两个样本卡方检验

这个问题来自范德法特（Van der Vaart）的书《渐近统计》（渐近统计）。253.＃3：

假设和是具有参数和独立多项式向量。在零假设下表明 $\mathbf{X}_m$ $\mathbf{Y}_n$ $(m,a_1,\ldots,a_k)$ $(n,b_1,\ldots,b_k)$ $a_i=b_i$

\sum_{i = 1}^{k} \frac{(X_{m, i} - m {\hat{c}}_{i})^{2}}{m {\hat{c}}_{i}} + \sum_{i = 1}^{k} \frac{(Y_{n, i} - n {\hat{c}}_{i})^{2}}{n {\hat{c}}_{i}}

$\sum_{i=1}^k \dfrac{(X_{m,i} - m\hat{c}_i)^2}{m\hat{c}_i} + \sum_{i=1}^k \dfrac{(Y_{n,i} - n\hat{c}_i)^2}{n\hat{c}_i}$ 具有分布。其中。

χ_{k - 1}^{2}

$\chi^2_{k-1}$

{\hat{c}}_{i} = (X_{m, i} + Y_{n, i}) / (m + n)

$\hat{c}_i = (X_{m,i} + Y_{n,i})/(m+n)$

我需要一些入门帮助。这里的策略是什么？我能够将两个求和数合并为：

\sum_{i = 1}^{k} \frac{(m Y_{n, i} - n X_{m, i})^{2}}{m n (m + n) {\hat{c}}_{i}}

$\sum_{i=1}^k \dfrac{(mY_{n,i} - nX_{m,i})^2}{mn(m+n)\hat{c}_i}$

但与CLT，因为它的加权组合这不会工作 $X_m$ 和 $Y_n$ 。不确定这是否是正确的路径。有什么建议么？

编辑：如果 $m=n$ 则很容易，因为我们得到

\begin{aligned} \frac{m Y_{n} - n X_{m}}{\sqrt{m n (m + n)}} & = \frac{Y_{n} - X_{m}}{\sqrt{(m + n)}} \end{aligned}

$\begin{align*} \dfrac{mY_{n} - nX_{m}}{\sqrt{mn(m+n)}} &= \dfrac{Y_{n} - X_{m}}{\sqrt{(m+n)}} \end{align*}$

其中分子可以看作是多项式变量的差之和，因此我们可以应用CLT，然后使用同一章的定理17.2结束它。但是，我无法弄清楚如何在这种情况下使用不同的样本量来解决这个问题。有什么帮助吗？ $(1,a_1,\ldots,a_k)$

链接到van der Vaart的 Google图书的第17章

— 布迪奥诺维奇
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首先是一些符号。令和表示与相关的分类序列和，即。令。考虑二元化其中是克罗内克的三角洲。所以我们有 $\left\{X_t\right\}_{1,\ldots,m}$ $\left\{Y_t\right\}_{1,\ldots,n}$ $\mathbf{X}_m$ $\mathbf{Y}_n$ $\text{Pr}\left\{X_t = i\right\} = a_i, \text{Pr}\left\{Y_t = i\right\} = b_i$ $N=n+m$

\begin{aligned} X_{i}^{*} & = (X_{1, i}^{*}, \dots, X_{N, i}^{*}) = (δ_{i, X_{1}}, \dots, δ_{i, X_{n}}, 0, \dots, 0) \\ Y_{i}^{*} & = (Y_{1, i}^{*}, \dots, Y_{N, i}^{*}) = (0, \dots, 0, δ_{i, Y_{1}}, \dots, δ_{i, Y_{n}}) \end{aligned}

$\begin{align*} \mathbf{X}_{i}^* &= (X^*_{1,i},\ldots,X_{N,i}^*) = (\delta_{i,X_1},\ldots,\delta_{i,X_n},0,\ldots,0)\\ \mathbf{Y}_{i}^* &= (Y^*_{1,i},\ldots,Y_{N,i}^*)= (0,\ldots,0,\delta_{i,Y_1},\ldots,\delta_{i,Y_n})\\ \end{align*}$

δ_{i, j} \equiv 1_{i = j}

$\delta_{i,j}\equiv \mathbf{1}_{i=j}$

X_{m, i} = \sum_{t = 1}^{N} X_{t, i}^{*} = \sum_{t = 1}^{m} δ_{i, X_{t}} Y_{n, i} = \sum_{t = 1}^{N} Y_{t, i}^{*} = \sum_{t = 1}^{n} δ_{i, Y_{t}}

