[proofplan] We compare the Ljung-Box statistic $Q_m^*$ with the Box-Pierce statistic $Q_m$. Since $m$ is fixed, the multiplicative correction factor $\frac{n+2}{n-k}$ tends uniformly to $1$ over $1\leq k\leq m$. The assumed Box-Pierce limit implies that $Q_m$ is bounded in probability, and this boundedness makes the total correction error vanish in probability. Therefore $Q_m^*$ and $Q_m$ have the same limiting null distribution. [/proofplan] [step:Rewrite the Ljung-Box statistic as a weighted Box-Pierce sum] For each integer $k$ with $1\leq k\leq m$, define the deterministic weight \begin{align*} a_{n,k} := \frac{n+2}{n-k}. \end{align*} Then \begin{align*} Q_m^* &= n(n+2)\sum_{k=1}^{m}\frac{\hat r_k^2}{n-k} \\ &= n\sum_{k=1}^{m}a_{n,k}\hat r_k^2. \end{align*} The Box-Pierce statistic is \begin{align*} Q_m = n\sum_{k=1}^{m}\hat r_k^2. \end{align*} Therefore \begin{align*} Q_m^* - Q_m = n\sum_{k=1}^{m}(a_{n,k}-1)\hat r_k^2. \end{align*} [guided] The point of this rewrite is to isolate the only difference between the two statistics. The Box-Pierce statistic weights every squared residual autocorrelation by $n$, while the Ljung-Box statistic uses the adjusted weight \begin{align*} \frac{n(n+2)}{n-k}. \end{align*} For each fixed lag $k$, define \begin{align*} a_{n,k} := \frac{n+2}{n-k}. \end{align*} Then the Ljung-Box statistic becomes \begin{align*} Q_m^* &= n(n+2)\sum_{k=1}^{m}\frac{\hat r_k^2}{n-k} \\ &= n\sum_{k=1}^{m}\frac{n+2}{n-k}\hat r_k^2 \\ &= n\sum_{k=1}^{m}a_{n,k}\hat r_k^2. \end{align*} Since \begin{align*} Q_m = n\sum_{k=1}^{m}\hat r_k^2, \end{align*} subtracting the two expressions gives \begin{align*} Q_m^* - Q_m = n\sum_{k=1}^{m}(a_{n,k}-1)\hat r_k^2. \end{align*} Thus the proof reduces to showing that the weighted error term on the right converges to $0$ in probability. [/guided] [/step] [step:Show that the Ljung-Box correction weights converge uniformly to one] For $1\leq k\leq m$ and $n>m$, \begin{align*} a_{n,k}-1 &= \frac{n+2}{n-k}-1 \\ &= \frac{k+2}{n-k}. \end{align*} Since $1\leq k\leq m$, we have $k+2\leq m+2$ and $n-k\geq n-m$. Hence \begin{align*} 0\leq a_{n,k}-1\leq \frac{m+2}{n-m}. \end{align*} Define \begin{align*} \delta_n := \max_{1\leq k\leq m}|a_{n,k}-1|. \end{align*} Then \begin{align*} 0\leq \delta_n \leq \frac{m+2}{n-m}, \end{align*} so $\delta_n\to 0$ as $n\to\infty$. [/step] [step:Bound the difference between the two statistics by the Box-Pierce statistic] Using the definition of $\delta_n$ and the non-negativity of $\hat r_k^2$, we obtain \begin{align*} |Q_m^* - Q_m| &= \left|n\sum_{k=1}^{m}(a_{n,k}-1)\hat r_k^2\right| \\ &\leq n\sum_{k=1}^{m}|a_{n,k}-1|\hat r_k^2 \\ &\leq \delta_n n\sum_{k=1}^{m}\hat r_k^2 \\ &= \delta_n Q_m. \end{align*} By hypothesis, \begin{align*} Q_m \xrightarrow{d} \chi^2_{m-p-q}. \end{align*} We use the following standard terminology. A sequence $(Y_n)_{n\geq 1}$ of real-valued random variables is bounded in probability if, for every $\varepsilon>0$, there exists $M>0$ such that $\mathbb{P}(|Y_n|>M)<\varepsilon$ for all sufficiently large $n$. Convergence $Y_n\xrightarrow{\mathbb P}0$ means that, for every $\varepsilon>0$, $\mathbb{P}(|Y_n|>\varepsilon)\to 0$ as $n\to\infty$. Since $\chi^2_{m-p-q}$ is a