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Can't seem to figure this one out by thinking it through. Let's say that the simple return $R_t=P_{t+1}/P_t -1$ is assumed to be $R_t \sim iid N(0,\sigma^2)$. Thus, a two period return would be $(1+R_t)(1+R_{t+1})-1$. Would the variance of the two period return be equal to $2\sigma^2 + \sigma^4$?

$$Var((1+R_t)(1+R_{t+1})-1)=Var(1+R_{t+1}+R_t+R_tR_{t+1})$$ $$ = 2\sigma^2 +Var(R_tR_{t+1}) = 2\sigma^2 + \sigma^4$$ since variance of two independent random variable products are just the product of both random variable variance (with $\mu=0$).

Under log returns, returns become additive and two period would be $log(1+R_t)+log(1+R_{t+1})$ and variance is equal to

$$Var(log(1+R_t)+log(1+R_{t+1})) = Var(log(1+R_t))+Var(log(1+R_{t+1}))=\sigma^2 + \sigma^2$$

Am i missing anything here?

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  • $\begingroup$ The first 2 line equation is wrong. Look at my answer to see why. $\endgroup$ – Ric Aug 28 '15 at 9:51
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If you use log-returns, then it is true that the return over n periods is the sum of the returns over each subperiod (e.g. the 10 day return is the sum of 10 1-day returns) $$ R = \sum_{i=1}^n r_i. $$ If we now look at the variance of $R$ then we get $$ VAR(R) = VAR( \sum_{i=1}^n r_i ), $$ if we assume that the returns are uncorrelated then we get $$ VAR(R) = \sum_{i=1}^n VAR(r_i ). $$ if we finally assume that $VAR(r_i) = \sigma^2$, i.e. it is the same for each day, then we get $$ VAR(R) = n \sigma^2. $$

EDIT: in you equastion above you have the expression $VAR(R_tR_{t+1})$. This is not (!) $\sigma^4$ as $R_t$ and $R_{t+1}$ are usually 2 different random variables. They might have the same variance $\sigma^2$ but this does not mean that $VAR(R_tR_{t+1}) = E[R_t^4]$.

This expression is rather: $$ VAR(R_tR_{t+1}) = E[(R_tR_{t+1})^2]-E[R_tR_{t+1}]^2, $$ and $E[R_tR_{t+1}]$ is connected to the lag one auto-correlation.

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  • $\begingroup$ My question was that, is this same concept applicable under simple/normal returns? Where $R$ is the product of all the subperiod returns. $\endgroup$ – Kevin Pei Aug 26 '15 at 19:38
  • $\begingroup$ The answer is simple: no. You need all of the three such that it is true: the time-additiviy of log-returns, uncorrelatedness and equal variances. $\endgroup$ – Ric Aug 27 '15 at 6:48
  • $\begingroup$ I took my result there from a math question: stats.stackexchange.com/questions/52646/… which assumes independence. Is that wrong? Furthermore, if your simple answer (no) is true, i see applications like quantopian zipline calculate it as (STD[simple daily returns]*number of trading days) $\endgroup$ – Kevin Pei Sep 2 '15 at 17:24
  • $\begingroup$ strictly speaking with simple returns it is wrong. Of course the difference for small returns is not big ... but you switch back and forth from simple returns to log-returns and use the one that fits .. perform some calculation and then back. $\endgroup$ – Ric Sep 3 '15 at 9:06
  • $\begingroup$ The formula for the variance of two independent random variables is true. But $r_t$ and $r_{t+1}$ are usually not independent. This can be seen in the case of volatility clustering where $r_t^2$ and $r_{t+1}^2$ are highly correlated which contradicts the former to be independent. $\endgroup$ – Ric Sep 3 '15 at 9:10
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You should clarify a bit your question.

  1. Your first computation with $$Var((1+R_t)(1+R_{t+1})-1)$$ is ambiguous. What do you mean by $Var$ here? You have time subscripts $t$ and $t + 1$ so unless you specify which filtration you compute your variance, it is unclear what you're computing.

  2. If you are computing at $t = 0$, then this is not a two period return (especially when $t > 1$).

  3. If you are computing at $\tau = t$ then the $t$-measurable random variables drop out of the variance computation.

  4. As to your log return computation, you are just abusing notations. Note that if $Z \sim N(\mu, \sigma)$, then it is definitely not true that $Var ( log (1 + Z) ) = \sigma^2$.

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  • $\begingroup$ I apologize for the lack of clarification. I'm not too sure of the formal way to impose the problem. To simplify, if we divide a one-period return into smaller subperiod returns, is the variance of the one period return equal to the sum of the variance of the subperiod returns under both simple and log returns? $\endgroup$ – Kevin Pei Aug 26 '15 at 19:40
  • $\begingroup$ @KevinPei Forget about returns and think back to basic statistics. If you have two random variables $X_1, X_2$ that are iid (not necessarily even normal). Is $Var(X_1 X_2)$ going to be equivalent to $Var( \log X_1 + \log X_2)$? The answer is going to be generically no. I think you need to first commit to the distribution form before you can answer your own question; are you going to model returns or log returns? And as a hint --- in finance we prefer modelling things in log-returns precisely of the additive nature that you'd already suggested. $\endgroup$ – user32416 Aug 26 '15 at 19:54
  • $\begingroup$ If we were to for now discard log-returns (where ln(1+r) is normally distributed), what would be the variance of a simple return divided up into two periods? I now understand that applying a log changes the entire distribution if the original variable was normally distributed. $\endgroup$ – Kevin Pei Aug 27 '15 at 0:54
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    $\begingroup$ I really think you need to clarify your question. 1. So if you say $log (1 + R)$ is $N(\mu, \sigma)$, then it implies that $R = e^Z - 1$, where $Z \sim N(\mu, \sigma)$. This already imposes a distribution assumption on the simple returns. 2. You need to be clearer on what you mean by "variance of simple return divided up into two periods". You are effectively thinking about the variance of your terminal wealth. So if $R_t$ is the period $t$ simple return, and your initial wealth is 1 then your $t = 2$ terminal wealth is $W_2 = 1 \times (1 + R_1) (1 + R_2)$. $\endgroup$ – user32416 Aug 27 '15 at 1:48
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    $\begingroup$ Then the holding period return for holding from $t = 0$ to $t = 2$ is $R_{0,2} = (W_2 - 1) / 1$. Hence, computing the variance of this holding period return $Var(R_{0,2})$ is clearly equivalent to just computing the variance of $W_2$ --- of which you'd already computed yourself. 3. Even without normality, it should be clear that if you have a random variable $X$, it is clear that $g(X)$ has a different distribution than $X$ (i.e. consider the Jacobian on the pdf). $\endgroup$ – user32416 Aug 27 '15 at 1:53
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Variance is additive also for any other distribution.

The reason that variance is additive is because of the assumption of independent increments - ie, the change in underlying value between 2 moments is assumed to be independent of the change in value between 2 later moments

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