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

Step 1: Recall definition of t-distribution.
If $Z \sim N(0,1)$ and $U \sim \chi^2_n$ independently, then $\dfrac{Z}{\sqrt{U/n}} \sim t_n$.

Step 2: Apply to given expression.
Here, $X_1 \sim N(0,1)$ and $(X_2^2 + X_3^2) \sim \chi^2_2$. Thus, \[ \frac{\sqrt{2} X_1}{\sqrt{X_2^2 + X_3^2}} = \frac{X_1}{\sqrt{(X_2^2 + X_3^2)/2}}. \] This matches the definition of $t_2$.

Step 3: Check other options.
(B) Denominator $|X_2 + X_3|$ does not follow $\chi^2_1$, so not $t_1$. (C) Ratio of quadratic terms does not simplify to $F_{1,1}$. (D) $\sum X_i^2 \sim \chi^2_3$, not $\chi^2_2$.

Step 4: Conclusion.
\[ \boxed{\frac{\sqrt{2} X_1}{\sqrt{X_2^2 + X_3^2}} \sim t_2.} \]