I want to prove the following statement:

Under $H_{0}$ the test statistic $Q=n(n+2)$ $\sum \limits_{k=1}^h \frac{\hat{p}_{k}^2}{n-k}$ follows a $\chi ^2(h)$ chi-squared distribution with $h$ degrees of freedom. $H_{0}$ is the hypothesis that all data points are independently distributed.

$n$: sample size

$\hat{p}_{k}$: sample autocorrelation at lag k

$h$: number of lags being tested

I know that if $H_{0}$ is valid it follows that $Q=0$ because $\hat{p}_{k}=0$ for all k.

But how do I conclude that $Q$ is chi-squared distributed?

  • $\begingroup$ You seem to have created two separate accounts which is limiting your ability to edit your own question: click here to see how to merge them. $\endgroup$
    – Silverfish
    Commented Dec 27, 2015 at 14:02
  • $\begingroup$ Incidentally, the statement that $Q=0$ under $H_0$ is false, because even if the true autocorrelations are zero, the sample autcorrelations are not, and small nonzero values get magnified by $n$ - so it would be better to write that $Q$ takes small values if $H_0$ is true (and small is precisely measured by its $\chi^2$ null distribution). $\endgroup$ Commented Dec 28, 2015 at 14:05

2 Answers 2


First, note that $$Q=n\sum_{k=1}^h \frac{(n+2)\hat{p}_{k}^2}{n-k}$$ and that $\frac{n+2}{n-k}\to1$ as $n\to\infty$, so that $Q$ will behave like $$\tilde Q=n\sum_{k=1}^h\hat{p}_{k}^2$$ asymptotically.

To show that $\tilde Q$ is $\chi^2(h)$ under $H_0$, consider the following intermediate result adapted from Brockwell and Davis (1991), Theorem 7.2.1: Let $\hat{p}=(\hat p_1,\ldots,\hat p_h)^\top$. For a white noise process (i.e., one for which the null is true) $$Y_t=\mu+\epsilon_t$$ with $E|\epsilon_t|^4<\infty$ it holds that $$ \sqrt{n}\hat{p}\to_d N(0,I_h) $$ Thus, the first $h$ sample autocorrelations are multivariate normal with expected value 0 each (the true autocorrelation of any order for a white noise process) and asymptotic covariance matrix equal to the identity matrix. Hence, they are asymptotically independent. This also implies that each autocorrelation is asymptotically standard normal. Next, observe that $$ \tilde Q=\sqrt{n}\hat{p}^\top\sqrt{n}\hat{p}$$ Now, we know that the sum of $h$ independent squared standard normal random variables is $\chi^2(h)$.

  • $\begingroup$ Are you refering to this book [books.google.de/…? I am not sure how you conclude that $\sqrt{n} \hat{p} \rightarrow N(0, I_{h})$. I understand the rest. $\endgroup$
    – sunny
    Commented Dec 27, 2015 at 20:49
  • $\begingroup$ I posted a second answer on this, as I felt it would clutter the first one on Ljung Box too much. $\endgroup$ Commented Dec 28, 2015 at 10:15

This second answer provides a more detailed derivation of why the asymptotic distribution is multivariate normal with identity covariance matrix for $Y_t$ white noise. (Notation differs slightly from the question and previous answer, in that autocorrelations are denoted by $\rho$ and that we look at $p$ of them, following Brockwell and Davis, Time Series - Theory and Methods more closely.)

