Generates variates of $F^s$ given a known algorithm for variates of $F$ Let rv $X$ have a distribution $F (t):={\rm Prob}(X <t)$ and rv $Y$ have a distribution $G (t):=(F(t))^s$ for some constant $s>0$. Suppose that I know an algorithm for generating random variates of $X$. Is there a general argument/theorem, which could help me to construct an algorithm for generating the variates of $Y$ using this information? 
 A: Here is an idea based on the rejection method that should work for real $s>1$ without requiring inversion.
The density of $Y$ is given by
$$
g(t)=sF(t)^{s-1}f(t)
$$
For $s>1$, we thus have
$$
g(t)\leq sf(t)
$$
as $F(t)\leq1$ and hence $F(t)^{s-1}\leq1$, too. 
Hence, we may set up a rejection sampler that treats a draw from $X$ accepted with probability 
$$
\frac{g(t)}{sf(t)}=F(t)^{s-1}
$$
as a draw from $Y$. The idea would not work for $s<1$ as $F(t)^{s-1}>1$ in that case. If $g$ has bounded support, you could use a standard rejection algorithm with a rectangular box around it.
We thus accept draws from $F$ if a uniform r.v.s is less than $g$ (red) divided by $s$ times their density (blue) at that draw. Here is a visualization for the normal case with $s=4$:

Here is an implementation in R for the known case of $\max(X_1,\ldots,X_s)$ for iid $X_i\sim N(0,1)$, that reveals that the algorithm works, but also that it is inefficient as it discards many draws, in particular negative ones. In particular, the unconditional acceptance probability of a draw is $1/s$, the relative area under the two curves.
s <- 4
draws <- 10000
plot(ecdf(replicate(draws,max(rnorm(s)))), lwd=2, col="darkgreen", xlim=c(-1,3)) # the empirical "target"
curve(pnorm(x)^s,-1,3, add=T, lwd=2, col="salmon") # the known theoretical cdf

rejection.method <- function(s,draws){
  X <- rnorm(draws)
  accept <- runif(draws) < pnorm(X)^(s-1)
  return(X[accept])
} 

Y.rejection.method <- rejection.method(s,draws)
curve(ecdf(Y.rejection.method)(x), -1, 3, add=T, lwd=2, col="lightblue") # empirical cdf of the sampler

length(Y.rejection.method)/draws # just around 25%


A: Given a CDF $P(x)$ and a RNG that generates uniform deviates $U \in (0, 1)$, the standard way of producing non-uniform deviates that follow the CDF is to take $P^{-1}(U)$. So you would generate $X = F^{-1}(U)$ and $Y = G^{-1}(U)$. $G(x) = [F(x)]^s$ implies $G^{-1}(u) = F^{-1}(u^{1/s})$, so
  $$Y = F^{-1}(U^{1/s}) = F^{-1}([F(X)]^{1/s})$$
