Question

In the situation of Exercise 9, suppose that a prior distribution is used forθwith p.d.f.ξ(θ)=0.1 exp(−0.1θ)forθ >0. (This is the exponential distribution with parameter 0.1.)

1. Prove that the posterior p.d.f. ofθgiven the data observed in Exercise 9 is

\({\bf{\xi }}\left( {{\bf{\theta }}\left| {\bf{x}} \right.} \right){\bf{ = }}\left\{ \begin{align}{l}{\bf{4}}{\bf{.122exp}}\left( {{\bf{ - 0}}{\bf{.1\theta }}} \right)\;{\bf{if}}\;{\bf{4}}{\bf{.8 < \theta < 5}}{\bf{.2}}\\{\bf{0}}\;{\bf{otherwise}}\end{align} \right.\)

2. Calculate the posterior probability \(\left| {{\bf{\theta - }}{{{\bf{\bar X}}}_{\bf{2}}}} \right|{\bf{ < 0}}{\bf{.1}}\), which \({{\bf{\bar X}}_{\bf{2}}}\)is the observed average of the datavalues.
3. Calculate the posterior probability thatθis in the confidence interval found in part (a) of Exercise 9.
4. Can you explain why the answer to part (b) is so close to the answer to part (e) of Exercise 9? Hint:Compare the posterior p.d.f. in part (a) to the function in Eq. (8.5.15).

Short answer

1. Proved
2. The posterior probability is 0.5
3. The posterior probability is 1.
4. Explained.

Step-by-step solution

Given information

In exercise 9, the random variable is observed from a uniform distribution on the interval \(\left( {\theta - \frac{1}{2},\theta + \frac{1}{2}} \right)\;,\;\left( { - \infty < \theta < \infty } \right)\). A prior distribution is used \(\theta \) with the probability density function \(\xi \left( \theta \right) = 0.1\exp \left( { - 0.1\theta } \right)\;for\;\theta > 0\).

The consistent with the observed data values\(\theta \)are between\(\theta - \frac{1}{2}\;and\;\theta + \frac{1}{2}\)

Therefore,

\(\begin{align}\left( {\theta - \frac{1}{2}} \right) &= 5.3 - \frac{1}{2}\\ &= 4.8\end{align}\)

And

\(\begin{align}\left( {\theta + \frac{1}{2}} \right) &= 4.7 + \frac{1}{2}\\ &= 5.2\end{align}\)

(a) Determine the posterior probability

Consider the prior distribution of \(\theta \).

\(\begin{align}\xi \left( \theta \right) = 0.1\exp \left( { - 0.1\theta } \right)\\\xi \left( \theta \right) \propto \exp \left( { - 0.1\theta } \right)\end{align}\)

So, the posterior probability distribution of\(\theta \)is,

\(\begin{align}\xi \left( {\theta \left| x \right.} \right) \propto \xi \left( {x\left| \theta \right.} \right)\xi \left( \theta \right)\\\xi \left( {\theta \left| x \right.} \right) \propto 1 \times \exp \left( { - 0.1\theta } \right)\\\xi \left( {\theta \left| x \right.} \right) \propto \exp \left( { - 0.1\theta } \right)\end{align}\)

Therefore, the posterior probability distribution of\(\theta \)is,

\(\xi \left( {\theta \left| x \right.} \right) = \left\{ \begin{align}{l}c\exp \left( { - 0.1\theta } \right)\;if\;4.8 < \theta < 5.2\\0\;otherwise\end{align} \right.\)

Now, consider the value of constant c,

\(\begin{align}\int_{4.8}^{5.2} {c\exp \left( { - 0.1\theta } \right) = 1} \\ \Rightarrow - 10c\left( {{e^{ - 0.1\theta }}} \right)_{4.8}^{5.2} = 1\\ \Rightarrow - 10c\left( {{e^{ - 0.52}} - {e^{ - 0.48}}} \right) = 1\\ \Rightarrow - 10c\left( { - 0.0243} \right) &= 1\\ \Rightarrow c = 4.122\end{align}\)

Thus, the posterior probability distribution is,

\(\xi \left( {\theta \left| x \right.} \right) = \left\{ \begin{align}{l}4.122\exp \left( { - 0.1\theta } \right)\;for\;4.8 < \theta < 5.2\\0\;otherwise\end{align} \right.\)

Hence, proved.

