Question

The study on acid concentration in cheese included a total of 30 lactic acid measurements, the 10 given in Example 8.5.4 on page 487 and the following additional 20:

1.68, 1.9, 1.06, 1.3, 1.52, 1.74, 1.16, 1.49, 1.63, 1.99, 1.15, 1.33, 1.44, 2.01, 1.31, 1.46, 1.72, 1.25, 1.08, 1.25.

a. Using the same prior as in Example 8.6.2 on page 498, compute the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) based on all 30 observations.

b. Use the posterior distribution found in Example 8.6.2 on page 498 as if it were the prior distribution before observing the 20 observations listed in this problem. Use these 20 new observations to find the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)and compare the result to the answer to part (a).

Short answer

1. \({\mu _1} = 1.427,{\lambda _1} = 31,{\alpha _1} = 15.5,{\beta _1} = 1.1405\)
2. \({\mu _1} = 1.872,{\lambda _1} = 41,{\alpha _1} = 82.5,{\beta _1} = 1.0839\)

Step-by-step solution

Given information

The 10 data points are: 0.86, 1.53, 1.57, 1.81, 0.99, 1.09, 1.29, 1.78, 1.29, 1.58.

The additional 20 data points are: 1.68, 1.9, 1.06, 1.3, 1.52, 1.74, 1.16, 1.49, 1.63, 1.99, 1.15, 1.33, 1.44, 2.01, 1.31, 1.46, 1.72, 1.25, 1.08, 1.25.

The prior in example 8.6.2 is \({\mu _0} = 1,{\lambda _0} = 1,{\alpha _0} = 0.5,{\beta _0} = 0.5\)

Define Normal-Gamma distribution

Let \(\mu \,\,{\rm{and}}\,\,\tau \) be random variables. Suppose that the conditional distribution of \(\mu \,\,{\rm{given}}\,\,\tau \) is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\tau \) .

Suppose also that the marginal distribution of \(\,\tau \) is the gamma distribution with parameters \({\alpha _0}\,\,{\rm{and}}\,\,\,{\beta _0}\).

Then we say that the joint distribution of\(\mu \,\,{\rm{and}}\,\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

(a) Define the posterior variables 

We know that,

\(\begin{align}{\mu _1} &= \frac{{{\lambda _0}{\mu _0} + n\overline {{x_n}} }}{{{\lambda _0} + n}}\\{\lambda _1} &= {\lambda _0} + n\\{\alpha _1} &= {\alpha _0} + \frac{n}{2}\\{\beta _1} &= {\beta _0} + \frac{{{s_n}^2}}{2} + \frac{{n{\lambda _0}{{\left( {\overline {{x_n}} - {\mu _0}} \right)}^2}}}{{2\left( {{\lambda _0} + n} \right)}}\end{align}\)

Also, the values of:

\(\begin{align}\overline {{x_n}} &= \frac{{0.86 + 1.53 + 1.57 + {\rm{ }} \ldots + 1.25 + {\rm{ }}1.08 + {\rm{ }}1.25}}{{30}}\\ &= 1.442\end{align}\)

\(\begin{align}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {\left( {{x_i} - \overline {{x_n}} } \right)} }}{{n - 1}}\\ &= \frac{{2.67108}}{{29}}\\ &= 0.0921\end{align}\)

Substitute the values

\(\begin{align}{\mu _1} &= \frac{{1 \times 1 + 30 \times 1.442}}{{1 + 30}}\\ &= 1.427\\{\lambda _1} &= 1 + 30\\ &= 31\\{\alpha _1} &= 0.5 + \frac{{30}}{2}\\ &= 15.5\\{\beta _1} &= 1 + \frac{{0.0921}}{2} + \frac{{30 \times 1 \times {{\left( {1.442 - 1} \right)}^2}}}{{2\left( {1 + 30} \right)}}\\ &= 1.1405\end{align}\)

Therefore, the answer is: \({\mu _1} = 1.427,{\lambda _1} = 31,{\alpha _1} = 15.5,{\beta _1} = 1.1405\)

(b) Changing the prior distribution and recomputing the posterior distribution

Now,\({\mu _0} = 1.345,{\lambda _0} = 11,{\alpha _0} = 5.5,{\beta _0} = 1.0484\), \(\overline {{x_n}} = 1.442\), \({s_n}^2 = 0.0921\).

\(\begin{align}{\mu _1} &= \frac{{11 \times 1.345 + 30 \times 1.442}}{{1 + 30}}\\ &= 1.872\\{\lambda _1} &= 11 + 30\\ = 41\end{align}\)

\(\begin{align}{\alpha _1} &= 5.5 + \frac{{30}}{2}\\ &= 82.5\\{\beta _1} &= 1 + \frac{{0.0921}}{2} + \frac{{30 \times 11 \times {{\left( {1.442 - 1.345} \right)}^2}}}{{2\left( {11 + 30} \right)}}\\ &= 1.0839\end{align}\)

Therefore, the answer is: \({\mu _1} = 1.872,{\lambda _1} = 41,{\alpha _1} = 82.5,{\beta _1} = 1.0839\)