Dari mana berasal dari teorema limit pusat (CLT)?

Versi yang sangat sederhana dari teorema terbatas pusat seperti di bawah ini yang merupakan Lindeberg-Lévy CLT. Saya tidak mengerti mengapa ada di sebelah kiri. Dan Lyapunov CLT mengatakan tetapi mengapa bukan ? Adakah yang bisa memberitahu saya apa saja faktor-faktor ini, seperti dan ? bagaimana kita mendapatkannya di teorema?

$$\sqrt{n}\bigg(\bigg(\frac{1}{n}\sum_{i=1}^n X_i\bigg) - \mu\bigg)\ \xrightarrow{d}\ \mathcal{N}(0,\;\sigma^2)$$ $\sqrt{n}$ $$\frac{1}{s_n} \sum_{i=1}^{n} (X_i - \mu_i) \ \xrightarrow{d}\ \mathcal{N}(0,\;1)$$ $\sqrt{s_n}$$\sqrt{n}$$\frac{1}{s_n}$

Pertanyaan yang bagus (+1) !!

Anda akan ingat bahwa untuk variabel acak independen dan Y , V a r ( X + Y ) = V a r ( X ) + V a r ( Y ) dan V a r ( a ⋅ X ) = a 2 ⋅ V a r ( X ) . Jadi varians dari Σ n i = 1 X i adalah$X$$Y$$Var(X+Y) = Var(X) + Var(Y)$$Var(a\cdot X) = a^2 \cdot Var(X)$$\sum_{i=1}^n X_i$ , dan varians ˉ X = $\sum_{i=1}^n \sigma^2 = n\sigma^2$adalahnσ2/n2=σ2/n.$\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$$n\sigma^2 / n^2 = \sigma^2/n$

Ini untuk varians . Untuk membakukan variabel acak, Anda membaginya dengan standar deviasi. Seperti yang Anda tahu, nilai yang diharapkan dari adalah μ , jadi variabel$\bar{X}$$\mu$

memiliki nilai yang diharapkan 0 dan varian 1. Jadi jika cenderung ke Gaussian, itu harus menjadi standar GaussianN(0

$$\frac{\bar{X} - E\left( \bar{X} \right)}{\sqrt{ Var(\bar{X}) }} = \sqrt{n} \frac{\bar{X} - \mu}{\sigma}$$ . Formulasi Anda dalam persamaan pertama adalah setara. Dengan mengalikan sisi kiri dengan σ Anda mengatur varians ke σ 2 .$\mathcal{N}(0,\;1)$$\sigma$$\sigma^2$

Mengenai poin kedua Anda, saya percaya bahwa persamaan yang ditunjukkan di atas menggambarkan bahwa Anda harus membaginya dengan dan bukan $\sigma$ untuk membakukan persamaan, menjelaskan mengapa Anda menggunakansn(penaksirσ)dan bukan$\sqrt{\sigma}$$s_n$$\sigma)$ .$\sqrt{s_n}$

Tambahan: @whuber menyarankan untuk mendiskusikan mengapa penskalaan oleh . Dia melakukannya disana, tetapi karena jawabannya sangat panjang, saya akan mencoba menangkap esensi argumennya (yang merupakan rekonstruksi pemikiran de Moivre).$\sqrt{n}$

Jika Anda menambahkan sejumlah besar dari +1 dan -1, Anda dapat memperkirakan probabilitas bahwa jumlah tersebut akan menjadi j dengan penghitungan dasar. Log probabilitas ini sebanding dengan - j 2 / n . Jadi jika kita ingin probabilitas di atas konvergen ke konstanta ketika n menjadi besar, kita harus menggunakan faktor normalisasi dalam O ( $n$$j$$-j^2/n$$n$.$O(\sqrt{n})$

Menggunakan alat matematika modern (post de Moivre), Anda dapat melihat perkiraan yang disebutkan di atas dengan memperhatikan bahwa probabilitas yang dicari adalah

$$P(j) = \frac{{n \choose n/2+j}}{2^n} = \frac{n!}{2^n(n/2+j)!(n/2-j)!}$$

yang kami perkirakan dengan rumus Stirling

$$P(j) \approx \frac{n^n e^{n/2+j} e^{n/2-j}}{2^n e^n (n/2+j)^{n/2+j} (n/2-j)^{n/2-j} } = \left(\frac{1}{1+2j/n}\right)^{n+j} \left(\frac{1}{1-2j/n}\right)^{n-j}.$$

$$\log(P(j)) = -(n+j) \log(1+2j/n) - (n-j) \log(1-2j/n) \\ \sim -2j(n+j)/n + 2j(n-j)/n \propto -j^2/n.$$

There is a nice theory of what kind of distributions can be limiting distributions of sums of random variables. The nice resource is the following book by Petrov, which I personally enjoyed immensely.