Brockwell and Davis, Thm. 7.2.1, states that the distribution of the correlation coefficients $\hat{\rho}=(\hat{\rho}_1,\ldots,\hat{\rho}_p)^\top$ if $Y_t$ is a general linear process $$ Y_t=\mu+\sum_{j=-\infty}^{\infty}\psi_j\epsilon_{t-j} $$ with $E|\epsilon_t|^4<\infty$ and $\sum_{j=-\infty}^{\infty}|\psi_j|<\infty$ is given by $$ \sqrt{n}(\hat{\rho}-\rho)=N(0,W) $$ where the elements of $W$ are given by $$ w_{ij}=\sum_{k=-\infty}^{\infty}\rho_{k+i}\rho_{k+j}+\rho_{k-i}\rho_{k+j}+2\rho_{i}\rho_{j}\rho_{k}^2-2\rho_{i}\rho_{k}\rho_{k+j}-2\rho_{j}\rho_{k}\rho_{k+i} $$ This can also be written as $$ w_{ij}=\sum_{k=1}^{\infty}\{\rho_{k+i}+\rho_{k-i}-2\rho_{i}\rho_{k}\}\{\rho_{k+j}+\rho_{k-j}-2\rho_{j}\rho_{k}\}\qquad(*) $$ To see this, write $w_{ij}=\sum_{k=-\infty}^{\infty}w_{ij,k}$ and decompose $w_{ij}$ as $$ w_{ij}=\sum_{k=-\infty}^{-1}w_{ij,k}+w_{ij,0}+\sum_{k=1}^{\infty}w_{ij,k} $$ Using $\rho_i=\rho_{-i}$ we directly verify that $w_{ij,0}=0$. Next, write \begin{eqnarray*} \sum_{k=-\infty}^{-1}w_{ij,k}&=&\sum_{k=1}^{\infty}w_{ij,-k}\\ &=&\sum_{k=1}^{\infty}\rho_{-k+i}\rho_{-k+j}+\rho_{-k-i}\rho_{-k+j}+2\rho_{i}\rho_{j}\rho_{-k}^2 -2\rho_{i}\rho_{-k}\rho_{-k+j}-2\rho_{j}\rho_{-k}\rho_{-k+i} \end{eqnarray*} Stationarity again implies that, e.g., $\rho_{-k+j}=\rho_{-(-k+j)}=\rho_{k-j}$. Hence, \begin{eqnarray*} \sum_{k=1}^{\infty}w_{ij,-k}&=&\sum_{k=1}^{\infty}\rho_{k-i}\rho_{k-j}+\rho_{k+i}\rho_{k-j}+2\rho_{i}\rho_{j}\rho_{k}^2 -2\rho_{i}\rho_{k}\rho_{k-j}-2\rho_{j}\rho_{k}\rho_{k-i} \end{eqnarray*} Hence, \begin{eqnarray*} \sum_{k=-\infty}^{\infty}w_{ij,k}&=&\sum_{k=1}^{\infty}\rho_{k+i}\rho_{k+j}+\rho_{k-i}\rho_{k+j}+2\rho_{i}\rho_{j}\rho_{k}^2 -2\rho_{i}\rho_{k}\rho_{k+j}-2\rho_{j}\rho_{k}\rho_{k+i}\notag\\ &&\;+\sum_{k=1}^{\infty}\rho_{k-i}\rho_{k-j}+\rho_{k+i}\rho_{k-j}+2\rho_{i}\rho_{j}\rho_{k}^2 -2\rho_{i}\rho_{k}\rho_{k-j}-2\rho_{j}\rho_{k}\rho_{k-i}\notag\\ &=&\sum_{k=1}^{\infty}\rho_{k+i}\rho_{k+j}+\rho_{k-i}\rho_{k+j}+4\rho_{i}\rho_{j}\rho_{k}^2 -2\rho_{i}\rho_{k}\rho_{k+j}-2\rho_{j}\rho_{k}\rho_{k+i}\notag\\ &&\qquad+\rho_{k-i}\rho_{k-j}+\rho_{k+i}\rho_{k-j}-2\rho_{i}\rho_{k}\rho_{k-j}-2\rho_{j}\rho_{k}\rho_{k-i}\qquad(**) \end{eqnarray*} Multiplying out (*) gives \begin{eqnarray*} w_{ij}&=&\sum_{k=1}^{\infty}\rho_{k+i}\rho_{k+j}+\rho_{k+i}\rho_{k-j}-2\rho_{k+i}\rho_{j}\rho_{k}+\rho_{k-i}\rho_{k+j}\\ &&\;+\rho_{k-i}\rho_{k-j}-2\rho_{k-i}\rho_{j}\rho_{k}-2\rho_{i}\rho_{k}\rho_{k+j}-2\rho_{i}\rho_{k}\rho_{k-j}+4\rho_{i}\rho_{k}^2\rho_{j} \end{eqnarray*} which is the same as (**).

To show the desired result for $Y_t$ white noise, note that the diagonal elements are \begin{eqnarray*} w_{ii}&=&\sum_{k=-\infty}^{\infty}\rho_{k+i}\rho_{k+i}+\rho_{k-i}\rho_{k+i}+2\rho_{i}\rho_{i}\rho_{k}^2 -2\rho_{i}\rho_{k}\rho_{k+i}-2\rho_{i}\rho_{k}\rho_{k+i}\\ &=&\sum_{k=-\infty}^{\infty}\rho_{k+i}^2+\rho_{k-i}\rho_{k+i}+2\rho_{i}^2\rho_{k}^2 -4\rho_{i}\rho_{k}\rho_{k+i} \end{eqnarray*} If $Y_t$ is white noise then $\rho_i=0$ for all $i\neq0$. Hence, $$\sum_{k=-\infty}^{\infty}\rho_{k+i}^2+\rho_{k-}\rho_{k+i}+2\rho_{i}^2\rho_{k}^2 -4\rho_{i}\rho_{k}\rho_{k+i}=1+0+0-0=1,$$ where the unit entry obtains when $k=-i$. The second term is always zero because the two terms cannot simultaneously have index zero. For $i\neq j$, note from $i,j,k>0$ that $$ w_{ij}=\sum_{k=1}^{\infty}0+0-2\cdot0+0+\rho_{k-i}\rho_{k-j}-2\cdot0-2\cdot0-2\cdot0+4\cdot0 $$ The only remaining entry must also be zero as $i\neq j$ such that not both $\rho_m$ can have index $m=0$.

  • $\begingroup$ Ok that makes sense. So now we know that $\sqrt{n}(\hat{p}-h)\rightarrow N(0, I_{h})$ but $\sqrt{n}(\hat{p}-h)^T\sqrt{n}(\hat{p}-h) \neq \tilde{Q}$ or can we ignore $h$ as it is a constant? $\endgroup$
    – sunny
    Commented Dec 28, 2015 at 11:24
  • $\begingroup$ No, we know that $\sqrt{n}(\hat p -p)\to_dN(0,I_h)$! So, the scaled difference between the first $h$ estimated and true autocorrelations. Now, the true autocorrelations in the white noise case ("under the null") are all zero, so we need not subtract the resulting zero vector when forming $\tilde Q$. $\endgroup$ Commented Dec 28, 2015 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.