Formula for the unconditional variance of the sum of observations from an autoregressive time series

Question

I have notes that say that we can make the following calculations. I'm a little confused about some of the calculations that are being made. What assumptions would I need to get the following results? Or are there errors? Specifically, I am confused by equation (1) below. In particular, this is strange to me because if I let $\rho = -1$, then equation (1) sometimes gives negative variance. (Perhaps the calculation is undefined when $\rho = -1$?)

Let $r_{t,t+n} = \sum_{i=1}^n r_{t+i}$. Suppose that $r_t$ is an AR(1) process (say with errors given by a mean zero Normal distribution with variance $\sigma^2$) where $$ \text{Cov}(r_t,r_{t+j}) = \rho^j \sigma^2 $$ and thus $$ \text{Corr}(r_t, r_{t+j}) = \rho^j. $$

The notes that I have say that $\text{Var}(r_{t,t+2}) = 2(1 + \rho) \sigma^2$ and that $$ \text{Var}(r_{t,t+n}) = \left( n + 2 \sum_{i=1}^{n-1} \rho^i (n-i) \right) \sigma^2. \tag{1} $$

(FWIW, this question deals with cumulative (log) returns.)

Answer 1

For an AR(1) process (I omit any drift), the coefficient on the lag is the 1st-order correlation coefficient,

$$r_{t+1} = \rho r_t + u_{t+1}$$

So

$$r_{t,t+2} = \sum_{i=1}^2 r_{t+i} = r_{t+1} + r_{t+2} = \rho r_t + u_{t+1} + \rho r_{t+1} + u_{t+2} $$

$$=\rho r_t + u_{t+1} + \rho \big(\rho r_t + u_{t+1}\big) + u_{t+2} = \rho(1+\rho)r_t+(1+\rho)u_{t+1} + u_{t+2}$$

and now the three components are independent. So

$${\rm Var}(r_{t,t+2}) = \rho^2(1+\rho)^2\frac {\sigma^2}{1-\rho^2}+ (1+\rho)^2\sigma^2 +\sigma^2$$

$$=\left( (1+\rho)^2\frac {\rho^2+1-\rho^2}{1-\rho^2}+1\right)\cdot \sigma^2$$

$$=\left( \frac {1+\rho}{1-\rho}+1\right)\cdot \sigma^2 = \frac {2}{1-\rho}\sigma^2$$

So we arrive at a different formula for ${\rm Var}(r_{t,t+2})$, compared to the one provided in these notes, and we will arrive also at a different expression for general $n$.

Note that my result is valid for $\rho =-1$, giving ${\rm Var}(r_{t,t+2}) = \sigma^2$ in this case since here

$$r_{t+1} = - r_t + u_{t+1}$$

and so

$$r_{t+1} + r_{t+2} = - r_t + u_{t+1}- r_{t+1} + u_{t+2} = - r_t + u_{t+1} - \big(- r_t + u_{t+1}\big) + u_{t+2} = u_{t+2}$$

Answer 2

Ok. I made some small computation errors and got confused here. The notes make sense with the following notational assumptions. If I write out the AR(1) process as follows (ignoring drift) $$ r_{t+1} = r_t + \epsilon_{t+1}, $$ then we have $\text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma_\epsilon^2$, where $\sigma^2_\epsilon := Var(\epsilon)$. The point that I should have caught on to in the notes was exactly the point that the variance is unconditional and so $\text{Var}(r_t) = \sigma^2 = \frac{1}{1 - \rho^2} \sigma_\epsilon^2$ is the unconditional variance (the time series is stationary only when $|\rho| < 1$, so beware when using $\rho = -1$ as mentioned above). Given this definition, everything works. First, note that $$ \text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma_\epsilon^2 = \rho^j \sigma^2, $$ as noted above. Then also notice that \begin{align} \text{Var}(r_{t,t+n}) &= \text{Var}\left (\sum_{i=1}^n r_{t+i} \right) \\ &= \sum_{j=1}^n \sum_{i=1}^n \text{Cov}(r_{t+1}, r_{t+j}) \\ &= \sum_{j=1}^n \sum_{i=1}^n \rho^{|i - j|} \sigma^2 \\ &= \left (n + 2 \sum_{i=1}^{n-1} \rho^j (n - i) \right) \sigma^2. \end{align} So, it all works out.