Question

Use the data in the Table \({\bf{11}}{\bf{.5}}\) on page \({\bf{699}}\) suppose that \({{\bf{y}}_{\bf{i}}}\) is the logarithm of pressure \({x_i}\)and is the boiling point for the I the observation \({\bf{i = 1,}}...{\bf{,17}}{\bf{.}}\) Use the robust regression scheme described in Exercise \({\bf{8}}\) to \({\bf{a = 5, b = 0}}{\bf{.1}}\,\,{\bf{and f = 0}}{\bf{.1}}{\bf{.}}\) Estimate the posterior means and standard deviations of the parameter \({{\bf{\beta }}_{\bf{0}}}{\bf{,}}{{\bf{\beta }}_{\bf{1}}}\,\) and n.

Short answer

Estimate the posterior means and standard deviations of the parameter.

Use the following values of a, b and f to obtain N samples.

\(a = 5,{\rm{ }}b = {\rm{ }}0.1,{\rm{ }}f = 0.1.\)

The estimate of the posterior means of\({\beta _0}\) is\( - 0.953\) ; of\({\beta _1}\,\) is\(0.021\) ; of v is \(0.0000113.\)

The estimate of the standard deviations of \({\beta _0}\)is \(0.001507\); of \({\beta _1}\,\)is \(0.00007395\); of v is \(0.000006724.\)

Step-by-step solution

Definition of the conditional distribution

A conditional distributionis a probabilistic distribution that only pertains to a portion of a population. In other words, it indicates the likelihood that a randomly selected item in a subpopulation possesses a characteristic of interest to you.

The data consists of \(17\) observations of boiling points and pressure.

The procedure of obtaining samples - the robust regression scheme is given in the previous exercise.

Use the following values of a, b and f to obtain N samples.

\(a = 5,{\rm{ }}b = {\rm{ }}0.1,{\rm{ }}f = 0.1.\)

The conditional distributions used for simulations are

For V is gamma distribution with parameters

\(\frac{{na + b}}{2}\,\,and\,\,\frac{1}{2}\left( {f + a\sum\nolimits_{i = 1}^n {{\tau _i}} } \right).\)

For \({\tau _i}\)is gamma distribution with parameters

\(\frac{{a + 1}}{2}\,\,and\,\,\frac{1}{2}\,\left( {av + {{\left( {{y_i} - {\beta _0} - {\beta _1}{x_i}} \right)}^2}} \right).\)

For\({\beta _0}\) is the normal distribution with parameters

\(mean = \frac{1}{{\sum\nolimits_{i = 1}^n {{\tau _i}} }}\,\sum\nolimits_{i = 1}^n {{\tau _i}} \left( {{y_i} - {\beta _1}{x_i}} \right)\)

\(precision = \sum\nolimits_{i = 1}^n {{\tau _i}} \)

Distribution parameter

A distribution's parametersare variables included in the density function that allows the distribution to be tailored to various situations.

For \({\beta _1}\,\)is the normal distribution with parameters

\(\begin{aligned}{l}mean &= \frac{1}{{\sum\nolimits_{i = 1}^n {{\tau _i}{x^2}_i} }}\,\sum\nolimits_{i = 1}^n {{\tau _i}{x_i}} \left( {{y_i} - {\beta _1}{x_i}} \right)\\precision &= \sum\nolimits_{i = 1}^n {{\tau _i}} {x^2}_i\end{aligned}\)

The estimate of the posterior mean of\({\beta _0}\) using \(N = 15000\)sample is\( - 0.953\) ; of\({\beta _1}\,\) is\(0.021\) ; and of v is \(0.0000113.\)

The estimate of the standard deviations of \({\beta _0}\)using \(N = 15000\)sample is \(0.001507\); of \({\beta _1}\,\)is \(0.00007395\); and of v is \(0.000006724.\)

Hence,

The estimate of the posterior mean of\({\beta _0}\) is\( - 0.953\) ; of\({\beta _1}\,\) is\(0.021\) ; of v is \(0.0000113.\)

The estimate of the standard deviations of \({\beta _0}\)is \(0.001507\); of \({\beta _1}\,\)is \(0.00007395\); of v is \(0.000006724.\)