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Given $X$ is a random variable and $Y = T(X)$ where $T$ is a deterministic function, then, $Y$ is a random variable. Consider the following sampling procedure: sample $x \sim X$, then apply $T$ on $x$ to get $T(x)$. Is this sampling procedure guaranteed to be that for sampling from $Y = T(X)$?

This question could be trivial (because it is intuitively correct that the sampling procedure above is indeed for $Y$) but I could not either find a counter-example against that or find some text that explicitly claims that.

Thank you.

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    $\begingroup$ Yes, $T(x)$ follows the distribution of $Y$ by definition. $\endgroup$
    – angryavian
    Jul 14, 2019 at 7:22
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    $\begingroup$ @angryavian Do you have any pointer to that definition? $\endgroup$
    – TNg
    Jul 14, 2019 at 7:35

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I agree with you and angryavian that the statement is trivially correct, since for every $A$ in the considered $\sigma$-algebra we have $$ P(T(x)\in A) \stackrel{x\sim X}{=} P(T(X)\in A) = P(Y\in A). $$

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