How can I get an expression of the probability mass function of: \begin{equation} Y_i=\sum_{k=1}^i f\left(\sum_{n=1}^{k} X_n\right) \end{equation} being $x_n, n=1,2,...$ iid random variables and $f(\cdot)$ some (one-to-one) function?
I have tried firstly to express the PMF of $Y_i$ as a function of the conditional probabilities as follows: $$ \begin{split} pmf(Y_i) =\sum_{\Omega_{Y_{i-1}}} pmf(Y_i | Y_{i-1}=y_{n-1})\cdot prob\left(Y_{i-1}=y_{n-1}\right) \end{split} $$ The expression above makes me think that I can model $Y_i$ as a Markov chain, where $pmf(Y_i | Y_{i-1}=y_{n-1})$ would give me the transition probability. Is this fact useful to find the distribution of $Y_i$ at all? If so, which concept related to Markov chains I am missing?
Edit. I have tried the following: I have used a linear approximation of the function $f(x)$ around the mean value at every point, like this: \begin{equation} Y_i = \sum_{k=1}^i \left(f\left(k\cdot E\left[ X\right]\right)+\left(\sum_{n=1}^{k} X_n-k\cdot E\left[ X\right]\right) \right) \frac{df\left(x\right)}{dx}\Big|_{x=kE\left[ X\right]} \end{equation} So, basically, now my problem is a weighted sum of iid discrete random variables like $\sum_{k=1}^i X_k \left(\sum_{n=k}^i \omega_n\right)$, with $\omega_n=\frac{df\left(x\right)}{dx}\Big|_{x=nE\left[ X\right]}$.
Any suggestion to get the PMF of $\sum_{k=1}^i X_k \left(\sum_{n=k}^i \omega_n\right)$? I have considered to use the probability generating function (scaling propertie with the weights $\omega_r$ and the product of probability generating functions of $X$ ($M_x(\sum_{n=1}^i\omega_nt)M_x((\sum_{n=2}^i\omega_nt)t)...$)), but is there any way to get analytical expression for it?