When Multi-Period Return Variance Adds Across Time

# Is variance additive only under Log-returns?


# Is variance additive only under Log-returns?












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?

## Answer by Richi Wa (score 1, accepted)

https://quant.stackexchange.com/a/19478

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.

## Answer by user32416 (score 1)

https://quant.stackexchange.com/a/19477

You should clarify a bit your question.

- 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.

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

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

- 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 $Var ( log (1 + Z) ) = \sigma^2$.

## Answer by Randor (score 0)

https://quant.stackexchange.com/a/19503

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