来自两个独立的伯努利人口的抽样分布

让我们假设我们有两个独立的伯努利随机变量的样本， $\mathrm{Ber}(\theta_1)$ 和 $\mathrm{Ber}(\theta_2)$ 。

我们如何证明吗？

\frac{({\bar{X}}_{1} - {\bar{X}}_{2}) - (θ_{1} - θ_{2})}{\sqrt{\frac{θ_{1} (1 - θ_{1})}{n_{1}} + \frac{θ_{2} (1 - θ_{2})}{n_{2}}}} \overset{d}{\to} N (0, 1)

$\frac{(\bar X_1-\bar X_2)-(\theta_1-\theta_2)}{\sqrt{\frac{\theta_1(1-\theta_1)}{n_1}+\frac{\theta_2(1-\theta_2)}{n_2}}}\xrightarrow{d} \mathcal N(0,1)$

假设。 $n_1\neq n_2$

distributions sampling bernoulli-distribution

— 一个老人在海中。
source

Z_i = X_1i-X_2i是具有有限均值和方差的iid rv序列。因此，它满足您的结果遵循的Levy-Linderberg中心极限定理。还是您要要求提供证明的证据？

— 3 Diag

@ThreeDiag您如何应用CLT的LL版本？我认为那是不对的。给我写一个答案以检查详细信息。

— 一位老人在海里。

所有详细信息已经存在。为了应用LL，您需要一个具有有限均值和方差的iid rv序列。变量Z_i = X_i1和X_i2满足所有三个要求。独立性来自于两个原始bernoulli var的独立性，您可以通过应用E和V的标准属性来看到E（Z_i）和V（Z_i）是有限的

— 3 Diag

“两个独立的伯努利随机变量的样本”-错误的表达。必须为：“来自伯努利分布的两个独立样本”。

— 维克多

请添加“ as

”。

n_{1}, n_{2} \to \infty

$n_1,n_2\to \infty$

— 维克多

Answers:

放， $a=\frac{\sqrt{\theta_1(1-\theta_1)}}{\sqrt{n_1}}$ ，，。我们有。在的特性功能方面则意味着 $b=\frac{\sqrt{\theta_2(1-\theta_2)}}{\sqrt{n_2}}$ $A=(\bar{X}_1-\theta_1)/a$ $B=(\bar{X}_2-\theta_2)/b$ $A\to_d N(0,1),\ B\to_d N(0,1)$ 我们要证明

ϕ_{A} (t) \equiv E e^{i t A} \to e^{- t^{2} / 2}, ϕ_{B} (t) \to e^{- t^{2} / 2} .

$\phi_A(t)\equiv {\bf E}e^{itA}\to e^{-t^2/2},\ \phi_B(t)\to e^{-t^2/2}.$

D := \frac{a}{\sqrt{a^{2} + b^{2}}} A - \frac{b}{\sqrt{a^{2} + b^{2}}} B \to_{d} N (0, 1)

$D:=\frac{a}{\sqrt{a^2+b^2}}A-\frac{b}{\sqrt{a^2+b^2}}B\to_d N(0,1)$

由于和是独立的，所以 $A$ $B$ 因为我们希望它是。

ϕ_{D} (t) = ϕ_{A} (\frac{a}{\sqrt{a^{2} + b^{2}}} t) ϕ_{B} (- \frac{b}{\sqrt{a^{2} + b^{2}}} Ť ） \to Ë^{- Ť^{2} / 2} ，

$\phi_D(t)=\phi_A\left(\frac{a}{\sqrt{a^2+b^2}}t\right)\phi_B\left(-\frac{b}{\sqrt{a^2+b^2}}t\right)\to e^{-t^2/2},$

该证明不完整。在这里，我们需要一些估计来统一特征函数。但是，在考虑中的情况下，我们可以进行显式计算。把。 $p=\theta_1,\ m=n_1$ 为。因此，对于固定的，

\begin{aligned} ϕ_{X_{1, 1}} (t) & = 1 + p (e^{i t} - 1), \\ ϕ_{{\bar{X}}_{1}} (t) & = (1 + p (e^{i t / m} - 1))^{m}, \\ ϕ_{{\bar{X}}_{1} - θ_{1}} (t) & = (1 + p (e^{i t / m} - 1))^{m} e^{- i p t}, \\ ϕ_{A} (t) & = (1 + p (e^{i t / \sqrt{m p (1 - p)}} - 1))^{m} e^{- i p t \sqrt{m} / \sqrt{p (1 - p)}} \\ = {((1 + p (e^{i t / \sqrt{m p (1 - p)}} - 1)) e^{- i p t / \sqrt{m p (1 - p)}})}^{m} \\ = {(1 - \frac{t^{2}}{2 m} + O (t^{3} m^{- 3 / 2}))}^{m} \end{aligned}

