The chi-square approximation of Pearson's statistic

Posted on July 12, 2019 by Stéphane Laurent
Tags: maths, statistics

Our goal is to derive the asymptotic distribution of Pearson’s statistic for goodness-of-fit testing. We follow the method given in David Williams’s book Weighing the odds, but we provide a bit more details.

Let \(Y_1\), \(\ldots\), \(Y_n\) be independent and identically distibuted random variables in \(\{1, \ldots, b\}\) whose common distribution is given by \[ \Pr(Y_m = k) = p_k \] with \(\sum_{k=1}^b p_k = 1\).

Denoting by \(N_k\) the number of \(Y_m\) equal to \(k\), and setting \[ W_k = \frac{N_k - np_k}{\sqrt{np_k}}, \] the Pearson statistic is \(\sum_{k=1}^b W_k^2\). To derive its asymptotic distribution, we firstly derive the one of the random vector \({(W_1, \ldots, W_k)}'\).

Define the random variables \[ X_k^{(m)} = \begin{cases} 1 & \text{if } Y_m = k \\ 0 & \text{if } Y_m \neq k \end{cases}, \] so that \[ N_k = X_k^{(1)} + \cdots + X_k^{(n)}. \]

For any \(m_1\), \(m_2\), the random vectors \((X_1^{(m_1)}, \ldots, X_b^{(m_1)})\) and \((X_1^{(m_2)}, \ldots, X_b^{(m_2)})\) have the same distribution. It is easy to see that \[ \mathbb{E}X_k^{(m)} = p_k, \quad \mathbb{E}X_k^{(m)}X_\ell^{(m)} = \begin{cases} p_k & \text{if } k = \ell \\ 0 & \text{if } k \neq \ell \end{cases}, \] leading to \[ V_{k\ell} := \text{Cov}\bigl(X_k^{(m)},X_\ell^{(m)}\bigr) = \begin{cases} p_k(1-p_k) & \text{if } k = \ell \\ -p_k p_\ell & \text{if } k \neq \ell \end{cases}. \]

One can write \[ \begin{pmatrix} N_1 \\ \vdots \\ N_b \end{pmatrix} = \begin{pmatrix} X_1^{(1)} \\ \vdots \\ X_b^{(1)} \end{pmatrix} + \cdots + \begin{pmatrix} X_1^{(n)} \\ \vdots \\ X_b^{(n)} \end{pmatrix}. \] and this is a sum of independent and identically distributed random vectors. Therefore we have, for large \(n\), \[ \begin{pmatrix} N_1 \\ \vdots \\ N_b \end{pmatrix} \approx \mathcal{M}\mathcal{N}\left(\begin{pmatrix} n p_1 \\ \vdots \\ n p_b \end{pmatrix}, n V\right) \] by the multivariate central limit theorem.

Recall that \[ W_k = \frac{N_k - np_k}{\sqrt{np_k}}. \] Then one has \[ \begin{pmatrix} W_1 \\ \vdots \\ W_b \end{pmatrix} \approx \mathcal{M}\mathcal{N}\left(\begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}, C\right) \] where \[ C_{k\ell} = \begin{cases} 1-p_k & \text{if } k = \ell \\ -\sqrt{p_k p_\ell} & \text{if } k \neq \ell \end{cases}. \]

Now we are going to derive the characteristic function of \(\mathcal{M}\mathcal{N}(\mathbf{0}, C)\).

The covariance matrix \(C\) is not a strictly positive definite matrix since \(\sum_{k=1}^b \sqrt{np_k} W_k = 0\). But let us firstly give an expression of \(\mathbb{E}\mathbf{e}^{-\alpha \sum_{k=1}^b W_k^2}\) when \({(W_1, \ldots, W_b)}' \sim \mathcal{M}\mathcal{N}(\mathbf{0}, C)\) in the case of a strictly positive definite covariance matrix \(C\). Then we will argue that this expression still holds for our \(C\). In the strictly positive case, by using the pdf of the multivariate normal distribuion, we get \[\begin{align} \mathbb{E}\mathbf{e}^{-\alpha \sum_{k=1}^b W_k^2} & = \frac{1}{{(2\pi)}^\frac{b}{2}\sqrt{\det(C)}} \int_{\mathbb{R}^b} \mathbf{e}^{-\frac12\mathbf{w}' (C^{-1} + 2 \alpha I) \mathbf{w}}\mathrm{d}\mathbf{w} \\ & = \frac{{\bigl(\det(C^{-1} + 2\alpha I)\bigr)}^{-\frac12}}{\sqrt{\det(C)}} \\ & = {\bigl(\det(I+2\alpha C)\bigr)}^{-\frac12}. \end{align}\] Now, by a continuity argument, this equality holds when \(C\) is non-negative definite. Let us detail this point. Let \(\mathbf{W} = {(W_1, \ldots, W_b)}' \sim \mathcal{M}\mathcal{N}(\mathbf{0}, C)\) where \(C\) is non-negative and not strictly positive. Let \(\mathbf{G} = {(G_1, \ldots, G_b)}'\) be a standard normal distribution on \(\mathbb{R}^b\) and let \(\epsilon > 0\). Then \[ \mathbf{W} + \sqrt{\epsilon}\mathbf{G} \sim \mathcal{M}\mathcal{N}(\mathbf{0}, C + \epsilon I), \] and since \(C + \epsilon I\) is strictly positive, we know by the previous result that \[ \mathbb{E}\mathbf{e}^{-\alpha \sum_{k=1}^b {(W_k + \sqrt{\epsilon}G_k)}^2} = {\Bigl(\det\bigl(I+2\alpha (C+\sqrt{\epsilon}I)\bigr)\Bigr)}^{-\frac12}, \] and we get the announced result by letting \(\epsilon \to 0\).

Thus we have to derive \(\det(I+2\alpha C)\) now. Observe that \[ I-C = \mathbf{u} \mathbf{u}' \] where \(\mathbf{u} = {\bigl(\sqrt{p_1}, \ldots, \sqrt{p_b}\bigr)}'\). Since \(\mathbf{u}' \mathbf{u} = 1\), \[ I-C = \mathbf{u} {\bigl(\mathbf{u}'\mathbf{u})}^{-1} \mathbf{u}' \] is the matrix of the orthogonal projection on the line directed by \(\mathbf{u}\), therefore \(C\) is the matrix of the orthogonal projection on \({[\mathbf{u}]}^\perp\).

Thus \(C\) has one eigenvalue \(0\), with associated eigenvector \(\mathbf{u}\), and \(b-1\) eigenvalues equal to \(1\). Consequently, for every real number \(\alpha\), the matrix \(I + 2\alpha C\) has one eigenvalue \(1\) and \(b-1\) others \(1+2\alpha\). Therefore \[ \det(I+2\alpha C) = {(1+2\alpha)}^{b-1}. \] We finally get \[ \mathbb{E}\mathbf{e}^{-\alpha \sum_{k=1}^b W_k^2} \approx {(1+2\alpha)}^{-\frac{b-1}{2}}, \] and we recognize the characteristic function of the \(\chi^2_{b-1}\) distribution.