Generating random variables with inverse df

Question

Assume a distribution $X$, and we know the idf $F^{-1}$ of $X$. Let $U \sim U(0,1)$.

Why is drawing an element from X according to $F^{-1}(u) = Z$ considered to be more random than just drawing an element from $X$? Edit: It's not.

My guess would be that in an unevenly distributed set, there are areas that are more concentrated, and thus have a higher chance of the item being picked from there. Violating the uniform distribution property.

By sampling from a uniformly distributed set $U$, you somehow bypass that. I get that for every sample from $U : u$, you get a corresponding minimal $x$ from $X$ where $P(X < x) > u$. But what I don't get is how we bypass the chance of drawing more elements from more concentrated areas.

Answer 1

Inverse transform sampling is not considered better and does not provide anyhow more "random" samples than direct sampling. This method is used when we cannot sample directly from the distribution of interest while knowing inverse cumulative distribution function. For some distributions it is possible to simulate their values using other methods, e.g. you can obtain samples from standard triangular distribution by sampling $U_1, U_2 \sim \mathcal{U}(-1, 1)$ and taking $X = (U_1+U_2)/2$, but for many other distributions we do not have good methods for sampling, so inverse transform sampling is used. Notice that inverse sampling is not the best method available and in some cases there exist better methods (example here). Luc Devroye (1986) wrote a book on non-uniform random sampling that goes into greater details.

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Devroye, L. (1986). Non-Uniform Random Variate Generation. New York: Springer-Verlag.