50% confidence interval for mean: 4.07 to 6.46, stddev: 2.15 to 4.74
90% confidence interval for mean: 0.98 to 9.55, stddev: 1.46 to 11.20
If there’s only one sample, the calculation fails due to division by n-1, so the frequentist says “no answer”. The Bayesian says the same if he used the improper prior Cyan mentioned.
The prior for variance that matches the frequentist conclusion isn’t flat. And even if it were, a flat prior for variance implies a non-flat prior for standard deviation and vice versa. :-)
Using the flat improper prior I was talking about before, when there’s only one data point the posterior distribution is improper, so the Bayesian answer is the same as the frequentist’s.
http://www.xuru.org/st/DS.asp
50% confidence interval for mean: 4.07 to 6.46, stddev: 2.15 to 4.74
90% confidence interval for mean: 0.98 to 9.55, stddev: 1.46 to 11.20
If there’s only one sample, the calculation fails due to division by n-1, so the frequentist says “no answer”. The Bayesian says the same if he used the improper prior Cyan mentioned.
Hm, should I understand it that the frequentist assumes normal distribution of the mean value with peak at the estimated 5.26?
If so, then frequentism = bayes + flat prior.
Improper priors are however quite tricky, they may lead to paradoxes such as the two-envelope paradox.
The prior for variance that matches the frequentist conclusion isn’t flat. And even if it were, a flat prior for variance implies a non-flat prior for standard deviation and vice versa. :-)
Of course, I meant flat distribution of the mean. The variance cannot be negative at least.
In this problem, yes. In the general case no one knows exactly what the flat prior is, e.g. if there are constraints on model parameters.
Using the flat improper prior I was talking about before, when there’s only one data point the posterior distribution is improper, so the Bayesian answer is the same as the frequentist’s.