Question

Let \( X_1, X_2, \dots, X_n \ (n \ge 2) \) be independent and identically distributed random variables with probability density function \[ f(x) = \begin{cases} \dfrac{1}{x^2}, & x \ge 1, \\ 0, & \text{otherwise.} \end{cases} \] Then, which of the following random variables has/have finite expectation?

• A. \(X_1\)
• B. \(\dfrac{1}{X_2}\)
• C. \(\sqrt{X_1}\)
• D. \(\min\{X_1, ..., X_n\}\)

Hint:

When testing for expectation convergence, check tail behavior using \(\int_a^\infty x f(x)\,dx\). Power-law tails like \(1/x^2\) yield convergence for exponents greater than 2.

Answer 1

The Correct Option is B, C, D

Step 1: Compute \(E(X_1)\).
\[ E(X_1) = \int_1^\infty x \cdot \frac{1}{x^2} dx = \int_1^\infty \frac{1}{x} dx, \] which diverges (logarithmic divergence). Hence \(E(X_1)\) is infinite.
Step 2: Compute \(E(1/X_2)\).
\[ E\left(\frac{1}{X_2}\right) = \int_1^\infty \frac{1}{x} \cdot \frac{1}{x^2} dx = \int_1^\infty \frac{1}{x^3} dx = \frac{1}{2}. \] This is finite.
Step 3: Compute \(E(\sqrt{X_1})\).
\[ E(\sqrt{X_1}) = \int_1^\infty \sqrt{x} \cdot \frac{1}{x^2} dx = \int_1^\infty x^{-3/2} dx = 2, \] which is finite. However, \(E(X_1)\) diverges, and we check for \(\min(X_1, ..., X_n)\).
Step 4: Expectation of \(\min(X_1, ..., X_n)\).
For \(X_i\) i.i.d. with \(P(X>x) = 1/x\) for \(x \ge 1\), \[ P(\min(X_1, ..., X_n)>x) = \left(\frac{1}{x}\right)^n. \] Hence, \[ E(\min(X_1, ..., X_n)) = \int_0^\infty P(\min(X_1, ..., X_n)>x) dx = 1 + \int_1^\infty \frac{1}{x^n} dx = 1 + \frac{1}{n-1}. \] This is finite for all \(n \ge 2\).
Step 5: Conclusion.
Finite expectations: \(E(1/X_2)\) and \(E(\min(X_1, ..., X_n))\). Final Answer: \[ \boxed{(B) \text{ and } (D)} \]