Démonstration du théorème

Pour montrer le point

, définissons, pour tout $t\in{\mathbb{Z}}$ ,

si $t=1,\ldots,N$ , et $X'_t=\bar X_N$ sinon. Soient $p\geq1$ , $t_1,\ldots,t_p\in{\mathbb{Z}}$ et ${\alpha}_1,\ldots,{\alpha}_p\in{\mathbb{C}}$ quelconques ; il s'agit de montrer que $\sum_{i,j=1}^p{\alpha}_i\bar{\alpha}_j\hat{\gamma}_X^{(N)}(t_i-t_j)$ est positif :

$\displaystyle \sum_{i,j=1}^p{\alpha}_i\bar{\alpha}_j\hat{\gamma}_X^{(N)}(\vert t_i-t_j\vert)$	$\displaystyle =$	$\displaystyle \sum_{i,j=1}^p\frac{{\alpha}_i\bar{\alpha}_j}{N}\left(\sum_{t=1}^... ...i-t_j\vert}(X_t-\overline{X}_N)(X_{t+\vert t_i-t_j\vert}-\overline{X}_N)\right)$
	$\displaystyle =$	$\displaystyle \sum_{i,j=1}^p\frac{{\alpha}_i\bar{\alpha}_j}{N}\left(\sum_{t\in{\mathbb{Z}}}(X'_t-\overline{X}_N)(X'_{t+\vert t_i-t_j\vert}-\overline{X}_N)\right)$
	$\displaystyle =$	$\displaystyle \sum_{i,j=1}^p\frac{{\alpha}_i\bar{\alpha}_j}{N}\left(\sum_{t\in{\mathbb{Z}}}(X'_{t+t_j}-\overline{X}_N)(X'_{t+t_i}-\overline{X}_N)\right)$
	$\displaystyle =$	$\displaystyle \frac{1}{N}\sum_{t\in{\mathbb{Z}}}\left(\sum_{i,j=1}^p{\alpha}_i\bar{\alpha}_j(X'_{t+t_j}-\overline{X}_N)(X'_{t+t_i}-\overline{X}_N)\right)$
	$\displaystyle =$	$\displaystyle \frac{1}{N}\sum_{t\in{\mathbb{Z}}}\left\vert\sum_{i=1}^p{\alpha}_i(X'_{t+t_i}-\overline{X}_N)\right\vert^2\geq0$

$\displaystyle N{\mathbb{E}}\left[\hat{\gamma}_X^{(N)}(h)-{\gamma}_X(h)\right]$	$\displaystyle =$	$\displaystyle \sum_{t=1}^{N-h}\left({\mathbb{E}}[X_tX_{t+h}]-{\mathbb{E}}[\bar ... ...t]-{\mathbb{E}}[\bar X_NX_{t+h}]+{\mathbb{E}}[\bar X_N^2]\right)-N{\gamma}_X(h)$
	$\displaystyle =$	$\displaystyle (N-h){\gamma}_X(h)-N{\mathbb{E}}[\bar X_N^2]$
		$\displaystyle +\sum_{t=N-h+1}^N{\mathbb{E}}[\bar X_NX_t]-N{\mathbb{E}}[\bar X_N... ...m_{t=1}^h{\mathbb{E}}[\bar X_NX_t]+(N-h){\mathbb{E}}[\bar X_N^2]-N{\gamma}_X(h)$
	$\displaystyle =$	$\displaystyle -h{\gamma}_X(h)-N{\mathbb{E}}[\bar X_N^2]+\left(\sum_{t=N-h+1}^N{... ... X_NX_t]+\sum_{t=1}^h{\mathbb{E}}[\bar X_NX_t]-h{\mathbb{E}}[\bar X_N^2]\right)$