Question

Let $X_1, X_2, X_3$ be i.i.d. $N(0, 1)$ random variables. Then which of the following statements is/are TRUE?

• A. $\dfrac{\sqrt{2} X_1}{\sqrt{X_2^2 + X_3^2}} \sim t_2$
• B. $\dfrac{\sqrt{2} X_1}{|X_2 + X_3|} \sim t_1$
• C. $\dfrac{(X_1 - X_2)^2}{(X_1 + X_2)^2} \sim F_{1, 1}$
• D. $\sum_{i=1}^{3} X_i^2 \sim \chi^2_2$

Hint:

A ratio of a standard normal to the square root of an independent chi-square variable divided by its degrees of freedom follows a $t$-distribution.

Answer 1

The Correct Option is A, B, C

Analyzing each option:

(A) $\frac{\sqrt{2}X_1}{\sqrt{X_2^2 + X_3^2}} \sim t_2$

The numerator: $\sqrt{2}X_1 \sim N(0,2)$, so $\frac{\sqrt{2}X_1}{\sqrt{2}} = X_1 \sim N(0,1)$

The denominator: $X_2^2 + X_3^2 \sim \chi^2_2$

The t-distribution with $n$ degrees of freedom has the form: $$T = \frac{Z}{\sqrt{V/n}}$$ where $Z \sim N(0,1)$ and $V \sim \chi^2_n$ are independent.

Rewrite: $$\frac{\sqrt{2}X_1}{\sqrt{X_2^2 + X_3^2}} = \frac{X_1}{\sqrt{(X_2^2 + X_3^2)/2}}$$

Since $X_1 \sim N(0,1)$ and $(X_2^2 + X_3^2) \sim \chi^2_2$:

This matches the form of $t_2$ distribution.

(A) is TRUE 

(B) $\frac{\sqrt{2}X_1}{|X_2 + X_3|} \sim t_1$

Note that $X_2 + X_3 \sim N(0,2)$, so $\frac{X_2 + X_3}{\sqrt{2}} \sim N(0,1)$

Therefore: $(X_2 + X_3)^2 \sim 2\chi^2_1$

Rewrite: $$\frac{\sqrt{2}X_1}{|X_2 + X_3|} = \frac{\sqrt{2}X_1}{\sqrt{(X_2 + X_3)^2}} = \frac{X_1}{\sqrt{(X_2 + X_3)^2/2}}$$

Since $(X_2 + X_3)^2/2 \sim \chi^2_1$ and $X_1 \sim N(0,1)$:

This matches the form of $t_1$ distribution.

(B) is TRUE 

(C) $\frac{(X_1 - X_2)^2}{(X_1 + X_2)^2} \sim F_{1,1}$

Note that:

• $X_1 - X_2 \sim N(0,2)$, so $(X_1 - X_2)^2 \sim 2\chi^2_1$
• $X_1 + X_2 \sim N(0,2)$, so $(X_1 + X_2)^2 \sim 2\chi^2_1$

Therefore: $$\frac{(X_1 - X_2)^2}{(X_1 + X_2)^2} = \frac{2\chi^2_1/1}{2\chi^2_1/1} = \frac{\chi^2_1/1}{\chi^2_1/1}$$

This has the form of $F_{1,1}$ distribution.

(C) is TRUE 

(D) $\sum_{i=1}^3 X_i^2 \sim \chi^2_2$

Since each $X_i \sim N(0,1)$, we have $X_i^2 \sim \chi^2_1$

The sum of independent chi-squared variables: $$\sum_{i=1}^3 X_i^2 \sim \chi^2_3$$

NOT $\chi^2_2$

(D) is FALSE 

Answer: (A), (B), and (C) are true