For a dsitribution $p(X)$, let $x_1,\ldots,x_n$ be an independent sample of $p(X)$. Consider the one to one transformation $h(\cdot)$ such that $Y = h(X)$. If we apply the transformation to each of the samples such that $y_1 = h(x_1), \ldots ,y_n = h(x_n)$. Will $y_1,\ldots , y_n$ be a sample of $p(Y)$, or do you have to be more clever and go via inverse transform method to generate a sample from Y?
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1$\begingroup$ The definition of a sample involves the distribution and independence. There's no question about the distribution--since the $x_i$ are identically distributed, obviously the $h(x_i)$ are identically distributed. The independence of the $h(x_i)$ is demonstrated in the duplicate. $\endgroup$– whuber ♦Commented Apr 8, 2020 at 15:39
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Inverse transform method is just a particular case of the method you proposed. So for inverse transform you have $X$ distributed uniformly on $[0,1]$. The only reason why inverse transform is used is because of the availability of random number generators for standard uniform distribution. If you have already an independent sample from a given distribution, than you can simply use it.
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1$\begingroup$ I'm not sure how this answers the actual question, which is essentially "Will $y_1, \dots, y_n$ be a sample (from) p(Y)$?" $\endgroup$– jbowmanCommented Apr 8, 2020 at 15:20