Question

If \(X_1, X_2, \dots, X_n\) is a random sample from the population \(f(x, \theta) = (\theta+1)x^\theta; 0<x<1; \theta>-1\) and \(Y = -\sum_{i=1}^{n} \log(x_i)\). Then \(E\left(\frac{1}{Y}\right)\) is

• A. \( \frac{\theta+1}{n} \)
• B. \( \frac{\theta+1}{n-1} \)
• C. \( \frac{\theta}{n} \)
• D. \( \frac{\theta}{n-1} \)

Hint:

The transformation \(W = -\log(X)\) is a classic one. If \(X\) follows a Beta(\(a, 1\)) distribution (which is the case here with \(a=\theta+1\)), then \(-\log(X)\) follows an Exponential distribution. Recognizing this pattern saves time in identifying the distribution of \(Y\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem requires finding the expected value of the reciprocal of a statistic \(Y\), where \(Y\) is the sum of transformed random variables. The key is to first determine the probability distribution of \(Y\) and then use the definition of expected value.

Step 2: Key Formula or Approach:
1. Find the distribution of a single transformed variable, \(W_i = -\log(X_i)\), using the change of variable technique for probability distributions. 2. Recognize that \(Y\) is the sum of \(n\) i.i.d. random variables. The distribution of a sum of certain i.i.d. variables (like Exponential) is a known distribution (Gamma). 3. Calculate \(E\left[\frac{1}{Y}\right]\) by integrating \( \frac{1}{y} \times f_Y(y) \) over the support of Y.

Step 3: Detailed Explanation:
Let's find the distribution of \(W = -\log(X)\). The 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 is the CDF of an Exponential distribution with rate parameter \( \lambda = \theta+1 \). So, \( W_i = -\log(X_i) \) 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 \(\lambda\). Thus, \( Y \sim \text{Gamma}(n, \theta+1) \). The PDF of \(Y\) is \( f_Y(y) = \frac{\lambda^n}{\Gamma(n)} y^{n-1} e^{-\lambda y} \) for \(y>0\), with \(\lambda = \theta+1\). Now, we 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 is related to the Gamma function. The Gamma function is defined as \( \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt \). Using the substitution \(t = \lambda y \implies y = t/\lambda, dy = dt/\lambda\): \[ \int_0^\infty y^{n-2} e^{-\lambda y} dy = \int_0^\infty \left(\frac{t}{\lambda}\right)^{n-2} e^{-t} \frac{dt}{\lambda} = \frac{1}{\lambda^{n-1}} \int_0^\infty t^{(n-1)-1} e^{-t} dt = \frac{\Gamma(n-1)}{\lambda^{n-1}} \] 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: Final Answer:
The expected value is \( \frac{\theta+1}{n-1} \).