Suppose that ${\bf y}\sim N_4({\mu},\sigma^2I)$, and $\sum^4_{i=1}\mu_i=0$. Find the distribution of $$\frac{2(y_1-y_3)^2}{(\sum^4_{i=1}y_i)^2}$$
I guess there missed something in the numerator, since if it is $$\frac{2[(y_1-y_3)-(\mu_1-\mu_3)]^2}{(\sum^4_{i=1}y_i)^2}$$ I can figure out it is $F_{1,1}$ distribution, because then the ratio is just the ratio of two independent (the independence can be done by checking the covariance) standard normal square.
Any comment? Thanks~