Test dwóch próbek chi do kwadratu

To pytanie pochodzi z książki Van der Vaarta Asymptotic Statistics, str. 253. # 3:

Załóżmy, że $\mathbf{X}_m$ i $\mathbf{Y}_n$ to niezależne wielomianowy wektorów parametrów $(m,a_1,\ldots,a_k)$ a $(n,b_1,\ldots,b_k)$ . Zgodnie z hipotezą zerową, że wskazują, że $a_i=b_i$

ma dystrybucji. gdzie.

\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}}

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

χ_{k - 1}^{2}

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

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

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

Potrzebuję pomocy na początku. Jaka jest tutaj strategia? Udało mi się połączyć dwa lata w:

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

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

ale przyzwyczajenie pracy z CLT ponieważ jego ważonej kombinacji i . Nie jestem pewien, czy to właściwa ścieżka. Jakieś sugestie? $X_m$ $Y_n$

EDYCJA: jeśli to jest dość łatwe, ponieważ otrzymujemy $m=n$

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

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

gdzie licznik może być postrzegana jako suma różnic wielomianu zmiennych więc możemy zastosować CLT a następnie zakończyć ją z twierdzenia 17.2 z tego samego rozdziału. Nie mogę jednak wymyślić, jak to zrobić w tej sytuacji przy różnych wielkościach próbek. Jakaś pomoc? $(1,a_1,\ldots,a_k)$

Link do rozdziału 17 książek Google van van Vaarta

self-study chi-squared multinomial central-limit-theorem bdeonovic
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Najpierw notacja. Niech , a oznacza kategoryczne sekwencję związanego z i , tj . Niech $\left\{X_t\right\}_{1,\ldots,m}$ $\left\{Y_t\right\}_{1,\ldots,n}$ $\mathbf{X}_m$ $\mathbf{Y}_n$ $\text{Pr}\left\{X_t = i\right\} = a_i, \text{Pr}\left\{Y_t = i\right\} = b_i$ $N=n+m$ . Rozważ binerizacje gdzieto Kronecker Delta. Mamy więc

\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}}

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

\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}

$\begin{align*} X_{m,i} - m\hat{c}_i &= \dfrac{(n+m)X_{m,i} - m(X_{m,i} + Y_{n,i})}{n+m}\\ &= \dfrac{nX_{m,i} - mY_{n,i}}{n+m}\\ Y_{n,i} - n\hat{c}_i &= \dfrac{(n+m)Y_{n,i} - n(X_{m,i} + Y_{n,i})}{n+m}\\ &= \dfrac{mY_{n,i} - nX_{m,i}}{n+m} \end{align*}$ So we can write the test statistic as

\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}

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

Next note that

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

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

\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*}$

and so by multivariate CLT we have

\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)

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

(i, j)

$(i,j)$ th element of

Σ

$\Sigma$ ,

σ_{i j} = a_{i} (δ_{i j} - a_{j})

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

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

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

\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}}^{'})

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

I_{k}

$\mathbf{I}_k$ is the

k \times k

$k\times k$ identity matrix,

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

$\sqrt{\mathbf{a}} = (\sqrt{a_1},\ldots,\sqrt{a_k})$ . Since

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

$\mathbf{I}_k - \sqrt{\mathbf{a}}\sqrt{\mathbf{a}}'$ has eigenvalue 0 of multiplicty 1 and eigenvalue 1 of multiplicity

k - 1

$k-1$ , by the continuous mapping theorem (or see Lemma 17.1, Theorem 17.2 of van der Vaart) we have

\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}

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

bdeonovic
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Test dwóch próbek chi do kwadratu

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