Let $X$ and $Y$ be independent normal distribution i.e $X \sim N(\mu,\sigma^{2})$, $Y \sim N(\mu,\sigma^{2})$. Show that $\frac{X+Y}{\left\vert X-Y \right\vert}$ has a non-central t-distrbution and what are the parmaters ?
The first things that I have done are finding the distribution of $W_{1}=X+Y$ and $W_{2}=X-Y$.
$W_{1} \sim N(2\mu, 2\sigma^{2})$ and $W_{2}\sim N(0, 2\sigma^{2})$, but I am not sure where to go from there. Thank you in advance.
We have
$$\frac {X + Y} {|X - Y|} = \frac {\frac {X + Y -2 \mu} {\sigma \sqrt 2} + \frac {\mu \sqrt 2} \sigma} {\sqrt {\left( \frac {X - Y} {\sigma \sqrt 2} \right)^2} }, \\ \begin{pmatrix} \frac {X + Y -2 \mu} {\sigma \sqrt 2} \\ \frac {X - Y} {\sigma \sqrt 2} \end{pmatrix} \sim \mathcal N \!\left( 0, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right).$$
Therefore, $((X - Y)/(\sigma \sqrt 2))^2$ follows the chi-squared distribution with one degree of freedom and we have the definition of the noncentral t-distribution with $\nu_t = 1, \,\mu_t = \mu \sqrt 2/\sigma$. The independence requirement holds because uncorrelated components of a jointly normal distribution are independent.