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I had a question in my mind , if a i.i.d distribution function follows central limit theorem , does that mean it will follow Strong law of large numbers also ??

Since in both cases sample means tends to population mean

Please explain .

Here i mean Lindberg levy's Clt .

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    $\begingroup$ What does it mean to follow the central limit theorem? $\endgroup$
    – Dave
    Oct 23, 2021 at 14:12
  • $\begingroup$ There several versions of CLT, LLN and SLLN. Which of them do you imply? With i.i.d r.v.? Or not? $\endgroup$
    – Botnakov N.
    Oct 23, 2021 at 14:17
  • $\begingroup$ @dave I meant that in the distribution function as n tends to infinity the distribution will tend to become normal ( $\mu , \sigma^2$) $\endgroup$
    – simran
    Oct 23, 2021 at 14:37
  • $\begingroup$ @BotnakovN. I have tried to answer your question $\endgroup$
    – simran
    Oct 23, 2021 at 14:39
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    $\begingroup$ It seems you have a common misconception about the central limit theorem (like most people, present company included, have at some point). $\endgroup$
    – Dave
    Oct 23, 2021 at 14:40

2 Answers 2

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In short: No.

A bit longer: Convergence in distribution does not directly imply, by any way, convergence almost-surely.

Much longer:

Given a random vector $X$, whose components are independent and identically distributed and its first two moments are finite - then the CLT says that asymptotically, the sample mean $\bar{X}_n$ converges to $\mu$ with rate $\sqrt{n}$ and asymptotic distribution $N(0,\sigma^2)$. That's convergence in distribution. It means that the CLT provides us with information regarding the rate in which the sample mean converges weakly to the population mean as sample size increases.

The Weak LLN says that the sample mean of a vector with a finite first moment, converges in probability to the population mean. That is, $\lim_{n\rightarrow\infty}P(\left| \bar{X}_n-\mu \right|>\epsilon)=0$. This means that no matter how small is the nonzero margin $\epsilon$ that we take, a sufficiently large sample would bring the difference between the sample mean and the population mean inside this margin. As you can see here, it is possible (under several conditions) that the CLT would imply the WLLN.

The Strong LLN says that the sample mean converges almost surely to the population mean. That is, $P\left(\lim_{n\rightarrow\infty}\bar{X}_n=\mu \right)=1$. This means that for a sufficiently large sample size, the probability of $\bar{X}_n$ not converging to $\mu$ is 0. That is a substantially stronger form of convergence, and cannot be directly implied from the CLT.

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As you suppose that $X_i$ are i.i.d. such that CLT holds then you suppose that $EX_1$ exists (because it is a term in CLT). Hence $X_i$ are i.i.d. with finite expectation and SLLN holds true.

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  • $\begingroup$ hey , than what spatzle said , that convergence in distribution doesn't imply almost sure convergence ,So it could be true but not necessarily , than how can one be sure that slln will hold ? $\endgroup$
    – simran
    Oct 23, 2021 at 15:16
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    $\begingroup$ Maybe Spätzle answered to the previous version of your question. My solution is connected with the fact that $X_i$ are i.i.d., and this fact in not mentioned by him. But it's important assumption. $\endgroup$
    – Botnakov N.
    Oct 23, 2021 at 15:21
  • $\begingroup$ Oh okay , thank you $\endgroup$
    – simran
    Oct 24, 2021 at 3:39

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