Discussion
Let \(X_{1}, X_{2}, X_{3}\) be three independent random variables such that \(X_{1} \sim N\left(0, \frac{1}{2}\right)\), \(X_{2} \sim N(0, 2)\), and \(X_{3} \sim N(0, 4)\). Consider the following statements.
P: \(\frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}}\) has \(F\) distribution with \(1\) and \(2\) degrees of freedom.
Q: \(\frac{2X_{1} - X_{2}}{2X_{1} + X_{2}}\) has standard Cauchy distribution.
Then
Answer: C
Both P and Q are correct.
P: standardise each variable: \(\sqrt{2}X_{1}\), \(\frac{X_{2}}{\sqrt{2}}\) and \(\frac{X_{3}}{2}\) are independent \(N(0, 1)\). So \(2X_{1}^{2} \sim \chi_{1}^{2}\) and, independently,
\[ \begin{aligned} \frac{X_{2}^{2}}{2} + \frac{X_{3}^{2}}{4} &= \frac{2X_{2}^{2} + X_{3}^{2}}{4} \\ &\sim \chi_{2}^{2} \end{aligned} \]
The F ratio divides each by its degrees of freedom:
\[ \begin{aligned} \frac{2X_{1}^{2}/1}{(2X_{2}^{2} + X_{3}^{2})/8} &= \frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}} \\ &\sim F_{1,2} \end{aligned} \]
Q: \(2X_{1}\) and \(X_{2}\) are independent \(N(0, 2)\), with equal variances. Then \(U = 2X_{1} - X_{2}\) and \(V = 2X_{1} + X_{2}\) are jointly normal with
\[ Cov(U, V) = Var(2X_{1}) - Var(X_{2}) = 2 - 2 = 0 \]
so they are independent \(N(0, 4)\) variables. The ratio of two independent zero-mean normals with the same variance is standard Cauchy, so \(\frac{U}{V}\) is standard Cauchy.
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