$X_{m,i} = \sum_{t=1}^{N} X_{t,i}^* = \sum_{t=1}^m \delta_{i,X_t} \qquad Y_{n,i} = \sum_{t=1}^{N} Y_{t,i}^* = \sum_{t=1}^n \delta_{i,Y_t}$

现在我们开始证明。首先，我们将检验统计量的两个合计相结合。请注意，因此我们可以将测试统计信息写为

\begin{aligned} X_{m, i} - m {\hat{c}}_{i} & = \frac{(n + m) X_{m, i} - m (X_{m, i} + Y_{n, i})}{n + m} \\ = \frac{n X_{m, i} - m Y_{n, i}}{n + m} \\ Y_{n, i} - n {\hat{c}}_{i} & = \frac{(n + m) Y_{n, i} - n (X_{m, i} + Y_{n, i})}{n + m} \\ = \frac{m Y_{n, i} - n X_{m, i}}{n + m} \end{aligned}

$\begin{align*} X_{m,i} - m\hat{c}_i &= \dfrac{(n+m)X_{m,i} - m(X_{m,i} + Y_{n,i})}{n+m}\\ &= \dfrac{nX_{m,i} - mY_{n,i}}{n+m}\\ Y_{n,i} - n\hat{c}_i &= \dfrac{(n+m)Y_{n,i} - n(X_{m,i} + Y_{n,i})}{n+m}\\ &= \dfrac{mY_{n,i} - nX_{m,i}}{n+m} \end{align*}$

\begin{aligned} S & = \sum_{i = 1}^{k} \frac{(X_{m, i} - m {\hat{c}}_{i})^{2}}{m {\hat{c}}_{i}} + \sum_{i = 1}^{k} \frac{(Y_{n, i} - n {\hat{c}}_{i})^{2}}{n {\hat{c}}_{i}} \\ = \sum_{i = 1}^{k} \frac{(n X_{m, i} - m Y_{n, i})^{2}}{(n + m)^{2} m {\hat{c}}_{i}} + \sum_{i = 1}^{k} \frac{(n X_{m, i} - m Y_{n, i})^{2}}{(n + m)^{2} n {\hat{c}}_{i}} \\ = \sum_{i = 1}^{k} \frac{(n X_{m, i} - m Y_{n, i})^{2}}{n m (n + m) {\hat{c}}_{i}} \end{aligned}

$\begin{align*} S &= \sum_{i=1}^k \dfrac{(X_{m,i} - m\hat{c}_i)^2}{m\hat{c}_i} + \sum_{i=1}^k \dfrac{(Y_{n,i} - n\hat{c}_i)^2}{n\hat{c}_i}\\ &= \sum_{i=1}^k \dfrac{(nX_{m,i} - mY_{n,i})^2}{(n+m)^2m\hat{c}_i} + \sum_{i=1}^k \dfrac{(nX_{m,i} - mY_{n,i})^2}{(n+m)^2n\hat{c}_i}\\ &= \sum_{i=1}^k \dfrac{(nX_{m,i} - mY_{n,i})^2}{nm(n+m)\hat{c}_i} \end{align*}$

接下来注意，与以下属性

n X_{m, i} - m Y_{n, i} = \sum_{t = 1}^{N} n X_{t, i}^{*} - m Y_{t, i}^{*} = Z_{i}

$nX_{m,i} - mY_{n,i} = \sum_{t=1}^N nX_{t,i}^* - mY_{t,i}^* = Z_{i}$

\begin{aligned} E [Z_{i}] & = n E [X_{m, i}] - m E [Y_{n, i}] \\ = n m a_{i} - n m a_{i} = 0 \\ Var [Z_{i}] & = Var [n X_{m, i} - m Y_{n, i}] \\ = n^{2} Var [X_{m, i}] - m^{2} Var [Y_{n, i}] Note X_{m, i} and Y_{n, i} are independent \\ = n^{2} m a_{i} (1 - a_{i}) + m^{2} n a_{i} (1 - a_{i}) \\ = n m (n + m) a_{i} (1 - a_{i}) \\ Cov [Z_{i}, Z_{j}] & = E [Z_{i} Z_{j}] - E [Z_{i}] E [Z_{j}] \\ = E [(n X_{m, i} - m Y_{n, i}) (n X_{m, j} - m Y_{n, j})] \\ = n^{2} (- m a_{i} a_{j} + m^{2} a_{i} a_{j}) - 2 n^{2} m^{2} a_{i} a_{j} + m^{2} (- n a_{i} a_{j} + n^{2} a_{i} a_{j}) \\ = - n m (n + m) a_{i} a_{j} \end{aligned}

