Ljung-Box Test in finite sample Proof

[mertcan]

Hi everyone, initially I have seen that in order to analyze residuals for finite sample, Ljung - Box is defined as $$n*(n+2)*\sum_{n=0}^h p_k^2/(n-k)$$ where n is the sample size, $$p_k$$ is the sample autocorrelation at lag k, and h is the number of lags being tested. Actually I know the proof of formula when sample size goes infinity but in finite sample case there is a little adjustment. Also no sufficient information exist for that adjustment. Could you provide me with the proof?

Regards;

 

[Valkarie]

The proof of the Ljung-Box statistic for finite sample sizes is based on the fact that the sample autocorrelations are estimated from a sample of finite size. Specifically, the formula for the finite sample Ljung-Box statistic is given by:$$n*(n+2)*\sum_{k=0}^h \frac{p_k^2}{n-k}$$where $n$ is the sample size, $p_k$ is the sample autocorrelation at lag k and h is the number of lags being tested.The derivation of this formula starts with the definition of the sample autocorrelation coefficient at lag k, which is given by:$$p_k = \frac{\sum_{i=1}^{n-k} (x_i - \bar x)(x_{i+k}-\bar x)}{\sum_{i=1}^n (x_i - \bar x)^2}$$where $x_i$ is the value of the series at time i and $\bar x$ is the mean of the series.The numerator of the above equation can be rewritten as follows:$$\sum_{i=1}^{n-k} (x_i - \bar x)(x_{i+k}-\bar x) = \sum_{i=1}^n (x_i - \bar x)(x_{i+k}-\bar x) - \sum_{i=n-k+1}^n (x_i - \bar x)(x_{i+k}-\bar x)$$The second term on the right-hand side is zero because $x_{i+k}$ is undefined for $i > n-k$. Thus, we have:$$\sum_{i=1}^{n-k} (x_i - \bar x)(x_{i+k}-\bar x) = \sum_{i=1}^n (x_i - \bar x)(x_{i+k}-\bar x)$$Substituting this into the formula for $p_k$ gives:$$p_k = \frac{\sum_{i=1}^