# Chi squared distribution

• May 4th 2008, 02:20 PM
Connemara Pony
Chi squared distribution
note that $Y = \frac{1}{\sigma^2} (X_i-\mu)^2$ has a chi-squared distribution of $n$ degrees of freedom since each term in the sum is a squared normal random variable and is independent of other random variables in the sum. Then show that $Y - \frac{(n-1)S^2}{\sigma^2} = \left(\frac{\bar{X} - \mu}{\sigma/ \sqrt{n}} \right)^2$. The RHS is a chi-squared random variable with one degree of freedom. Since df's add, this implies that $\frac{(n-1)S^2}{\sigma^2}$ is a chi-squared distribution with $n-1$ degrees of freedom provided that $\frac{(n-1)S^2}{\sigma^2}$ and $\left(\frac{\bar{X} - \mu}{\sigma/ \sqrt{n}} \right)^2$ are independent.