중앙 한계 정리와 다수의 법칙에 동의하지 않는 경우

이것은 본질적으로 math.se에서 찾은 질문 의 복제이며 , 원하는 답변 을 얻지 못했습니다.

보자 $\{ X_i \}_{i \in \mathbb{N}}$ 함께 독립적으로 동일하게 분산 된 랜덤 변수들의 시퀀스 일 $\mathbb{E}[X_i] = 1$ 및 $\mathbb{V}[X_i] = 1$ .

의 평가를 고려

$$\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)$$

불평등 사건의 양쪽이 무한대 인 경향이 있기 때문에이 표현은 조작되어야한다.

A) SUBTRACTION 시도

제한 문구를 고려하기 전에 양쪽에서 을 빼십시오 .$\sqrt{n}$

$$\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i -\sqrt{n} \leq \sqrt{n}-\sqrt{n} \right) = \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq 0\right) \\ = \Phi(0) = \frac{1}{2}$$

CLT에 의한 마지막 동등성. 여기서 는 표준 정규 분포 함수입니다.$\Phi()$

B) 멀티 플레이 시도

양변에 \ lim_ {n \ to \ infty} \ mathbb {P} \ left (\ frac {1} {\ sqrt {n}} \ cdot \ frac {1} {\ sqrt를 곱하십시오. {n}} \ sum_ {i = 1} ^ n X_i \ leq \ frac {1} {\ sqrt {n}} \ cdot \ sqrt {n} \ right) = \ lim_ {n \ to \ infty} \ mathbb {P} \ left (\ frac {1} {n} \ sum_ {i = 1} ^ n X_i \ leq 1 \ 오른쪽)$1/\sqrt{n}$

$$\lim_{n \to \infty} \mathbb{P}\left(\frac {1}{\sqrt{n}}\cdot \frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \frac {1}{\sqrt{n}}\cdot\sqrt{n} \right) = \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i \leq 1\right)$$

$$= \lim_{n \to \infty} \mathbb{P}\left(\bar X_n \leq 1\right) = \lim_{n \to \infty}F_{\bar X_n}(1) = 1$$

여기서 은 표본 평균 의 분포 함수이며 , LLN에 의해 ​​확률 (및 분포도)이 상수 수렴 되므로 마지막 평등입니다.$F_{\bar X_n}()$$\bar X_n$$1$

따라서 결과가 상충됩니다. 어느 것이 맞습니까? 그리고 왜 다른 사람이 잘못 되었습니까?

이 오류는 여기에 다음과 같은 사실 가능성이 높습니다 : 유통의 융합은 암시 적으로 그 가정 로 수렴을 F ( X ) 에서 의 연속성의 점 F ( X ) . 한계 분포는 상수 랜덤 변수 이므로 x = 1 에서 점프 불연속성을 가지므로 CDF가 F ( x ) = 1로 수렴한다고 결론을 내릴 수 없습니다 . $F_n(x)$$F(x)$ $F(x)$$x=1$$F(x)=1$

IID 랜덤 변수 가진 E [ X I ] = VAR ( X I ) = 1 정의 Z의 $X_i$$E[X_i]= \operatorname{var}(X_i)=1$ 이제 CLT는 모든고정실수z에 대해limn→∞FZn(z)=Φ(z-1)라고 말합니다. OP는 CLT를 적용하여Limn→∞P(Zn≤1을 평가

$$\begin{align}Z_n &= \frac{1}{\sqrt{n}}\sum_{i=1}^n X_i,\\ Y_n &= \frac{1}{{n}}\sum_{i=1}^n X_i. \end{align}$$ $z$$\lim_{n\to\infty} F_{Z_n}(z) = \Phi(z-1)$ $$\lim_{n\to\infty}P\left(Z_n \leq \frac{1}{\sqrt{n}}\right) = \Phi(0) = \frac 12.$$

OP의 질문에 대한 다른 의견과 여러 의견이 지적했듯이 OP의 평가는 입니다. iid X i 가 확률이 1 인 값 0 과 2 를 취하는 이산 랜덤 변수 인 특수한 경우를 고려하십시오.$\lim_{n\to\infty} P(Y_n \leq 1)$$X_i$$0$$2$ . 이제Σ N 난 = 1 Xi가취할 수있는 모든에서도 정수 값[0,2N]등 때를N, 홀수 Σ N 난 = 1 X전값에 대해 수행 할 수없음그러므로 및YN=$\frac 12$$\sum_{i=1}^n X_i$$[0,2n]$$n$$\sum_{i=1}^n X_i$$n$는 값1을취할 수 없습니다. 또한,Yn의 분포는약1이므로 대칭이므로 P(Yn≤1)=FYn(1)의값은$Y_n = \frac 1n \sum_{i=1}^n X_i$ $1$$Y_n$$1$$P(Y_n \leq 1) = F_{Y_n}(1)$n이 홀수 일 때마다 2 따라서,시퀀스번호의 P(Y1≤1),P(Y2≤1),...,P(YN≤1),... 포함 된서브 시퀀스P(Y1≤1),P가(Y3≤1),…,P($\frac 12$$n$

