This question is a little more specific than the title. Here I use the same notation that is set forth in this other question about cumulative returns (the sum of return observations). That is, let $r_{t,t+n} = \sum_{i=1}^n r_{t+i}$. Suppose that $r_t$ is an AR(1) process (say with Normally distributed errors) where $$ \text{Cov}(r_t,r_{t+j}) = \rho^j \sigma^2 $$ and thus $$ \text{Corr}(r_t, r_{t+j}) = \rho^j. $$ In the same notes references in the other question I see a claim that when $\rho = -1$ that the cumulative return is riskless, i.e., $$ \text{Var}(r_{t,t+n}) = 0. $$ How can this be? For example, if I let define the AR(1) process as follows $$ r_{t+1} = - r_t + \epsilon_{t+1}, $$ then consider the following example: \begin{align} \text{Var}(r_{t,t+2}) &= \text{Var}(r_{t+1} + r_{t+2}) \\ &= \text{Var}(\epsilon_{t+2}), \end{align} or \begin{align} \text{Var}(r_{t,t+4}) &= \text{Var}(\epsilon_{t+2} + \epsilon_{t+4}). \end{align} So, this doesn't appear to work.

If I define $\sigma^2$ as the unconditional, long-run variance of an observation, $\text{Var}(r_{t+1}) = \sigma^2$ as I did in the other question (linked), then when $\rho = -1$ this variance doesn't is undetermined/infinite. Am I missing something? Can I reconcile the claim made in these notes?


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