Quick Expectation Check

Hey all,

If it isn't too much trouble, can someone verify that this works? It seems nifty, but I feel suspicious that there's a small mistake somewhere. Suppose $X \sim N(\theta, \sigma^2 \mbox{I}_p)$, with $\theta = (\theta_1, ..., \theta_p)$. Let $j = (1, 1, ..., 1)^T$. Then

$E(X - \theta)'(\bar{X}j - \theta)
= E \sum_{i = 1} ^ p (X_i - \theta_i)(\bar{X} - \theta_i)
= E \sum_i (X_i - \theta_i)(\frac{\sum_j X_j}{p} - \theta_i)$
$
= \frac{1}{p} E \sum_i (X_i - \theta_i) (\sum_j X_j - \theta_i)
= \frac{1}{p} E \sum_i \sum_j (X_i - \theta_i) (X_j - \theta_i)$
$
= \frac{1}{p} E \sum_i \sum_j \left((X_i - \theta_i)(X_j - \theta_j) + (X_i - \theta_i)(\theta_j - \theta_i)\right)
= \frac{1}{p} \sum_i \sum_j Cov(X_i, X_j) + 0$
$
= \frac{1}{p} \sum_i V(X_i)
= \sigma^2$