# Sampling and/or asymptotic distribution of a function

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 (it is actually the portfolio's annualized rate of return at time $t$)

$Y_t$ for $t = 1, 2, \cdots, T$ are random variables (it is actually the risk-free rate at time $t$)

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

What is the sampling and/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 (fixed) $T$ 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:

Similar to above, now assume $T$ isn't fixed, then clearly the solution $p$ to $f(p)=0$ implicitly depends on $T$, then what is the asymptotic distribution of $p$ as $T \rightarrow \infty$?

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.

3) Any assumption you see fit to yield a solution.

This is what I've tried so far.

If we let $Z_t = \frac {1 + X_t} {1 + Y_t}$, then we have $\displaystyle f(p) = \frac {1} {(1 - p)d} \ln\left(\frac {1} {T} \sum_{t=1}^T Z_t^{1-p} \right)$. Setting this equation to $0$ and rearranging, we have: $$\sum_{t=1}^T Z_t^{1-p} = T$$

The my question becomes:

1) Sampling distribution: For a fixed $T$, what is the distribution of $p$?

2) Asymptotic distribution: For $T \rightarrow \infty$, what is the distribution of $p$?

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So basically you are looking for a random variable $p$ that will give you the following result $\sum_{t=1}^T Z_t(\omega)^{1-p(\omega)}=T$ for all $\omega \in \Omega$ ? – Probilitator Mar 3 '14 at 10:13
could $p$ be a vector of length $T$ ? – Probilitator Mar 3 '14 at 10:31
also is the functional form of $f(p)$ mandatory or could one add some weights ? – Probilitator Mar 3 '14 at 14:14
@Probilitator Apologies for the late reply, I've been overseas (and still am, but finally managed to get internet =]). 1st, I am not sure what you mean by the notation $\omega$, could you expand on that? 2nd, $p$ is just a (single) numeric value. E.g., Let $T$ be fixed to be 3, then we have: $Z_1^{1-p} + Z_2^{1-p} + Z_3^{1-p}=3$, then for each realization of the random variable $Z_t$, i.e., $\{Z_1, Z_2, Z_3\}$, we can solve for a unique $p$. Hence as $Z_t$ changes, $p$ changes. 3rd, $f(p)$ is mandatory in that form, but sure, do what is necessary to get some results first. – siTTmo Mar 7 '14 at 3:24
yes thus $p$ is also a random variable one of the form $p=g(Z_1,Z_2,Z_3)$ with $g$ being some function that solves the corresponding equation ;) – Probilitator Mar 7 '14 at 6:55