How can I find $c$ such that $c(X_1-X_2)/ \sqrt{X_3^2 +X_4^2 + X_5^2}$ forms a t distribution?

Question

Let $X_1,\cdots,X_5$ be iid random variables from $N(0,\sigma^2)$. Find $c$ such that $c(X_1-X_2)/ \sqrt{X_3^2 +X_4^2 + X_5^2}$ forms a t distribution.

My work:

A distribution $W$ is a t distribution iff $W=U/\sqrt{V/p}$, where $U\sim N(0,1)$ and $V \sim \mathcal{X}^2_p$ with $p$ being the degrees of freedom.

$(X_1-X_2)/\sqrt{2\sigma^2} \sim N(0,1)$, which partially takes care of the numerator in $c(X_1-X_2)/ \sqrt{X_3^2 +X_4^2 + X_5^2}$.

Note that $(X_i - 0)^2 / \sqrt{\sigma^2} \sim \mathcal{X}^2_1$ for $i=1,2,3,4,5$. Since $X_1,X_2,\cdots,X_5$ are independent, then $X_1^2,X_2^2,\cdots,X_5^2$ are, too. A linear combination of chi-squared distributions is also a chi-squared distribution:

$(X_3^2+X_4^2+X_5^2)/(\sigma^2) \sim \mathcal{X}_3^2$

So, we have

$\frac{(X_1-X_2)/\sqrt{2\sigma^2}}{\sqrt{(X_3^2+X_4^2+X_5^2)/(3\cdot\sigma^2)}} \sim t_3$

(Note that the $3$ in the denominator is the degree of freedom for the $\mathcal{X}_3^2$ distribution.)

After some algebraic manipulation, you will get that $c=1/\sigma^2 \cdot \sqrt{3/2}$

Answer 1

A distribution $W$ is a t distribution iff $W=U/\sqrt{V/p}$, where $U\sim N(0,1)$ and $V \sim \mathcal{X}^2_p$ with $p$ being the degrees of freedom.

$(X_1-X_2)/\sqrt{2\sigma^2} \sim N(0,1)$, which partially takes care of the numerator in $c(X_1-X_2)/ \sqrt{X_3^2 +X_4^2 + X_5^2}$.

Note that $(X_i - 0)^2 / \sqrt{\sigma^2} \sim \mathcal{X}^2_1$ for $i=1,2,3,4,5$. Since $X_1,X_2,\cdots,X_5$ are independent, then $X_1^2,X_2^2,\cdots,X_5^2$ are, too. A linear combination of chi-squared distributions is also a chi-squared distribution:

$(X_3^2+X_4^2+X_5^2)/(\sigma^2) \sim \mathcal{X}_3^2$

So, we have

$\frac{(X_1-X_2)/\sqrt{2\sigma^2}}{\sqrt{(X_3^2+X_4^2+X_5^2)/(3\cdot\sigma^2)}} \sim t_3$

(Note that the $3$ in the denominator is the degree of freedom for the $\mathcal{X}_3^2$ distribution.)

After some algebraic manipulation, you will get that $c=\sqrt{3/2}$