当中心极限定理和大数定律不一致时

从本质上讲，这是我在math.se上发现的一个问题的复制，但没有得到我所希望的答案。

令为一系列独立的，均布的随机变量，其中和。$\{ X_i \}_{i \in \mathbb{N}}$$\mathbb{E}[X_i] = 1$$\mathbb{V}[X_i] = 1$

考虑对

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

由于不平等事件的两面都趋于无穷大，因此必须对此表达式进行操作。

A）尝试减法

在考虑限制语句之前，请从两侧减去$\sqrt{n}$：

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

CLT的最后一个等式，其中$\Phi()$是标准正态分布函数。

B）尝试乘法

将两侧乘以$1/\sqrt{n}$

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

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

其中$F_{\bar X_n}()$是样本均值\ bar X_n的分布函数$\bar X_n$，通过LLN可以将概率（以及分布）收敛到常数$1$，因此是最后的等式。

因此，我们得到了矛盾的结果。 哪个是正确的？又为什么另一个是错误的？

这里的错误很可能会在以下的事实：在分配收敛隐含的假设是收敛于˚F （X ）在连续性点˚F （X ）。由于极限分布是一个恒定的随机变量，因此它在x = 1处具有跳跃不连续性，因此无法得出CDF收敛至F （x ）= 1是不正确的。 $F_n(x)$$F(x)$ $F(x)$$x=1$$F(x)=1$

对于iid随机变量其中E [ X i ] = var （X i）= 1，定义 Z $X_i$$E[X_i]= \operatorname{var}(X_i)=1$ 现在，CLT说，对于每一个固定实数Ž，LIMÑ→交通∞˚FžÑ（ż）=Φ（ž-1）。的OP应用CLT来评估 LIMÑ→交通∞P（żÑ≤

$$\begin{align}Z_n &= \frac{1}{\sqrt{n}}\sum_{i=1}^n X_i,\\ Y_n &= \frac{1}{{n}}\sum_{i=1}^n X_i. \end{align}$$ $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.$$

至于其他的答案，以及一些在OP的问题的评论所指出的，它是对OP的评估这是犯罪嫌疑人。考虑特殊情况，即iid X i是离散随机变量，其概率为1且取值0和$\lim_{n\to\infty} P(Y_n \leq 1)$$X_i$$0$$2$。现在，Σ Ñ 我= 1 X我可以采取所有甚至整数值在[0，2Ñ]，所以当Ñ为奇数时， Σ Ñ 我= 1 X我不能承担值Ñ因此ÿÑ=$\frac 12$$\sum_{i=1}^n X_i$$[0,2n]$$n$$\sum_{i=1}^n X_i$$n$不能取值1。另外，由于分布ÿÑ是对称的约1，我们有 P（ÝÑ≤1）=˚FÝÑ（1）具有值$Y_n = \frac 1n \sum_{i=1}^n X_i$ $1$$Y_n$$1$$P(Y_n \leq 1) = F_{Y_n}(1)$每当n为奇数时为 2。因此，序列号的 P（Ý1≤1），P（ÿ2≤1），...，P（ÿÑ≤1），... 包含序列P（Ý1≤1），P（ÿ3≤1），…，P（$\frac 12$$n$

$$P(Y_1 \leq 1), P(Y_2 \leq 1), \ldots, P(Y_n \leq 1), \ldots$$ ，其中所有的术语都具有值 $$P(Y_1 \leq 1), P(Y_3 \leq 1), \ldots, P(Y_{2k-1} \leq 1), \ldots$$ 。在另一方面，所述亚序列P（Ý2≤1），P（ÿ4≤1），...，P（ÿ2ķ≤1），... 被会聚到1。因此，LIMÑ→交通∞P（ÝÑ≤1）不存在和收敛的权利要求P（ÝÑ≤$\frac 12$ $$P(Y_2 \leq 1), P(Y_ 4\leq 1), \ldots, P(Y_{2k} \leq 1), \ldots$$ $1$$\lim_{n\to\infty} P(Y_n \leq 1)$必须非常怀疑地看待1。$P(Y_n\leq 1)$

您的第一个结果是正确的结果。您的错误发生在第二部分的以下错误语句中：

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

此语句为假（右侧应为），并且不遵循所主张的大数定律。大数定律（您引用）说：$\tfrac{1}{2}$

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

For all $\varepsilon > 0$ the condition $|\bar{X}_n - 1| \leqslant \varepsilon$ spans some values where $\bar{X}_n \leqslant 1$ and some values where $\bar{X}_n > 1$. Hence, it does not follow from the LLN that $\lim_{n \rightarrow \infty} \mathbb{P} ( \bar{X}_n \leqslant 1 ) = 1$.

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.