$\begin{align*} \text{E}[Z_{i}] &= n\text{E}[X_{m,i}] - m\text{E}[Y_{n,i}]\\ &= nma_i - nma_i = 0\\ \text{Var}[Z_{i}] &= \text{Var}[nX_{m,i} - mY_{n,i}]\\ &= n^2\text{Var}[X_{m,i}] - m^2\text{Var}[Y_{n,i}] \qquad\text{Note $X_{m,i}$ and $Y_{n,i}$ are independent}\\ &= n^2ma_i(1-a_i) + m^2na_i(1-a_i)\\ &= nm(n+m)a_i(1-a_i)\\ \text{Cov}[Z_{i},Z_{j}] &= \text{E}[Z_{i}Z_{j}] - \text{E}[Z_{i}]\text{E}[Z_{j}]\\ &= \text{E}[(nX_{m,i} - mY_{n,i})(nX_{m,j} - mY_{n,j})]\\ &= n^2(-ma_ia_j + m^2a_ia_j) - 2n^2m^2a_ia_j + m^2(-na_ia_j+n^2a_ia_j)\\ &= -nm(n+m)a_ia_j \end{align*}$

因此，通过多元CLT，我们有，其中的的第i个元素，。由于通过Slutsky，我们有其中是单位矩阵，

\frac{1}{\sqrt{n m (n + m)}} Z = \frac{n X_{m} - m Y_{n}}{\sqrt{n m (n + m)}} \overset{D}{\to} N (0, Σ)

$\dfrac{1}{\sqrt{nm(n+m)}}\mathbf{Z} = \dfrac{n\mathbf{X}_m - m \mathbf{Y}_n}{\sqrt{nm(n+m)}}\overset{D}{\to} \text{N}(\mathbf{0},\Sigma)$

(i, j)

$(i,j)$

Σ

$\Sigma$

σ_{i j} = a_{i} (δ_{i j} - a_{j})

$\sigma_{ij} = a_i(\delta_{ij} - a_j)$

\hat{c} = ({\hat{c}}_{1}, \dots, {\hat{c}}_{k}) \overset{p}{\to} (a_{1}, \dots, a_{k}) = a

$\hat{\mathbf{c}} = (\hat{c}_1,\ldots,\hat{c}_k) \overset{p}{\to} (a_1,\ldots,a_k)=\mathbf{a}$

\frac{n X_{m} - m Y_{n}}{\sqrt{n m (n + m)} \hat{c}} \overset{D}{\to} N (0, I_{k} - \sqrt{a} {\sqrt{a}}^{'})

$\dfrac{n\mathbf{X}_m - m \mathbf{Y}_n}{\sqrt{nm(n+m)}\hat{\mathbf{c}}}\overset{D}{\to} \text{N}(\mathbf{0},\mathbf{I}_k - \sqrt{\mathbf{a}}\sqrt{\mathbf{a}}')$

I_{k}

$\mathbf{I}_k$

k \times k

$k\times k$

\sqrt{a} = (\sqrt{a_{1}}, \dots, \sqrt{a_{k}})

$\sqrt{\mathbf{a}} = (\sqrt{a_1},\ldots,\sqrt{a_k})$ 。由于具有乘数1的特征值0和乘数特征值1（通过连续映射定理（或参见范德瓦尔特的引理17.1，定理17.2）有

I_{k} - \sqrt{a} {\sqrt{a}}^{'}

$\mathbf{I}_k - \sqrt{\mathbf{a}}\sqrt{\mathbf{a}}'$

k - 1

$k-1$

\sum_{i = 1}^{k} \frac{(n X_{m, i} - m Y_{n, i})^{2}}{n m (n + m) {\hat{c}}_{i}} \overset{D}{\to} χ_{k - 1}^{2}

$\sum_{i=1}^k \dfrac{(nX_{m,i} - mY_{n,i})^2}{nm(n+m)\hat{c}_i} \overset{D}{\to} \chi^2_{k-1}$

— 布迪奥诺维奇
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