Deriving the sampling (or asymptotic) distribution

Assume we have the following function:

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


\(d\) 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.


TS Contributor
Usually when you are talking about the asymptotics of a certain statistic, it should be a sequence indexed by the sample size \( n \). However, now it seems that the \( n \) group you defined are independent to each other so that \( p_i \) are just a sequence of i.i.d. random variables which does not related to any asymptotic result. Am I missing anything?

Besides, if we define \( Z_t = \frac {1 + X_t} {1 + Y_t} \)

then the function becomes \( f(p) = \frac {1} {(1 - p)d} \ln\left(\frac {1} {T} \sum_{i=1}^T Z_t^{1-p} \right) \)

So what why do we need to write it out like that?

Last but not least, \( \frac {1} {a} \ln b = 0 \iff \ln b = 0 \iff b = 1 \)

So the actually relation can be simplified to \( \sum_{i=1}^T Z_t^{1-p} = T \) ?
You are quite right. Yes so the relation to which you have simplified to is the one I am after. Assume we fix \(T\), how can we derive the distribution of \(p\)?