$$P(Y_1 \leq 1), P(Y_2 \leq 1), \ldots, P(Y_n \leq 1), \ldots$$ 모든 항의 값이 $$P(Y_1 \leq 1), P(Y_3 \leq 1), \ldots, P(Y_{2k-1} \leq 1), \ldots$$ . 한편,서브 시퀀스P(Y2≤1),P(Y4≤1),...,P(Y2K≤1),... 되는집광에1. 따라서,limn→∞P(Yn≤1)가 존재하지 않으며P의 수렴 주장(Yn≤$\frac 12$ $$P(Y_2 \leq 1), P(Y_ 4\leq 1), \ldots, P(Y_{2k} \leq 1), \ldots$$ $1$$\lim_{n\to\infty} P(Y_n \leq 1)$ 부터 1까지는 많은 의심으로보아야합니다.$P(Y_n\leq 1)$

첫 번째 결과는 맞습니다. 다음과 같은 잘못된 설명에서 두 번째 부분에서 오류가 발생합니다.

$$\lim_{n \rightarrow \infty} F_{\bar{X}_n}(1) = 1.$$

이 진술은 허위입니다 (오른쪽은 1 이어야합니다) ) 그것은주장 된많은 수의법칙을따르지 않습니다. 많은 수의 약한 법칙 (당신이 부르는)은 다음과 같이 말합니다 :$\tfrac{1}{2}$

$$\lim_{n \rightarrow \infty} \mathbb{P} \Big( |\bar{X}_n - 1| \leqslant \varepsilon \Big) = 1 \quad \quad \text{for all } \varepsilon > 0.$$

$\varepsilon > 0$$|\bar{X}_n - 1| \leqslant \varepsilon$$\bar{X}_n \leqslant 1$$\bar{X}_n > 1$$\lim_{n \rightarrow \infty} \mathbb{P} ( \bar{X}_n \leqslant 1 ) = 1$

Convergence in probability implies convergence in distribution. But... what distribution? If the limiting distribution has a jump discontinuity then the limits become ambiguous (because multiple values are possible at the discontinuity).

where $F_{\bar X_n}()$ is the distribution function of the sample mean $\bar X_n$, which by the LLN converges in probability (and so also in distribution) to the constant $1$,

This is not right, and it is also easy to show that it can not be right (different from the disagreement between CLT and LLN). The limiting distribution (which can be seen as the limit for a sequence of normal distributed variables) should be:

$$F_{\bar{X}_\infty}(x) = \begin{cases} 0 & \text{for } x<1 \\ 0.5& \text{for } x=1\\ 1 & \text{for } x>1 \end{cases}$$

for this function you have that, for any $\epsilon>0$ and every $x$, the difference $|F_{\bar{X}_n}(x)-F_{\bar{X}_\infty}(x)|<\epsilon$ for sufficiently large $n$. This would fail if $F_{\bar{X}_\infty}(1)=1$ instead of $F_{\bar{X}_\infty}(1)=0.5$

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Limit of a normal distribution

It may be helpful to explicitly write out the sum used to invoke the law of large numbers.

$$\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i \sim N(1,\frac{1}{n})$$

^{The limit $n\to \infty$ for $\hat{X}_n$ is actually equivalent to the Dirac Delta function when it is represented as the limit of the normal distribution with the variance going to zero.}

Using that expression it is more easy to see what is going on under the hood, rather than using the ready-made laws of the CLT an LLN which obscure the reasoning behind the laws.

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Convergence in probability

The law of large numbers gives you 'convergence in probability'

$$\lim_{n \to \infty} P(|\bar{X}_n-1|>\epsilon) =0$$

with $\epsilon > 0$

^{An equivalent statement could be made for the central limit theorem with $\lim_{n \to \infty} P(|\frac{1}{\sqrt{n}}\sum \left( X_i-1 \right)|>\frac{\epsilon}{n}) =0$}

It is wrong to state that this implies

$$\lim_{n \to \infty} P(|\bar{X}_n-1|>0) =0$$

^{It is less nice that this question is cross-posted so early (confusing, yet interesting to see the different discussions/approaches math vs stats, so not that too bad). The answer by Michael Hardy on the math stackexchange deals with it very effectively in terms of the strong law of large numbers (the same principle as the accepted answer from drhab in the cross posted question and Dilip here). We are almost sure that a sequence $\bar{X}_1, \bar{X}_2, \bar{X}_3, ... \bar{X}_n$ converges to 1, but this does not mean that $\lim_{n \to \infty} P(\bar{X}_n = 1)$ will be equal to 1 (or it may not even exist as Dilip shows). The dice example in the comments by Tomasz shows this very nicely from a different angle (instead of the limit not existing, the limit goes to zero). The mean of a sequence of dice rolls will converge to the mean of the dice but the probability to be equal to this goes to zero.}

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Heaviside step function and Dirac delta function

The CDF of $\bar{X}_n$ is the following:

$$F_{\bar{X}_n}(x) = \frac{1}{2} \left(1 + \text{erf} \frac{x-1}{\sqrt{2/n}} \right)$$

with, if you like, $\lim_{n \to \infty} F_{\bar{X}_n}(1) = 0.5$ (related to the Heaviside step function, the integral of the Dirac delta function when viewed as the limit of normal distribution).

