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?