Question

If $X_{1}, X_{2}, X_{3}$ are independent and identically distributed standard normal variates and let $U=\frac{\sqrt{2}X_{3}}{\sqrt{X_{1}^{2}+X_{2}^{2}}}$ then $U^{2}$ follows

• A. $t_{4}$
• B. $t_{2}$
• C. $F(1,2)$
• D. $F(1,4)$

Hint:

Remember that $t_n^2 = F(1, n)$. In this problem, $U$ itself actually follows a Student's $t$-distribution with 2 degrees of freedom ($t_2$), so its square naturally follows $F(1, 2)$.

Answer 1

The Correct Option is C

To determine the distribution of $U^2$, we must analyze the components of the expression for $U$ based on the properties of standard normal variables.

Step 1: \color{redIdentify the Distribution of the Numerator and Denominator
We are given $X_1, X_2, X_3 \sim i.i.d. N(0,1)$.
The numerator contains $X_3^2$ when we square $U$. Since $X_3 \sim N(0,1)$, $X_3^2 \sim \chi^2_1$ (Chi-square distribution with 1 degree of freedom).
The denominator contains $X_1^2 + X_2^2$. Since $X_1, X_2$ are independent standard normal variables, the sum of their squares follows a Chi-square distribution with degrees of freedom equal to the number of variables:
$X_1^2 + X_2^2 \sim \chi^2_2$.

Step 2: \color{redDefine the Ratio for F-distribution
An $F$-distribution $F(n_1, n_2)$ is defined as the ratio of two independent Chi-square variables, each divided by its respective degrees of freedom:
$F = \frac{\chi^2_{n_1} / n_1}{\chi^2_{n_2} / n_2}$.

Step 3: \color{redTransform U into U²
Given $U = \frac{\sqrt{2}X_3}{\sqrt{X_1^2+X_2^2}}$
Squaring both sides gives:
$U^2 = \frac{2 X_3^2}{X_1^2 + X_2^2}$.

Step 4: \color{redMap to the F-distribution Form
Rewrite $U^2$ to match the $F$-distribution definition:
$U^2 = \frac{X_3^2 / 1}{(X_1^2 + X_2^2) / 2}$.
Here, the numerator is a $\chi^2_1$ variable divided by $1$, and the denominator is a $\chi^2_2$ variable divided by $2$.
Therefore, $U^2 \sim F(1, 2)$.
Thus, the distribution is $F(1,2)$.