Step 1: Concept Overview:
The problem aims to calculate the expected value of the reciprocal of a statistic \(Y\), where \(Y\) represents the sum of transformed random variables. The core idea involves determining the probability distribution of \(Y\) and then applying the definition of expected value.
Step 2: Methodology:
1. Determine the distribution of a single transformed variable, \(W_i = -\log(X_i)\), using the change of variables technique.2. Recognize \(Y\) as the sum of \(n\) independent and identically distributed (i.i.d.) random variables. The sum of i.i.d. Exponential variables results in a Gamma distribution.3. Compute \(E\left[\frac{1}{Y}\right]\) by integrating \( \frac{1}{y} \times f_Y(y) \) over the appropriate range of Y.
Step 3: Detailed Solution:
First, derive the distribution of \(W = -\log(X)\).The cumulative distribution function (CDF) of \(X\) is \( F_X(x) = \int_0^x (\theta+1)t^\theta dt = [t^{\theta+1}]_0^x = x^{\theta+1} \) for \(0<x<1\).The CDF of \(W\) for \(w>0\) is:\( F_W(w) = P(W \le w) = P(-\log(X) \le w) = P(\log(X) \ge -w) = P(X \ge e^{-w}) \)\( = 1 - P(X<e^{-w}) = 1 - F_X(e^{-w}) = 1 - (e^{-w})^{\theta+1} = 1 - e^{-(\theta+1)w} \).This corresponds to the CDF of an Exponential distribution with rate parameter \( \lambda = \theta+1 \).Therefore, \( W_i = -\log(X_i) \) are i.i.d. Exponential(\(\theta+1\)) random variables.\( Y = \sum_{i=1}^n W_i \). The sum of \(n\) i.i.d. Exponential(\(\lambda\)) variables follows a Gamma distribution with shape parameter \(k=n\) and rate parameter \(\lambda\).Hence, \( Y \sim \text{Gamma}(n, \theta+1) \).The probability density function (PDF) of \(Y\) is \( f_Y(y) = \frac{\lambda^n}{\Gamma(n)} y^{n-1} e^{-\lambda y} \) for \(y>0\), where \(\lambda = \theta+1\).Now, calculate \(E\left[\frac{1}{Y}\right]\):\[ E\left[\frac{1}{Y}\right] = \int_0^\infty \frac{1}{y} f_Y(y) dy = \int_0^\infty \frac{1}{y} \frac{\lambda^n}{\Gamma(n)} y^{n-1} e^{-\lambda y} dy \]\[ = \frac{\lambda^n}{\Gamma(n)} \int_0^\infty y^{n-2} e^{-\lambda y} dy \]The integral relates to the Gamma function, defined as \( \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt \).Substitute \(t = \lambda y \implies y = t/\lambda, dy = dt/\lambda\):\[ \int_0^\infty y^{n-2} e^{-\lambda y} dy = \int_0^\infty \left(\frac{t}{\lambda}\right)^{n-2} e^{-t} \frac{dt}{\lambda} = \frac{1}{\lambda^{n-1}} \int_0^\infty t^{(n-1)-1} e^{-t} dt = \frac{\Gamma(n-1)}{\lambda^{n-1}} \]Substitute this back into the expectation formula:\[ E\left[\frac{1}{Y}\right] = \frac{\lambda^n}{\Gamma(n)} \frac{\Gamma(n-1)}{\lambda^{n-1}} \]Using the property \( \Gamma(n) = (n-1)\Gamma(n-1) \):\[ E\left[\frac{1}{Y}\right] = \frac{\lambda^n}{(n-1)\Gamma(n-1)} \frac{\Gamma(n-1)}{\lambda^{n-1}} = \frac{\lambda}{n-1} \]Since \( \lambda = \theta+1 \), the final result is:\[ E\left[\frac{1}{Y}\right] = \frac{\theta+1}{n-1} \]
Step 4: Solution:
The expected value is \( \frac{\theta+1}{n-1} \).