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    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)

    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.

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    Re: Deriving the sampling (or asymptotic) distribution

    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 ?

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  4. #3
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    Re: Deriving the sampling (or asymptotic) distribution

    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?

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    Re: Deriving the sampling (or asymptotic) distribution


    Any further ideas?

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