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Assume that we have a set of n data samples from a Normal distribution with unknown mean m and unknown standard deviation s. We would like to estimate the mean together with the appropriate level of uncertainty. A Normal distribution can have a mean anywhere in [-∞, +∞], so we could use a Uniform improper prior p(m) = k. The uninformed prior for the standard deviation should be p(s) = 1/s to ensure invariance under a linear transformation. The likelihood function is given by the Normal distribution density function:

 


i(X|\mu ,\sigma)=\frac{1}{(2\pi \sigma ^{2})^{n/2}}exp\Big(-\frac{1}{2 \sigma ^{2}}\displaystyle\sum_{i-1}^{n}(x_{i}-\mu)^{2}\Big)


 

Multiplying the priors together with the likelihood function and integrating over all possible values of s, we arrive at the posterior distribution for m:

 

f(\mu)\propto[(n-1)\hat{\sigma} ^{2}+n(\bar{x}-\mu)^{2}]^{-n/2}

 

      

f(\mu)\propto \Big[1+\frac{n(\bar{x} -\mu)^{2}}{(n-1)\hat{\sigma }^{2}}\Big]^{-n/2}

                            

                                                                                 (1)

 

 

where x and  are the mean and sample standard deviation of the data values. The Student-t distribution with n degrees of freedom has probability density:

 

       

f(y)\propto\Big[1+\frac{y^{2}}{\nu}\Big]^{-\big(\frac{\nu+1}{2}\big)}

                                  

                                                                                  (2)

 

Equation 1 and 2 are the same functions if we set n = n – 1 and   y=\frac{\sqrt{n}(\bar{x}-\mu )}{\hat{\sigma}}.  

 

Thus:

 

\mu =t(n-1)\frac{\hat{\sigma} }{\sqrt{n}}+\bar{x}

 

where t(n-1) represents the Student-t (0,1,n-1) distribution with (n-1) degrees of freedom. Therefore, with Crystal Ball 7.0+, we can make the following model as also shown in the model Estimate_mean_and_stdev_for_Normal_distribution_when_neither_known

 

m = Student (x, s/SQRT(n), n-1)

 

This is the same result used in classical statistics.

 

 

 


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