It turns out, that if you are investigating limits of this type

$$\frac{1}{a_n}\sum_{i=1}^nX_n-b_n, \quad (1)$$ where $X_i$ are independent random variables, the distributions of limits are only certain distributions.

There is a lot of mathematics going around then, which boils to several theorems which completely characterizes what happens in the limit. One of such theorems is due to Feller:

Theorem Let $\{X_n;n=1,2,...\}$ be a sequence of independent random variables, $V_n(x)$ be the distribution function of $X_n$, and $a_n$ be a sequence of positive constant. In order that

$$\max_{1\le k\le n}P(|X_k|\ge\varepsilon a_n)\to 0, \text{ for every fixed } \varepsilon>0$$

and

$$\sup_x\left|P\left(a_n^{-1}\sum_{k=1}^nX_k<x\right)-\Phi(x)\right|\to 0$$

it is necessary and sufficient that

$$\sum_{k=1}^n\int_{|x|\ge \varepsilon a_n}dV_k(x)\to 0 \text{ for every fixed }\varepsilon>0,$$

$$a_n^{-2}\sum_{k=1}^n\left(\int_{|x|<a_n}x^2dV_k(x)-\left(\int_{|x|<a_n}xdV_k(x)\right)^2\right)\to 1$$

and

$$a_n^{-1}\sum_{k=1}^n\int_{|x|<a_n}xdV_k(x)\to 0.$$

This theorem then gives you an idea of what $a_n$ should look like.

The general theory in the book is constructed in such way that norming constant is restricted in any way, but final theorems which give necessary and sufficient conditions, do not leave any room for norming constant other than $\sqrt{n}$.

s$_n$ represents the sample standard deviation for the sample mean. s$_n$$^2$ is the sample variance for the sample mean and it equals S$_n$$^2$/n. Where S$_n$$^2$ is the sample estimate of the population variance. Since s$_n$ =S$_n$/√n that explains how √n appears in the first formula. Note there would be a σ in the denominator if the limit were

N(0,1) but the limit is given as N(0, σ$^2$). Since S$_n$ is a consistent estimate of σ it is used in the secnd equation to taken σ out of the limit.

Intuitively, if $Z_n \to \mathcal N(0, \sigma^2)$ for some $\sigma^2$ we should expect that $\mbox{Var}(Z_n)$ is roughly equal to $\sigma^2$; it seems like a pretty reasonable expectation, though I don't think it is necessary in general. The reason for the $\sqrt n$ in the first expression is that the variance of $\bar X_n - \mu$ goes to $0$ like $\frac 1 n$ and so the $\sqrt n$ is inflating the variance so that the expression just has variance equal to $\sigma^2$. In the second expression, the term $s_n$ is defined to be $\sqrt{\sum_{i = 1} ^ n \mbox{Var}(X_i)}$ while the variance of the numerator grows like $\sum_{i = 1} ^ n \mbox{Var}(X_i)$, so we again have that the variance of the whole expression is a constant ($1$ in this case).

Essentially, we know something "interesting" is happening with the distribution of $\bar X_n := \frac 1 n \sum_i X_i$, but if we don't properly center and scale it we won't be able to see it. I've heard this described sometimes as needing to adjust the microscope. If we don't blow up (e.g.) $\bar X - \mu$ by $\sqrt n$ then we just have $\bar X_n - \mu \to 0$ in distribution by the weak law; an interesting result in it's own right but not as informative as the CLT. If we inflate by any factor $a_n$ which is dominated by $\sqrt n$, we still get $a_n(\bar X_n - \mu) \to 0$ while any factor $a_n$ which dominates $\sqrt n$ gives $a_n(\bar X_n - \mu) \to \infty$. It turns out $\sqrt n$ is just the right magnification to be able to see what is going on in this case (note: all convergence here is in distribution; there is another level of magnification which is interesting for almost sure convergence, which gives rise to the law of iterated logarithm).