Question

Let \(X_1,X_2,X_3\) be three independent random variables such that \(X_1\sim N\left(0,\frac12\right)\), \(X_2\sim N(0,2)\), and \(X_3\sim N(0,4)\). Consider the following statements.

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

An \(F\)-distribution is formed by the ratio of two independent chi-square variables divided by their degrees of freedom, and the ratio of two independent standard normal variables follows a standard Cauchy distribution.

Answer 1

The Correct Option is C

Step 1: Standardize the random variables.
Since
\[ X_1\sim N\left(0,\frac12\right), \] we can write
\[ Z_1=\sqrt{2}X_1\sim N(0,1) \] Similarly,
\[ Z_2=\frac{X_2}{\sqrt{2}}\sim N(0,1) \] and
\[ Z_3=\frac{X_3}{2}\sim N(0,1) \] Also, \(Z_1,Z_2,Z_3\) are independent.

Step 2: Check statement \(P\).
Now,
\[ X_1^2=\frac{Z_1^2}{2} \] \[ X_2^2=2Z_2^2 \] and
\[ X_3^2=4Z_3^2 \] Therefore,
\[ \frac{16X_1^2}{2X_2^2+X_3^2} = \frac{16\cdot \frac{Z_1^2}{2}}{2(2Z_2^2)+4Z_3^2} \] \[ = \frac{8Z_1^2}{4Z_2^2+4Z_3^2} \] \[ = \frac{2Z_1^2}{Z_2^2+Z_3^2} \] This can be written as
\[ \frac{Z_1^2/1}{(Z_2^2+Z_3^2)/2} \] Since
\[ Z_1^2\sim \chi_1^2 \] and
\[ Z_2^2+Z_3^2\sim \chi_2^2, \] we get
\[ \frac{Z_1^2/1}{(Z_2^2+Z_3^2)/2}\sim F_{1,2} \] Hence, statement \(P\) is correct.

Step 3: Check statement \(Q\).
Let
\[ U=2X_1-X_2 \] and
\[ V=2X_1+X_2 \] Since \(X_1\) and \(X_2\) are independent normal random variables, \(U\) and \(V\) are also normal random variables.
Now,
\[ Var(2X_1)=4Var(X_1) \] \[ =4\cdot \frac12=2 \] and
\[ Var(X_2)=2 \] So,
\[ Var(U)=Var(2X_1-X_2)=2+2=4 \] and
\[ Var(V)=Var(2X_1+X_2)=2+2=4 \] Also,
\[ Cov(U,V)=Cov(2X_1-X_2,2X_1+X_2) \] \[ =Var(2X_1)-Var(X_2) \] \[ =2-2=0 \] Since \(U\) and \(V\) are jointly normal and uncorrelated, they are independent.
Thus,
\[ U\sim N(0,4), \qquad V\sim N(0,4) \] and \(U,V\) are independent.
Therefore,
\[ \frac{U}{V} = \frac{2X_1-X_2}{2X_1+X_2} \] is the ratio of two independent normal random variables with mean \(0\) and same variance.
Hence,
\[ \frac{2X_1-X_2}{2X_1+X_2} \] has standard Cauchy distribution.
Therefore, statement \(Q\) is correct.

Step 4: Final conclusion.
Both statements \(P\) and \(Q\) are correct.
Hence,
\[ \boxed{(C)} \]