Reply With QuoteIf you extend the Fisher-Neyman factorization theorem to two parameters you can obtain:Originally Posted by jareddm
and
as the joint sufficient statistics. And, thus, the single-valued functions are also joint sufficient statistics, namely,
and
.
I'm trying to work out the sufficient statistic for a normal distribution. I've figured out the statistic for when mu is unknown as that's simply x-bar. But when I thought that just S^2 was the sufficient statistic for sigma^2, I was told that was incorrect. Additionally, when both are unknown it's not just the two sufficient statistics of the constituents like it is with a Gamma distribution.
I know I'm missing something here but I'm not sure what...
If you extend the Fisher-Neyman factorization theorem to two parameters you can obtain:
as the joint sufficient statistics. And, thus, the single-valued functions are also joint sufficient statistics, namely,
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