Deriving a statistic of the t distribution with 2 degrees of freedom

Question

Assume I have an independent $X_{1}, X_{2}, \ldots, X_{n}$ with $X_{i} \sim N(i,i^{2})$ and I want to find a statistic that has a t distribution with 2 degrees of freedom.

How would I go about showing that? I don't think t distribution is one that is related to the normal or the F, but I would like someone to help clarify the steps and methodology towards tackling problems like these.

Answer 1

How do you define $t_n$ distribution? By $Z/\sqrt{\chi^2_n /n}$ where $Z\sim N(0,1)$ indept of $\chi^2_n$.

Note, $X_i/i \overset{iid}{\sim} N(1,1)$ or

$Y_i=\frac{X_i}{i}-1 \overset{iid}{\sim} N(0,1)$. Consider $\frac{\sqrt{2} Y_1}{\sqrt{Y_2^2+Y_3^2}}$

Answer 2

Suppose $X_1,\ldots,X_n \sim\ \text{i.i.d.}\ N(\mu,\sigma^2)$ and $$ \bar X = \frac{X_1+\cdots+X_n} n $$ and $$ S^2 = \frac 1 {n-1} \left( (X_1 - \bar X)^2 + \cdots + (X_n - \bar X)^2 \right). $$ Then $$ \frac{ \bar X - \mu}{\sigma/\sqrt n} \sim N(0,1) $$ and $$ \frac{\bar X - \mu}{S/\sqrt n} \sim t_{n-1}. $$ If you want $2$ degrees of freedom, just use $X_1,X_2,X_3$.