Elianna Lawrence

2022-07-19

Assume I have an independent ${X}_{1},{X}_{2},\dots ,{X}_{n}$ with ${X}_{i}\sim N\left(i,{i}^{2}\right)$ and I want to find a statistic that has a t distribution with $2$ degrees of freedom.
How can we show this?

Do you have a similar question?

Jeroronryca

Expert

Suppose and
$\overline{X}=\frac{{X}_{1}+\cdots +{X}_{n}}{n}$
and
${S}^{2}=\frac{1}{n-1}\left(\left({X}_{1}-\overline{X}{\right)}^{2}+\cdots +\left({X}_{n}-\overline{X}{\right)}^{2}\right).$
Then
$\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\sim N\left(0,1\right)$
and
$\frac{\overline{X}-\mu }{S/\sqrt{n}}\sim {t}_{n-1}.$
If you want $2$ degrees of freedom, just use ${X}_{1},{X}_{2},{X}_{3}$.

Still Have Questions?

Faith Welch

Expert

How do you define ${t}_{n}$ distribution? By $Z/\sqrt{{\chi }_{n}^{2}/n}$ where $Z\sim N\left(0,1\right)$ indept of ${\chi }_{n}^{2}$.
Note, ${X}_{i}/i\stackrel{iid}{\sim }N\left(1,1\right)$ or
${Y}_{i}=\frac{{X}_{i}}{i}-1\stackrel{iid}{\sim }N\left(0,1\right)$ . Consider $\frac{\sqrt{2}{Y}_{1}}{\sqrt{{Y}_{2}^{2}+{Y}_{3}^{2}}}$

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