2.3 Horizon scaling and the variance ratio

Variance of a sum of uncorrelated terms is the sum of variances.∎

$\operatorname{Var}(\sum_{i=0}^{k-1} r_{t-i}) = \sum_{i=0}^{k-1}\sum_{l=0}^{k-1} \operatorname{Cov}(r_{t-i}, r_{t-l}) = \sigma^2 \sum_{i,l} \rho_{|i-l|}$. In the $k \times k$ matrix $(\rho_{|i-l|})$, the main diagonal has $k$ entries equal to $\rho_0 = 1$, and each of the two diagonals at distance $j \ge 1$ has $k - j$ entries equal to $\rho_j$.∎

Stationarity and $\operatorname{Cov}(r_t, r_{t-j}) = \phi \operatorname{Cov}(r_{t-1}, r_{t-j})$ for $j \ge 1$ give $\rho_j = \phi^j$. Substitute in (2.6) and evaluate $S_k = \sum_{j=1}^{k-1}(k-j)\phi^j$. Write $S_k = k\sum_{j=1}^{k-1}\phi^j - \sum_{j=1}^{k-1} j\phi^j$. With $\sum_{j=1}^{k-1}\phi^j = \phi(1-\phi^{k-1})/(1-\phi)$ and $\sum_{j=1}^{k-1} j\phi^j = \phi\,[1 - k\phi^{k-1} + (k-1)\phi^k]/(1-\phi)^2$,

after expanding $k(1-\phi)(1-\phi^{k-1}) = k - k\phi - k\phi^{k-1} + k\phi^k$ and collecting terms. Check $k = 2$: $S_2 = \phi[2 - 2\phi - 1 + \phi^2]/(1-\phi)^2 = \phi$. Equation (2.9) is (2.8) divided by $q\sigma^2$; as $q \to \infty$ the bracket is dominated by $k + 2\phi k/(1-\phi) = k(1+\phi)/(1-\phi)$.∎

To first order, $\widehat{VR}(q) - 1 = 2\sum_{j=1}^{q-1}(1 - j/q)\hat\rho_j + o_p(T^{-1/2})$, which is (2.7) with sample autocorrelations. (a) Under iid returns, $\sqrt{T}(\hat\rho_1, \dots, \hat\rho_{q-1})$ is asymptotically $N(0, I)$ (Bartlett's formula with all $\rho_j = 0$). Hence the asymptotic variance is

(b) Under martingale differences, $\sqrt{T}\hat\rho_j$ is still asymptotically normal with mean 0, but its variance is $\delta_j = E[(r_t-\mu)^2(r_{t-j}-\mu)^2]/\sigma^4$, which exceeds 1 when volatility clusters (large $|r_{t-j}|$ predicts large $|r_t|$). Under Assumption H the $\hat\rho_j$ are asymptotically uncorrelated across $j$, so the variance becomes $\sum_j [2(1-j/q)]^2\delta_j = \theta(q)$, and $\hat\delta_j$ in (2.12) is its sample analogue.∎