# Deriving the sampling (or asymptotic) distribution

#### TrueTears

##### New Member
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.

#### BGM

##### 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 $$\sum_{i=1}^T Z_t^{1-p} = T$$ ?

#### TrueTears

##### New Member
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 $$p$$?