Help for $U = T^2$ has an $F$ distribution with 1 numerator and $v$ denominator degrees of freedom. If $T$ has a $t$ distribution with $v$ degrees of freedom, then $U = T^2$ has an $F$ distribution with 1 numerator and $v$ denominator degrees of freedom.
First, I set $$T = \frac{Z}{\sqrt{W/v}}$$ where $T$ is t distribution with $v$ df, $W$ is a chi-squared distributed variable with $v$ df, and $Z$ has a standard normal distribution.
Then, I set $$T^2 = \frac{Z^2}{(W/v)}$$...
My question is that how do we know or show $Z$ and $W$ are independent?
Thanks a lot
 A: In this link (see Wikipedia) you can see that the random variables $Z$ and $W$ are independent by definition. That is, Student's t-distribution with $v$ degrees of freedom can be defined as the distribution of the random variable T with 
    $$T=\frac{Z}{\sqrt{W/v}}$$
where


*

*$Z$ is normally distributed with expected value $0$ and variance $1$, (i.e standard normal distribution)

*$W$ has a chi-squared distribution with $v$ degrees of freedom, and

*$Z$ and $W$ are independent.
So, this is established by definition and you do not have to show it.This is an assumption, so that $T$ has the $t-$distribution. Were they not independent then the above random variable $T$ would not have the $t-$distribution.
Another link that states the above fact can be found here. According to it, 
A random variable X has a standard Student's t distribution with $v$ degrees of freedom if it can be written as a ratio $$T=\frac{N(0,1)}{\sqrt{\chi^2_v/v}}$$between a standard normal random variable $Z\sim N(0,1)$ and ... an independent of $Z$ Chi-square random variable with $v$ degrees of freedom divided by $v$.
