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Suppose $X$ is a random variable with a density $f(x)$ such that $f(x)$ is a convolution of some density $g$ with some other density $q$: $$ f = g\ast q. $$

Under what conditions does $X=h(Y)$, where $Y\sim g(y)$ and $h$ is some function? Under what conditions does there exist a function $h$ such $h(Y)$ has the same density as $X$, where $Y\sim g(y)$?

This problem is equivalent to the following: Given independent random variables $(Y,U)$, when does there exist an $h$ such that $h(Y)\sim U+Y$? (In this re-formulation, $U\sim q(u)$.)

In my use case, $Y$ is simple (e.g. a Gaussian), $f$ and $q$ are very general (but smoothness/regularity conditions are fine if necessary), and $h$ is allowed to be general as well. So this precludes counterexamples like $X=Y$, etc.

Edit: Clarified that I did not mean $h(Y)=U+Y$, which of course implies $U$ is a.s. constant.

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    $\begingroup$ Did you mean $U+Y=h(Y)$, which is impossible for independent variables unless $U$ is constant, or that $ U+Y$ has the same distribution as $h(Y)$? $\endgroup$
    – Yuval Peres
    Commented Jul 21, 2022 at 17:06
  • $\begingroup$ Yes, thank you! Sorry for the confusion. I tried to clarify this above. $\endgroup$
    – edgar314
    Commented Jul 21, 2022 at 17:46
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    $\begingroup$ In general, changing the question after it has been answered is frowned upon. In this case, it was clear to me that you meant equality in distribution, and this is the natural meaning of the first form of the question. The edited answer by I. Pinelis now addresses both variants. $\endgroup$
    – Yuval Peres
    Commented Jul 21, 2022 at 18:25

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Suppose that $Y,U$ are independent random variables (r.v.'s) such that $h(Y)=U+Y$ for some (Borel-measurable) function $h$. Then $U=Z:=h(Y)-Y$ and $U,Z$ are independent. So, $U$ is independent of itself. So, $U$ is constant almost surely (a.s.): $P(U=u)=1$ for some real $u$. (Indeed, if $U$ is not constant a.s., then the support of the distribution of $U$ contains at least two real numbers $a,b$ such that $a<b$. Take any $c\in(a,b)$. Then the independence of $U$ of itself implies $0=P(U<c,U>c)=P(U<c)P(U>c)>0$, a contradiction.)

Vice versa, if $P(U=u)=1$ for some real $u$, then $h(Y)=U+Y$ a.s. with $h(Y):=u+Y$.

Thus, if $Y,U$ are independent, then there exists a (Borel-measurable) function $h$ such that $h(Y)=U+Y$ if and only if $U$ is constant a.s.

Unfortunately, the OP has changed the question, thus invalidating the above answer. The changed question admits a trivial answer as well, though, and after the change the "almost never" answer becomes "always".

Indeed, the r.v. $Y$ is assumed to have a density $g$. So, the distribution of $Y$ is atomless. So, $$T:=F_Y(Y)$$ has the uniform distribution on the interval $(0,1)$, where $F_V$ is the cdf of a r.v. $V$. So, $Z:=U+Y$ equals $F_Z^{-1}(T)$ in distribution. That is, $U+Y$ equals $h(Y):=F_Z^{-1}(F_Y(Y))$ in distribution, as desired.

(As usual, $F^{-1}(t):=\inf\{x\in\mathbb R\colon F(x)\ge t\}.$)

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  • $\begingroup$ See the comment from @YuvalPeres above. $\endgroup$
    – edgar314
    Commented Jul 21, 2022 at 17:46
  • $\begingroup$ @edgar314 : The OP should not change a question so as to invalidate a valid answer. $\endgroup$
    – Iosif Pinelis
    Commented Jul 21, 2022 at 18:26
  • $\begingroup$ @IosifPinelis My apologies that my query led to a change in the question after you had answered it. Fortunately, the revised question was just as easy to answer. $\endgroup$
    – Yuval Peres
    Commented Jul 21, 2022 at 18:27
  • $\begingroup$ @YuvalPeres : I understand the desire to help the OP to make the question more meaningful. But there is also another party to this, the answerer. I think in this case your comment did not hurt anyone, after all. :-) $\endgroup$
    – Iosif Pinelis
    Commented Jul 21, 2022 at 18:31
  • $\begingroup$ @edgar314 : Do you have a further response to this answer? $\endgroup$
    – Iosif Pinelis
    Commented Jul 22, 2022 at 19:14

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