(b) Determine the posterior probability

Consider the value of \({\bar X_2}\)

\(\begin{align}{{\bar X}_2} &= \frac{{4.8 + 5.2}}{2}\\ &= 5\end{align}\)

So, the posterior probability that \(\left| {\theta - {{\bar X}_2}} \right| < 0.1\;\)is

\(\begin{align}\Pr \left( {\left| {\theta - 5} \right| < 0.1} \right) &= \int_{5 - 0.1}^{5 + 0.1} {4.122\exp \left( { - 0.1\theta } \right)} \\ &= - 41.22\left( {{e^{ - 0.1\theta }}} \right)_{4.9}^{5.1}\\ &= - 41.22\left( {{e^{ - 0.51}} - {e^{ - 0.49}}} \right)\\ &= - 41.22\left( { - 0.012} \right)\\ = 0.5\end{align}\)

Thus, the required probability is 0.5.

(c) Calculate the posterior probability

Consider the posterior probability that \(\theta \) is in the confidence interval (4.7,5.3). Now there is already concluded that the entire dataset of possible values of\(\theta \)is contained in the interval.

Therefore, the required posterior probability that \(\theta \) is in the confidence interval is 1.

(d) Compare the answers

The interval of possible values is consistent with the observed data values in part b of exercise 9. The experimental data values \({X_1} = 4.7\;and\;{X_2} = 5.3\) and the interval was (4.8,5.2). That is, the posterior probability distribution of\(\theta \)is constant over the break (4.8,5.2).

So, there can be concluded that for \(c \le 0.2\)

\(\begin{align}\Pr \left( {\left| {\theta - 5} \right| < c} \right) &= \int_{5 - c}^{5 + c} {4.122\exp \left( { - 0.1\theta } \right)} \\ &= - 41.22\left( {{e^{ - 0.1\theta }}} \right)_{4 + c}^{5 + c}\\ &= - 41.22{e^{ - 0.5}}\left( {{e^{ - 0.1c}} - {e^{ - 0.1c}}} \right)\\ &= 5c\end{align}\)

Now, the conditional probability that\({{\bf{\bar X}}_{\bf{2}}}\)is close to\({\bf{\theta }}\)given Z=z is,

\(\)\({\bf{Pr}}\left( {\left| {{{{\bf{\bar X}}}_{\bf{2}}}{\bf{ - \theta }}} \right|{\bf{ < c}}\left| {{\bf{Z = z}}} \right.} \right){\bf{ = }}\left\{ \begin{align}{l}\frac{{{\bf{2c}}}}{{{\bf{1 - z}}}}\;{\bf{if}}\;{\bf{c}} \le \frac{{{\bf{1 - z}}}}{{\bf{2}}}\\{\bf{1}}\;{\bf{if}}\;{\bf{c > }}\frac{{{\bf{1 - z}}}}{{\bf{2}}}\end{align} \right.\)

So, the conditional probability is

\(\begin{align}\Pr \left( {\left| {{{\bar X}_2} - 5} \right| < c\left| {Z = 0.6} \right.} \right) &= \frac{{2c}}{{1 - z}}\\ \Rightarrow 5c &= \frac{{2c}}{{1 - z}}\\ \Rightarrow 5 &= \frac{2}{{1 - 0.6}}\\ \Rightarrow 5 &= 5\end{align}\)

Thus, that’s why the answer of part b is close to the response of part e of exercise 9.