Question

Let \(X\) and \(Y\) be independent \(N(0,1)\) random variables and \(Z = \left|\frac{X}{Y}\right|\). Then, which of the following expectations is finite?

• A. \(E\left(\frac{1}{\sqrt{Z}}\right)\)
• B. \(E(Z\sqrt{Z})\)
• C. \(E(Z)\)
• D. \(E\left(\frac{1}{Z\sqrt{Z}}\right)\)

Hint:

The ratio of two independent standard normal variables follows a Cauchy distribution; only negative powers of \(Z\) less than 1 yield finite expectations.

Answer 1

The Correct Option is A

Step 1: Recall the distribution of \(Z\).
If \(X, Y \sim N(0,1)\) are independent, then \(\frac{X}{Y}\) follows a standard Cauchy distribution. Hence, \(Z = \left|\frac{X}{Y}\right|\) follows a half-Cauchy distribution with pdf \[ f_Z(z) = \frac{2}{\pi(1 + z^2)}, \quad z>0 \]
Step 2: Check finiteness of each expected value.
We must check whether \(\int_0^\infty g(z) f_Z(z)\, dz\) converges for each function \(g(z)\). - For \(E(Z)\): \[ \int_0^\infty z \frac{2}{\pi(1+z^2)}\,dz \] diverges because for large \(z\), the integrand behaves like \(\frac{1}{z}\). - For \(E(Z\sqrt{Z}) = E(Z^{3/2})\): \[ \int_0^\infty z^{3/2} \frac{2}{\pi(1+z^2)}\,dz \] also diverges since \(z^{3/2-2} = z^{-1/2}\) diverges at infinity. - For \(E\left(\frac{1}{Z\sqrt{Z}}\right) = E(Z^{-3/2})\): This diverges near \(z=0\) because \(z^{-3/2}\) becomes unbounded. - For \(E\left(\frac{1}{\sqrt{Z}}\right) = E(Z^{-1/2})\): \[ \int_0^\infty z^{-1/2}\frac{2}{\pi(1+z^2)}\,dz \] This converges since it is finite near both \(z=0\) and \(z\to\infty\).
Step 3: Conclusion.
Only \(E(Z^{-1/2}) = E\left(\frac{1}{\sqrt{Z}}\right)\) is finite. Final Answer: \[ \boxed{E\left(\frac{1}{\sqrt{Z}}\right)} \]