Answer to Question #181937 in Statistics and Probability for ANJU JAYACHANDRAN

Question

If a random variable u has t -distribution with n degree of freedom, find the 

distribution of u^2.

Answer 1

Let s shows the sample standard deviation, $\mu$ shows the populaiton mean, $\bar{x}$ shows the sample mean, $\sigma$ is population standard deviation. and n+1 is sample size.

Chi sqaure distribution with (n-1) degree of freedom will be

$\chi^{2}_{(n-1)}\sim \frac{s^{2}(n-1)}{\sigma^{2}}$

Here we will use following relationships between the variables:

Relation between standard normal distribution Z and chi-sqaure distribution:

$Z^{2}\sim \chi^{2}_{1}$

Relation between F distribution and chi-sqaure distribution:

$F_{a,b}\equiv \frac{\chi^{2}_{a}/a}{\chi^{2}_{b}/b}$

The t-statistics with n degree of freedom will be

$t_{n}=\frac{\bar{x}-\mu}{s/\sqrt{n+1}}=\frac{\bar{x}-\mu}{\sigma/\sqrt{n+1}}\cdot \frac{\sigma}{s}=Z\cdot \frac{\sigma}{s}=\frac{Z}{\sqrt{\frac{s^{2}}{\sigma^{2}}}}=\frac{Z}{\sqrt{\frac{s^{2}(n)}{\sigma^{2}(n)}}}=\frac{Z}{\sqrt{\frac{\chi^{2}_{n}}{n}}}$

 $=\frac{\sqrt{\frac{\chi^{2}_{1}}{1}}}{\sqrt{\frac{\chi^{2}_{n}}{n}}}$

Now squaring both sides gives:

$U=t^{2}_{n}=\frac{\frac{\chi^{2}_{1}}{1}}{\frac{\chi^{2}_{n}}{n}}\sim F_{1,n}$

Hence, $u^2$ has F distribution.