Assume we have the following function:

$$f(p) = \frac{1}{(1-p)d}\ln\left(\frac{1}{T}\sum_{t=1}^{T}\left[\frac{1+X_t}{1+Y_t} \right]^{1-p} \right)$$

where

$d$ is a constant

$T$ is a constant

$X_t$ for $t = 1, 2, \cdots, T$ are random variables

$Y_t$ for $t = 1, 2, \cdots, T$ are random variables

$p$ is defined such that it is value that satisfies $f(p) = 0$

I wish to derive a sampling (***or*** asymptotic) distribution for the statistic $p$.

By sampling distribution I mean the following:

The solution to $f(p) = 0$ doesn't have a closed-form solution, but it is obvious that the resulting value of $p$ depends on $X_t$ and $Y_t$, so $p$ can be treated as a random variable that depends on the random variables $X_t$ and $Y_t$. Then for every $T$ observations of $X_t$ and $Y_t$, we have a corresponding value $p$ that satisfies $f(p) = 0$, what is the sampling distribution of $p$?

By asymptotic distribution I mean the following:

Assume we have $n$ instances of $X_t$ and $Y_t$, that is, $n$ groups of $\{X_1, X_2, \cdots, X_T\}$ and $\{Y_1, Y_2, \cdots, Y_T\}$. Then we solve $f(p)=0$ and have $n$ observations of $p$, that is, $\{p_1, p_2, \cdots, p_n\}$. What is the distribution of $p$ as $n \rightarrow \infty$?

Also assume you are allowed the following assumptions to achieve the above:

1) You can make any distributional assumptions regarding $X_t$ and $Y_t$, e.g., $X_t$ and $Y_t$ are independent from each other, also $X_t$, $Y_t$ for $t = 1, 2, \cdots, T$ are independently and identically distributed.

2) Rather than making distributional assumptions about $X_t$ and $Y_t$, assume you can make some assumptions about the processes $\{X_t\}$ and $\{Y_t\}$, e.g., both processes are stationary (or weakly stationary) etc.