We have continuous random variables $X$ and $Y$. Our goal is to find the distribution of $X$, such that for any suitable function $g$, the distributions $g(X)$ and $Y$ have the same distribution. We could solve
$$E(\phi(g(X)))=\int \phi(g(x))f_X(x)dx = \int \phi(y)f_Y(y)dy=E(\phi(Y))$$
for $f_X$ with $\phi$ being an arbitrary suitable function. Substitution $g(x)=y$ gives us that the pdf of $X$ is
$$f_X(x)=f_Y(g(x))|g'(x)|.$$
The question is, what if $X$ and $Y$ are discrete? What would be the alternative to the change of variable to find the pmf of $X$? Could we use the same approach or is it fundamentally a different problem?
Discrete random variables are equal
discrete mathematicsprobability distributionsprobability theory
Best Answer
Let me try to answer one interpretation of your question. This should at least be useful to clarify your requirements. Let $Y$ be a discrete random variable with range $y_1,\ldots,y_m$ and $P(Y=y_j)=p_j$, where $p_j>0.$ Let $g$ be the "suitable function". For consistency you need $X=g^{-1}(Y)$.
Now define for each $y_j$ $n_j$ disjoint sets $x_{i,j}$ inside the domain of $g$ such that: $$ \bigcup_{i \leq n_j} x_{i,j}=g^{-1}(y_j).$$ Next choose any set of positive numbers $q_{i,j}$ with $$ \sum_{i\leq n_j} q_{i,j}=p_j \text{ for all } j=1\ldots, m.$$
Each $q_{i,j}$ defines the mass of the component $x_{i,j}$, i.e. $P(x_{i,j})=q_{i,j}$ and you can distribute the mass within each $x_{i,j}$ as you wish.
Following this recipe ensures: $$ P(g(X)=y_j)=P(X=g^{-1}(y_j))=P(\bigcup_{i \leq n_j} x_{i,j})=\sum_{i\leq n_j} P(x_{i,j})=p_j.$$