finite real-valued random variable, the standard tightness consequence of convergence in distribution implies that the sequence represented by $Q_m$ is bounded in probability. Fix $\varepsilon>0$ and $\eta>0$. Choose $M>0$ such that $\mathbb{P}(Q_m>M)<\eta$ for all sufficiently large $n$. Since $\delta_n\to 0$, for all sufficiently large $n$ we also have $\delta_n M<\varepsilon$. Therefore \begin{align*} \mathbb{P}(|Q_m^* - Q_m|>\varepsilon) &\leq \mathbb{P}(\delta_n Q_m>\varepsilon) \\ &\leq \mathbb{P}(Q_m>M) \\ &< \eta. \end{align*} Because $\eta>0$ was arbitrary, this proves \begin{align*} Q_m^* - Q_m \xrightarrow{\mathbb P} 0. \end{align*} [guided] The previous step showed that all correction weights $a_{n,k}$ are uniformly close to $1$. We now turn that deterministic fact into a probabilistic comparison of $Q_m^*$ and $Q_m$. By the triangle inequality and the definition of $\delta_n$, we have \begin{align*} |Q_m^* - Q_m| &= \left|n\sum_{k=1}^{m}(a_{n,k}-1)\hat r_k^2\right| \\ &\leq n\sum_{k=1}^{m}|a_{n,k}-1|\hat r_k^2 \\ &\leq \delta_n n\sum_{k=1}^{m}\hat r_k^2 \\ &= \delta_n Q_m. \end{align*} This estimate is useful because it separates the error into two factors: $\delta_n$, which is deterministic and tends to $0$, and $Q_m$, whose null asymptotics are already assumed. The hypothesis says \begin{align*} Q_m \xrightarrow{d} \chi^2_{m-p-q}. \end{align*} We now spell out the probabilistic notation. A sequence $(Y_n)_{n\geq 1}$ of real-valued random variables is bounded in probability when, for every $\varepsilon>0$, there is a threshold $M>0$ such that $\mathbb{P}(|Y_n|>M)<\varepsilon$ for all sufficiently large $n$. Also, $Y_n\xrightarrow{\mathbb P}0$ means that $\mathbb{P}(|Y_n|>\varepsilon)\to 0$ for every $\varepsilon>0$. The chi-squared distribution is finite-valued on $[0,\infty)$, so convergence in distribution to $\chi^2_{m-p-q}$ implies tightness of the sequence represented by $Q_m$. In the preceding terminology, $Q_m$ is bounded in probability. To see how this combines with $\delta_n\to0$, fix $\varepsilon>0$ and $\eta>0$. Choose $M>0$ such that $\mathbb{P}(Q_m>M)<\eta$ for all sufficiently large $n$. Then choose $n$ large enough that $\delta_nM<\varepsilon$. For such $n$, \begin{align*} \mathbb{P}(|Q_m^* - Q_m|>\varepsilon) &\leq \mathbb{P}(\delta_nQ_m>\varepsilon) \\ &\leq \mathbb{P}(Q_m>M) \\ &< \eta. \end{align*} Since $\eta>0$ was arbitrary, this is exactly \begin{align*} Q_m^* - Q_m \xrightarrow{\mathbb P}0. \end{align*} [/guided] [/step] [step:Apply Slutsky's theorem to transfer the Box-Pierce limiting distribution] We have proved \begin{align*} Q_m^* - Q_m \xrightarrow{\mathbb P}0. \end{align*} The assumed Box-Pierce limit is \begin{align*} Q_m \xrightarrow{d} \chi^2_{m-p-q}. \end{align*} Slutsky's theorem applies to the decomposition $Q_m^*=Q_m+(Q_m^*-Q_m)$ because the first summand converges in distribution to $\chi^2_{m-p-q}$ and the second summand converges in probability to the constant $0$. Hence \begin{align*} Q_m^* \xrightarrow{d} \chi^2_{m-p-q}. \end{align*} Thus the Ljung-Box statistic has the same large-sample null distribution as the Box-Pierce statistic, namely the chi-squared distribution with $m-p-q$ degrees of freedom. [/step]