What is the distribution of the ratio of sums of squared normal random variables? Given that $Y_1, \ldots, Y_6$ are independent normal random variables with mean $0$ and variance $\sigma^2$. Find the distribution of the following
$ T=\displaystyle\frac{Y^2_{1}+Y^2_{2}+Y^2_{3}}{Y^2_{4}+Y^2_{5}+Y^2_{6}} $
 A: To close this one:
We can write
$$T= \frac {\sigma^2}{\sigma^2} \cdot\frac{Y^2_{1}+Y^2_{2}+Y^2_{3}}{Y^2_{4}+Y^2_{5}+Y^2_{6}} = \frac{(Y_{1}/\sigma)^2+(Y_{2}/\sigma)^2+(Y_{3}/\sigma)^2}{(Y_{4}/\sigma)^2+(Y_{5}/\sigma)^2+(Y_{6}/\sigma)^2}$$
All these are now independent squared standard normals. The sum of such is a chi-square random variable with degrees of freedom equal to the number of the squared standard normals summed. So both the numerator and the denominator are chi-squares each with $3$ degrees of freedom.  
So (abusing notation)
$$T= \frac {\mathcal \chi^2_3}{\mathcal \chi^2_3} = \frac {\mathcal \chi^2_3/3}{\mathcal \chi^2_3/3}$$
The ratio of independent, as is the case here, chi-squares, each divided by its degrees of freedom, is an $F$-random variable that inherits the degrees of freedom, as its own parameters. So 
$$T \sim F(d_1=3,d_2=3)$$
Note that for "denominator" ($d_2$) degrees of freedom equal or lower than $4$, as is the case here, the variance of the distribution is not defined, and one has to use other measures of dispersion.
