Basu's theorem for normal sample mean and variance

Solution 1:

I hope I got your question correctly:

The way you explained is correct we show the joint pdf in the form of exponential family and suffice that the summation of Xi's is Complete and sufficient statistics for unknown u, however, in order to show this we need to fix our variance to some known arbitrary constant sigma not square.

And hence by Basu's theorem Mean(complete statistics) and S^2(Ancillary statistics) are independent for fixed sigma square. And it will also be true for all sigma square values as sigma square is positive, thus mean and variance are independent for all u and sigma square values.

mean/sigma is independent of S^2/sigma square. and mean/sigma follows standard normal N(0,1).

As sigma is not random, we have mean and S^2 are independent.

I didn't use latex hope I didn't confuse you further