why do you think that $\sum_{k\geq 3}\frac{C}{2k \ln^2(k)}=\infty$? This sum behaves like $$\int_3^\infty \frac{C}{2x \ln^2(x)}dx=\int_3^\infty \frac{C}{2 \ln^2(x)}d\ln x = -\frac{C}{2 \ln (x)}\Bigm|_3^\infty <\infty$$ $\endgroup$
    – NCh
    Jan 17, 2020 at 3:16
  • $\begingroup$ Thanks a lot! Can you give me the suggestion for these problem. Whether they meet the reuqirement of SLLN or WLLN. $\endgroup$
    – Olivia
    Jan 18, 2020 at 16:49
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$
    – NCh
    Jan 20, 2020 at 13:38

2 Answers 2

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This is the standard example of a sequence that satisfies the hypotheses of WLLN but not of SLLN, so you are correct. Also note that WLLN says something about convergence in probability, while SLLN says something about convergence almost surely, which is stronger.

If the hypotheses of SLLN would always be satisfied if those of WLLN are satisfied, WLLN would be a useless result after all. This exercise shows that this needs not be the case.

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  • $\begingroup$ Thanks! Why this question satisfies the WLLN but not of SLLN. It seems that $|X_i|<\infty$, which is the requirement of both WLLN and SLLN. How about the question (2) and (3)? I tried to work out question (3). $\endgroup$
    – Olivia
    Jan 16, 2020 at 21:09
  • $\begingroup$ Thanks a lot! Can you give me the suggestion for these problem. Whether they meet the reuqirement of SLLN or WLLN? $\endgroup$
    – Olivia
    Jan 18, 2020 at 16:50
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Sequence (1) satisfy WLLN and SLLN since i.i.d.r.v's satisfy $\mathbb E[|X|]<\infty$.

Consider sequence (2). Since $\mathbb E[|X|]=\infty$, SLLN is not fulfilled. This is a consequence of the SLLN:

Let $X_1,X_2,\ldots$ be i.i.d. and $S_n=X_1+\ldots+X_n$. Then $\frac{S_n}{n}\to a$ a.s. iff exists $\mathbb E[X_1]=a$.

And the WLLN can be fulfilled. You can find the following theorem in W.Feller's book "An Introduction to Probability Theory and its Applications", in Vol.2, Chapter VII, paragraph 7:

Theorem. Let $X_1,X_2,\ldots$ be independent with a common distribution $F$. In order that there exist constants $\mu_n$ such that for each $\epsilon>0$ $$\mathbb P\left(\left|\frac{S_n}{n}-\mu_n\right|>\epsilon\right)\to 0\tag{1}$$ it is necessary and sufficient that $$n \mathbb P(|X_1|\geqslant n)\to 0 \text{ at } n\to\infty. \tag{2}$$ In this case (1) holds with $$\mu_n =\mathbb E[X_1\mathbb 1_{\{|X_1|\leq n\}}].$$

Since our r.v.'s have symmetric distribution, we can take $\mu_n=\mu=0$ and check only (2). $$ n \mathbb P(|X_1|\geqslant n) = n\sum_n^\infty \frac{C}{k^2 \ln k}\leqslant \frac{n}{\ln n}\sum_n^\infty \frac{C}{k^2} \sim \frac{n}{\ln n} \int_n^\infty \frac{C}{x^2}dx = \frac{C}{\ln n}\to 0. $$ Therefore WLLN fulfilled.

For (3) both WLLN and SLLN are not fulfilled. You can check it by the same way. Note that the values $X_3,X_4,\ldots $ are i.i.d. so you can apply both theorems. $X_1$ and $X_2$ don't matter.

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  • $\begingroup$ Thanks a lot! I will update this question latter by your suggestion! $\endgroup$
    – Olivia
    Jan 20, 2020 at 2:08
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    $\begingroup$ @EpsilonDelta Thank you, I was not familiar with this result of Etemadi (1981). $\endgroup$
    – NCh
    Jan 20, 2020 at 4:34

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