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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 $c=1/\sigma^2 \cdot \sqrt{3/2}$

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1 Answer 1

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$\begingroup$

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 $c=\sqrt{3/2}$

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    $\begingroup$ Are you sure about the $\sigma^2$ term on $c$ there? $\endgroup$
    – Glen_b
    Nov 18, 2019 at 1:17
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    $\begingroup$ Check a little more carefully. $\endgroup$
    – Glen_b
    Nov 18, 2019 at 2:05
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    $\begingroup$ The expression in question clearly is invariant under a change of scale and therefore cannot depend on $|\sigma|$ in any way. That insight might help you check your work and identify where the mistake occurs. It also shows why if there is any correct answer, there must be two of them. $\endgroup$
    – whuber
    Nov 18, 2019 at 17:42
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    $\begingroup$ Consider whuber's statement. And then pay careful attention to your handling of $\sigma^2$ terms. Justify every step. $\endgroup$
    – Glen_b
    Nov 18, 2019 at 23:46
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    $\begingroup$ @whuber thank you for your comment. I realize now that $c=\sqrt{3/2}$ $\endgroup$
    – Ron Snow
    Nov 19, 2019 at 2:09

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