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Given two Random Variables: $X$ with probability density $f_X(x)$ and $Y$ with probability density $f_Y(y)$, I want to understand the random variable $Z$ such that $Z$ has probability density $f_X(z) + f_Y(z)$. Here, I'm allowing for negative probability densities, and $f_X$ and $f_Y$ do not necessarily integrate to 1.

NB: I'm not looking for the formula for the pdf of $Z = X+Y$, which was answered many times in other posts.

I want to know if there is a way to express $Z$ in terms of $X$ and $Y$. For example, can I find constants $c_{ij}$ to say $Z = \sum_{ij} c_{ij} X^iY^j$?

Thank You!

EDIT: After reading some comments, it seems like posing this question in terms of $X$ and $Y$ as random variables doesn't really make sense if $f_X(x)$ and $f_Y(y)$ are not always positive and do not integrate to 1. So I want to ask the simpler version of this question. If $f_X(x)$ and $f_Y(y)$ are proper probability density functions, and we have some random variable $Z$ with pdf: $pf_X(z) + (1-p)f_Y(z)$ where $p \in [0,1]$, is there a way to express $Z$ in terms of $X$ and $Y$ (like $Z = \sum_{ij} c_{ij} X^iY^j$)?

($X$ and $Y$ are independent)

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  • $\begingroup$ My apologies if I misunderstand your question, but if you don’t require that the individual densities of $X$ and $Y$ each integrate to $1$, then they are, by definition, not probability densities functions... $\endgroup$ Nov 25 '19 at 20:33
  • $\begingroup$ I don't think you misunderstood, I'm just coming from a physics background and don't have the language quite right. I heard of this idea of using negative probabilities, and It seemed like it was acceptable so long as the sum of (fake) probability density functions is a real pdf (integrated to 1). The idea is that while X and Y may not integrate to 1, Z would. $\endgroup$ Nov 25 '19 at 20:38
  • $\begingroup$ I see. Another comment I have, but this is mostly to do with terminology, is that PDFs do not represent probabilities: there exist valid PDFs (i.e. satisfying the two properties: $f(x)\geq 0$ and integrating to $1$) that may exceed $1$ for some input values. Consider exponential PDF with $\lambda>1$, ie $f(x)=\lambda e^{-\lambda x}$ at $x=0$. Probability density functions represent exactly that: probabilities per unit; a density but not probabilities proper. $\endgroup$ Nov 25 '19 at 20:43
  • $\begingroup$ Right. Yes I'll make sure the language is clear. Is there a nice word for a generalized pdf, where $f(x)$ can be negative, and does not have to integrate to 1? $\endgroup$ Nov 25 '19 at 20:52
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    $\begingroup$ If $f_X$ and $f_Y$ are not probability densities, then there are no random variables $X$ and $Y$, so what are you looking for? $\endgroup$
    – zhoraster
    Nov 25 '19 at 21:05
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The edited question is amenable to standard probability theory: $Z$ is a mixture of $X$ and $Y$. More precisely, let $B$ be Bernoulli with parameter $0<p<1$, independent of $X$ and $Y$. If $X$ and $Y$ have PDFs $f_X$ and $f_Y$ respectively, then $Z=BX+(1-B)Y$ has PDF $f(z)=pf_X(z)+(1-p)f_Y(z)$.

(Though it may not be what you desired to have the coefficients $c_{ij}$ themselves random...)

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    $\begingroup$ There seems to be an implicit assumption that $X$ and $Y$ are independent. $\endgroup$ Nov 25 '19 at 22:06
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    $\begingroup$ @herbsteinberg: No, independence is not needed here. $\endgroup$ Nov 25 '19 at 22:09
  • $\begingroup$ Thank You! Coefficients being random variables isn't awesome but it is certainly a step in the right direction. $\endgroup$ Nov 25 '19 at 22:16
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Note that the distribution or density functions of a function of your $\ X\ $ and $\ Y\ $ will, in general, depend not merely on their individual density functions, $\ f_X\ $ and $\ f_Y\ $, but on their joint distribution or density function, $\ f_{XY}\ $, assuming that this latter exists. With that qualification, if $\ \varphi:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\ $ is any function at all such that $\ \varphi(X,Y)\ $ has a density $\ f_\varphi\ $, then the random variable $$ \int_{-\infty}^{\varphi(X,Y)}f_\varphi(t)dt $$ will be uniformly distributed.

I have to confess that I have no idea what it means for a random variable $\ Z\ $ to have a density function $\ f_X+f_Y\ $ which integrates to a value of $2$, rather than $1$, but if you define $$ F_\text{zinv}(y)=\inf \left\{x\left| \int_{-\infty}^x\left(f_X(t)+f_Y(t)\right) dt\ge 2y\right.\right\}\ , $$ then the random variable $$ Z=F_\text{zinv}\left(\int_{-\infty}^{\varphi(X,Y)}f_\varphi(t)dt\right) $$ will have density function $\ \frac{f_X+f_Y}{2}\ $.

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