Question

Let X₁, X₂ and X₃ be three independent and identically distributed random variables having N(0, 1) distribution. If
\(U=\frac{2X^2_1}{(X_2+X_3)^2}\) and \(V=\frac{2(X_2-X_3)^2}{2X^2_1+(X_2+X_3)^2},\)
then which of the following statements is/are true ?

• A. U has F_1,1 distribution and V has F_1,2 distribution
• B. U has F_1,1 distribution and V has F_2,1 distribution
• C. U and V are independent
• D. \(\frac{1}{2}V(1+U)\) has F_2,3 distribution

Answer 1

The Correct Option is A, C

To solve this problem, we need to determine the distributions of the variables \(U\) and \(V\), and then check their independence.

**Step 1: Distribution of \(U\)**

Given that \(X_1, X_2, X_3\) are independent and identically distributed (i.i.d.) random variables with a standard normal distribution \(N(0, 1)\):

\(U = \frac{2X_1^2}{(X_2 + X_3)^2}.\)

\(X_1^2 \sim \chi^2_1\) because the square of a standard normal variable follows a chi-squared distribution with 1 degree of freedom. Similarly, \((X_2 + X_3)\) is normally distributed (\(N(0, 2)\)) since both \(X_2\) and \(X_3\) are \(N(0, 1)\) and independent. It follows that \(\frac{(X_2 + X_3)^2}{2} \sim \chi^2_1\).

Thus, \(U = \frac{2 \cdot \chi^2_1}{\chi^2_1}\), which is an F-distribution with degrees of freedom (1, 1), \(F_{1, 1}\).

**Step 2: Distribution of \(V\)**

The expression for \(V\) given is:

\(V = \frac{2(X_2 - X_3)^2}{2X_1^2 + (X_2 + X_3)^2}.\)

Notice that \(X_2 - X_3\) follows a distribution \(N(0, 2)\), thus \(\frac{(X_2 - X_3)^2}{2} \sim \chi^2_1\). Combining terms in the denominator, \(2X_1^2 + (X_2 + X_3)^2\), sums the independent chi-squared distributions which results in a total degrees of freedom of 2+1=3: \(2 \chi^2_1 + \chi^2_1 \sim \chi^2_3\).

Thus, the distribution for \(V\) can be shown as:

\(V = \frac{\chi^2_1}{\chi^2_2}\). This follows an \(F_{1, 2}\) distribution.

**Step 3: Independence of \(U\) and \(V\)**

The statistics \(U\) and \(V\) are formed from independent components \(X_1, X_2, X_3\) and are therefore independent, provided transformation dependency conditions hold, which they do under these settings.

**Conclusion**

The correct statements based on our findings are:

• \(U\) has \(F_{1, 1}\) distribution and \(V\) has \(F_{1, 2}\) distribution.
• \(U\) and \(V\) are independent.