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I believe that this view intuitively resolves your question regarding 'show that it is wrong' or at least it shows that the question about understanding the cause of this disagreement of CLT and LLN is equivalent to the question of understanding the integral of the Dirac delta function or a sequence of normal distributions with variance decreasing to zero.

I believe it should be clear by now that "the CLT approach" gives the right answer.

Let's pinpoint exactly where the "LLN approach" goes wrong.

Starting with the finite statements, it is clear then that we can equivalently either subtract $\sqrt{n}$ from both sides, or multliply both sides by $1/\sqrt{n}$. We get

$$\mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)=\mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n(X_i-1) \leq 0\right) = \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^nX_i \leq 1\right)$$

So if the limit exists, it will be identical. Setting $Z_n = \frac{1}{\sqrt{n}} \sum_{i=1}^n(X_i-1)$, we have, using distribution functions

$$\mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)= F_{Z_n}(0) = F_{\bar X_n}(1)$$

...and it is true that $\lim_{n\to \infty}F_{Z_n}(0)= \Phi(0) = 1/2$.

The thinking in the "LLN approach" goes as follows: "We know from the LLN that $\bar X_n$ converges in probabililty to a constant. And we also know that "convergence in probability implies convergence in distribution". So, $\bar X_n$ converges in distribution to a constant". Up to here we are correct.
Then we state: "therefore, limiting probabilities for $\bar X_n$ are given by the distribution function of the constant at $1$ random variable",

$$F_1(x) = \cases {1 \;\;\;\;x\geq 1 \\ 0 \;\;\;\;x<1} \implies F_1(1) = 1$$

... so $\lim_{n\to \infty} F_{\bar X_n}(1) = F_1(1) = 1$...

...and we just made our mistake. Why? Because, as @AlexR. answer noted, "convergence in distribution" covers only the points of continuity of the limiting distribution function. And $1$ is a point of discontinuity for $F_1$. This means that $\lim_{n\to \infty} F_{\bar X_n}(1)$ may be equal to $F_1(1)$ but it may be not, without negating the "convergence in distribution to a constant" implication of the LLN.

And since from the CLT approach we know what the value of the limit must be ($1/2$). I do not know of a way to prove directly that $\lim_{n\to \infty} F_{\bar X_n}(1) = 1/2$.

Did we learn anything new?

I did. The LLN asserts that

$$\lim_{n \rightarrow \infty} \mathbb{P} \Big( |\bar{X}_n - 1| \leqslant \varepsilon \Big) = 1 \quad \quad \text{for all } \varepsilon > 0$$

$$\implies \lim_{n \rightarrow \infty} \Big[ \mathbb{P} \Big( 1-\varepsilon <\bar{X}_n \leq 1\Big) + \mathbb{P} \Big( 1 <\bar{X}_n \leq 1+\varepsilon\Big)\Big] = 1$$

$$\implies \lim_{n \rightarrow \infty} \Big[ \mathbb{P} \Big(\bar{X}_n \leq 1\Big) + \mathbb{P} \Big( 1 <\bar{X}_n \leq 1+\varepsilon\Big)\Big] = 1$$

The LLN does not say how is the probability allocated in the $(1-\varepsilon, 1+\varepsilon)$ interval. What I learned is that, in this class of convergence results, the probability is at the limit allocated equally on the two sides of the centerpoint of the collapsing interval.

The general statement here is, assume

$$X_n\to_p \theta,\;\;\; h(n)(X_n-\theta) \to_d D(0,V)$$

where $D$ is some rv with distribution function $F_D$. Then

$$\lim_{n\to \infty} \mathbb P[X_n \leq \theta] = \lim_{n\to \infty}\mathbb P[h(n)(X_n-\theta) \leq 0] = F_D(0)$$

...which may not be equal to $F_{\theta}(0)$ (the distribution function of the constant rv).

Also, this is a strong example that, when the distribution function of the limiting random variable has discontinuities, then "convergence in distribution to a random variable" may describe a situation where "the limiting distribution" may disagree with the "distribution of the limiting random variable" at the discontinuity points. Strictly speaking, the limiting distribution for the continuity points is that of the constant random variable. For the discontinuity points we may be able to calculate the limiting probability, as "separate" entities.