$\begin{align} \phi_{X_{1,1}}(t) &= 1+p(e^{it}-1), \\ \phi_{\bar X_{1}}(t) &= (1+p(e^{it/m}-1))^m, \\ \phi_{\bar X_{1}-\theta_1}(t) &= (1+p(e^{it/m}-1))^m e^{-ipt}, \\ \phi_{A}(t) &= (1+p(e^{it/\sqrt{mp(1-p)}}-1))^m e^{-ipt\sqrt{m}/\sqrt{p(1-p)}} \\[5pt] &= \left( \left(1+p(e^{it/\sqrt{mp(1-p)}}-1)\right)e^{-ipt/\sqrt{mp(1-p)}}\right)^m \\[5pt] &=\left( 1-\frac{t^2}{2m}+O(t^3m^{-3/2}) \right)^m \end{align}$

t^{3} m^{- 3 / 2} \to 0

$t^3m^{-3/2}\to 0$

t

$t$

ϕ_{D} (t) = {(1 - \frac{a^{2} t^{2}}{2 (a^{2} + b^{2}) n_{1}} + O (n_{1}^{- 3 / 2}))}^{n_{1}} {(1 - \frac{b^{2} t^{2}}{2 (a^{2} + b^{2}) n_{2}} + O (n_{2}^{- 3 / 2}))}^{n_{2}} \to e^{- t^{2} / 2}

$\phi_D(t)=\left( 1-\frac{a^2t^2}{2(a^2+b^2)n_1}+O(n_1^{-3/2}) \right)^{n_1} \left( 1-\frac{b^2t^2}{2(a^2+b^2)n_2}+O(n_2^{-3/2}) \right)^{n_2} \to e^{-t^2/2}$

a \to 0

$a\to 0$

b \to 0

$b\to 0$

| e^{- y} - (1 - y / m)^{m} | \leq y^{2} / 2 m

$\left|e^{-y}-(1-y/m)^m\right|\le {y^2}/{2m}\$ when

y / m < 1 / 2

$\ y/m<1/2$ (see /math/2566469/uniform-bounds-for-1-y-nn-exp-y/ ).

Note that similar calculations may be done for arbitrary (not necessarily Bernoulli) distributions with finite second moments, using the expansion of characteristic function in terms of the first two moments.

— Viktor
source

This seems correct. I'll get back to you later on, when I have time to check everything. ;)

— An old man in the sea.

-1

Proving your statement is equivalent to proving the (Levy-Lindenberg) Central Limit Theorem which states

If $\{Z_i\}_{i=1}^n$ is a sequence of i.i.d random variable with finite mean $\mathbb{E}(Z_i) = \mu$ and finite variance $\mathbb{V}(Z_i) = \sigma^2$ then

\sqrt{n} (\bar{Z} - μ) \to^{d} N (0, σ^{2})

$\sqrt{n}(\bar{Z} - \mu) \to^d N(0,\sigma^2)$

Here $\bar{Z} = \sum_i Z_i/n$ that is the sample variance.

Then it is easy to see that if we put

Z_{i} = X_{1} i - X_{2} i

$Z_i = X_1i - X_2i$ with

X_{1 i}, X_{2 i}

$X_{1i}, X_{2i}$ following a

B e r (θ_{1})

$Ber(\theta_1)$ and

B e r (θ_{2})

$Ber(\theta_2)$ respectively the conditions for the theorem are satisfied, in particular

E (Z_{i}) = θ_{1} - θ_{2} = μ

$\mathbb{E}(Z_i) = \theta_1 - \theta_2 = \mu$

and

V (Z_{i}) = θ_{1} (1 - θ_{1}) + θ_{2} (1 - θ_{2}) = σ^{2}

$\mathbb{V}(Z_i)= \theta_1(1-\theta_1) +\theta_2(1-\theta_2)= \sigma^2$

(There's a last passage, and you have to adjust this a bit for the general case where $n_1 \neq n_2$ but I have to go now, will finish tomorrow or you can edit the question with the final passage as an exercise )

— Three Diag
source

I could not obtain what I wanted exactly because of the possibility of

n_{1} \neq n_{2}

$n_1\neq n_2$

— An old man in the sea.

I will show later if you can't get it. Hint: compute the variance of the sample mean of Z and use that as the variable in the theorem

— Three Diag

Three, could you please add the details for when

n_{1} \neq n_{2}

$n_1 \neq n_2$ ? Thanks

— An old man in the sea.

Will do as soon as find a little timr. There was in fact a subtlety that prevents from using LL clt without adjustment. There are three ways to go, the simplest of which is invoking the fact that for large n1 and n2, X1 and X2 go in distribution to normals, then a linear combination of normal is also normal. This is a property of normals that you can take as given, otherwise you can prove it by characteristic functions.

— Three Diag

其他两个或者需要不同的CLT（李雅普诺夫可能）或可替代地对待N1 = i和N2 = I + K。然后对于大号i，您基本上可以忽略k，您可以返回以应用LL（但仍然需要谨慎以确保正确的方差）

— 3 Diag