The distribution of functions of random variables is well-studied for various different and general cases, but I didn't find much result for the reverse.



> 
> Suppose that $X\_1, X\_2$ are (probably independent) random variables and we have $X\_1\sim f(x\_1)$ and $X\_2\sim g(x\_2)$. Construct a new distribution for the new random variable $X$ as follows:
>  $$X\sim p(x)=\frac{1}{\kappa}(f(x)+g(x))$$
>  where $\kappa$ is a normalization factor such that the function is a density function. 
>  Is it possible to express $X$ as a function of $X\_1$ and $X\_2$ in a way that their PDF is $p(x)$?
> 
> 
> More formally, find a function $G$ such that 
>  $$X=G(X\_1, X\_2)\sim p(x)$$
> 
> 
> 


Note: "$\sim$" denotes the probability density function of a random variable.


Note 2: If we had $p(x)=f(x)\*g(x)$, where $\*$ is the sign for convolution, then I think we may write $G(X\_1, X\_2) = X\_1+X\_2$.


Note 3: I have asked the same question [here](https://math.stackexchange.com/q/3588454/255648) for product of density functions.



Random variable corresponding to sum of density functions
I assume $X\_1 \sim f(x\_1)$ means that that the distribution of $X\_1$ has density function $f$.


Note first that your $\kappa$ can only be $2$, otherwise the integral of $p$ will not equal $1$. Then $p$ is the density of a [mixture](https://en.wikipedia.org/wiki/Mixture_distribution) of $X\_1, X\_2$; it corresponds to flipping a coin to decide whether to take $X\_1$ or $X\_2$. In notation, if $A$ is an event of probability $1/2$ that is independent of $(X\_1, X\_2)$, then $p$ is the density of the random variable $X = 1\_A X\_1 + 1\_{A^c} X\_2$. You can also think of it as $X = Z X\_1+ (1-Z) X\_2$ where $Z$ is a Bernoulli(1/2) random variable independent of $(X\_1, X\_2)$.


In general such a random variable cannot be written as a function of $X\_1, X\_2$ alone.


