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Assume $X_1$ and $X_2 \sim N(0,1)$ independent and identically distributed random variables. Determine the distribution of $T_3=\frac{(X_2+X_1)^2}{\sqrt{(X_2-X_1)^2}}$

We know $(\frac{X_2+X_1} {\sqrt{2}})^2 \sim \chi^2_{(1)}$, the same for $(\frac{X_2-X_1} {\sqrt{2}})^2 \sim \chi^2_{(1)}$.

My problem is that I don´t know how to handle the square root, I think the distribution looks like a Student´s T or an F one. How can I proceed?

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    $\begingroup$ This can't be a $t$-distribution since $T_3 \ge 0$. $\endgroup$ – angryavian Oct 6 '18 at 2:44
  • $\begingroup$ For any $k>0$, the probability that $T_3>k$ equals the probability that $|X_2+X_1|^2 > k|X_1-X_2|$. If $(a,b)$ is a random point in the square $[0,1]\times[0,1]$, what is area of the region such that $(a+b)^2>k|a-b|$? Differentiate such thing with respect to $k$ and you'll essentially have your distribution. $\endgroup$ – Jack D'Aurizio Oct 6 '18 at 2:45
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    $\begingroup$ Try deriving the distribution of $\frac{U^2}{|V|}$ where $(U,V)\stackrel{\text{ i.i.d}}{\sim}N(0,2)$. $\endgroup$ – StubbornAtom Oct 6 '18 at 5:13

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