iid time series through different time scales How can I show that if a time serie is iid at the scale 1 day, it is also the case at the scale 2 days?
More precisely, I am considering financial returns.
I have iid daily returns $$r_{t,1d}=\frac{p_t}{p_{t-1}}-1$$, how can I justify that the 2 days returns such as $$r_{t,2d}=\frac{p_t}{p_{t-2}}-1=(1+r_{t-1,1d})(1+r_{t,1d})-1$$ are iid?
 A: "IID" means independent and identically distributed. Not all two-day returns are independent; but they are all identically distributed.
They are identically distributed because the independence of the one-day returns $r_1(t)$ implies the bivariate random variables $(r_1(t-1), r_1(t))$ all have the same (2D) distributions: they have the same marginals (due to the "identical" i of iid) and their independence assures all the joint distributions are the same.  Consequently any other random variable constructed from these bivariate variables, such as $r_2(t)=(1+r_1(t-1))(1+r_1(t))-1,$ will be identically distributed.
See the left-hand plot for a scatterplot of $(r_1(t-1), r_1(t))$ in simulated data.  The red line is the least-squares fit; it is essentially zero, showing all lack of correlation.

$r_2(t)$ and $r_2(s)$ will also be independent when the times $s$ and $t$ are sufficiently separated, because when $|s-t|\gt 1$ they are formed from four separate independent variables.  See the right-hand plot in the figure which shows the case $|s-t|=2$ for simulated data.
However, any neighboring two-day returns $r_2(t)$ and $r_2(t+1)$ have something in common: namely, the price change $r_1(t).$  Indeed, when the common distribution of the $r_1(t)$ has a finite nonzero variance $\sigma^2,$ we may compute
$$\eqalign{
\operatorname{Cov}(r_2(t), r_2(t+1)) &= \operatorname{Cov}\left((1+r_1(t-1))(1+r_1(t))-1, (1+r_1(t))(1+r_1(t+1))-1\right)\\
&= \sigma^2 \ne 0.
}$$
This value comes from the covariance of the $r_1(t)$ terms appearing in both of the $r_2(t)$ and $r_2(t+1)$ variables.  The nonzero covariance shows that adjacent terms in the $r_2$ series have nonzero correlation, whence they cannot be independent.  The middle figure illustrates this case with the simulated data: the lack of independence is obvious.

Here is the R code to produce the simulated returns and the figures.
n <- 1e4    # Length of simulated price series
rho <- 0.16 # Annual mean return

set.seed(17)
p <- exp(cumsum(rnorm(n, 0, sqrt(rho^2/365)))) # Simulated price series
r <- p[-1] / p[-n] - 1                         # Daily returns

par(mfrow=c(1,3))
plot(r[-1], r[-(n-1)], asp=1, pch=16, col="#00000004", cex.main=0.9,
     main="One-day return dependence",
     xlab=expression(r[1](t-1)), ylab=expression(r[1](t)))
abline(lm(r[-(n-1)] ~ r[-1]), lwd=2, col="Red")

r2 <- p[-(1:2)] / p[-((n-1):n)] - 1
plot(r2[-1], r2[-(n-2)], asp=1, col="#00000004", cex.main=0.9,
     main="Two-day return dependence (lag 1)",
     xlab=expression(r[2](t-1)), ylab=expression(r[2](t)))
abline(lm(r2[-(n-2)] ~ r2[-1]), lwd=2, col="Red")

plot(r2[-(1:2)], r2[-((n-3):(n-2))], asp=1, col="#00000004", cex.main=0.9,
     main="Two-day return dependence (lag 2)",
     xlab=expression(r[2](t-2)), ylab=expression(r[2](t)))
abline(lm(r2[-((n-3):(n-2))] ~ r2[-(1:2)]), lwd=2, col="Red")
par(mfrow=c(1,1))

