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My question is as follows. Suppose you have two variables $X$, $Y$. If I pick a random sample $x$ from the distribution over $X$ that has probability density function $f_X$, and another random sample $y$ from the distribution over $Y$ that has probability density function $f_Y$, is $x-y$ a random sample of the distribution over $X-Y$ that has probability density function $f_{X-Y}(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(x-z)dx$? In other words, by taking two random samples $x$ and $y$ and setting $z=x-y$, have I taken a random sample from the distribution over $Z=X-Y$?

It seems like it should be the case by definition, but I haven't been able to convince myself beyond doubt. Can anyone help?

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    $\begingroup$ What does „be a sample“ mean for you? I.e. given some numbers $x_1,...,x_n$ when is that a sample of a series of random variables $X_1,...,X_n$? $\endgroup$ – Fabian Werner Sep 12 '19 at 11:16
  • $\begingroup$ I mean a sample picked at random from that distribution. I've edited the question to clarify. $\endgroup$ – user259707 Sep 12 '19 at 11:24
  • $\begingroup$ Please explain what you mean by "$x-y.$" Are you thinking of samples as being ordered sequences of values, so that you can subtract them component by component; or are you randomly subtracting elements of one sample from elements of another; or are you forming all possible differences of elements between the two samples? The answer depends on your meaning. $\endgroup$ – whuber Sep 12 '19 at 12:20
  • $\begingroup$ I was thinking of examples like $X\sim Lap(\mu_1,b_1)$ and $Y\sim Lap(\mu_2,b_2)$, in which case one can sample $x$ from X and $y$ from Y then easily take $x-y$ as both $x$ and $y$ are just real numbers. But I'm interested in every case, if there's a difference between them. $\endgroup$ – user259707 Sep 12 '19 at 13:22