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I have the following problem: let's say I have a function $y=f(x)$. Let $f$ be defined for all $x$ but it it might not be invertible. Further assume $x \sim p(x)$ with some probability density $p(x)$.

Can I claim that $p(y)$ exists and that $y \sim p(y)$ if I draw a sample from $p(x)$ and compute $y=f(x)$?

Let's say further I could draw samples from the conditional probability $p(x | z)$. If I use samples from this distribution and evaluate $y=f(x)$, will $y \sim p(y|z)$?

I'm asking because I'm trying to wrap my head around Thompson sampling using Bayesian Neural Networks and I was wondering if we actually sample from the posterior distribution of the expected reward by sampling the posterior distribution of the Q-network parameters.

Edit: A short explaination of the answer(s) would be greatly appreciated. Thanks!

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If $X$ is a random variable, values of $X$ appear with probabilities $p(x)$. If you transform those values using a function $y = f(x)$, still each of the transformed values will appear with some probability, i.e. $Y$ will be a random variable as well. $Y$ may not have "nice" form for the probability distribution, but it will have some probability distribution.

The easiest way to sample from $Y$ and then transform the values, if you don't know the distribution of $Y$.

If there is $p(X|Z)$, then there also is $p(f(X)|Z)$, i.e. $p(Y|Z)$, since $Y$ is transformed $X$.

One example of such transformed random variable is $\chi^2$ distribution. If $Z \sim \mathcal{N}(0, 1)$, then $Q = Z^2 \sim \chi^2_1$. Of course, this is an example of a distribution that has "nice", closed-form solution.

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