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.)

  • $\begingroup$ Is $$r_{t,t+n} = \sum_{i=1}^n r_{t+i}$$ or should it be $$r_{t,t+n} = \sum_{i=0}^n r_{t+i}$$? $\endgroup$ Jan 11, 2015 at 7:29
  • $\begingroup$ Also, please write explicitly the AR(1) process. $\endgroup$ Jan 11, 2015 at 7:35
  • $\begingroup$ Also the sum in $(1)$ runs from $i=1$ ? (you write $i+1$ below the sum, which is confusing). $\endgroup$ Jan 11, 2015 at 7:37
  • $\begingroup$ a) It is correct as written. I agree that it might be a little weird, but this is the convention used. b) The process defined enough for the purposes of the question. This is as explicit as it is defined in these notes. c) You're right. That's a typo. Fixed it. $\endgroup$
    – jmbejara
    Jan 12, 2015 at 0:03

2 Answers 2


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

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


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

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

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

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

  • $\begingroup$ It's worth noting that it appears like the notes are conditioning on $r_{t-1} = 0$ so that $Var(r_t) = \sigma^2$. That is how we would arrive at $\text{Cov}(r_t,r_{t+j}) = \rho^j \sigma^2$. Thanks for the answer. $\endgroup$
    – jmbejara
    Jan 12, 2015 at 5:11
  • $\begingroup$ That's confusing - so all the formulas in the question are formulas conditional on $r_{t-1}$? Because the title of the question explicitly mentions "unconditional". $\endgroup$ Jan 12, 2015 at 6:03
  • $\begingroup$ You're right. I think it's just worth pointing out that that is perhaps another error. Otherwise it seem like $\text{Cov}(r_t, r_{t+j}) = \frac{\rho^j}{1 - \rho^2} \sigma^2$. $\endgroup$
    – jmbejara
    Jan 12, 2015 at 6:12
  • $\begingroup$ Yes, that's the unconditional autocovariance function of an AR(1). $\endgroup$ Jan 12, 2015 at 6:37

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.

  • 1
    $\begingroup$ (+1) for working this out, and posting it. $\endgroup$ Jan 12, 2015 at 20:01

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