91影视

It is known that the time in minutes, X, required to serve a customers at a certain facility is exponentially distributed with unknown parameter\(\theta \). It is known that prior distribution of \(\theta \) is gamma with mean \({\mu _0} = 2\) and standard deviation\({\nu _0} = 1\) . The average time required to serve a random sample of \(n = 20\) customers is \({\overline x _n} = 3.8\) minutes.

Consider that prior distribution of \(\theta \) is gamma with parameters \(\alpha \) and \(\beta \) . Thus, we understand that mean of the prior distribution is:

\(\begin{array}{c}{\mu _0} = \frac{\alpha }{\beta }\\0.2 = \frac{\alpha }{\beta }\\\alpha = 0.2\beta \end{array}\)

鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹��. (1)

Similarly, standard deviation of prior distribution is:

\(\begin{array}{c}\sigma = \frac{{\sqrt \alpha }}{\beta }\\1 = \frac{{\sqrt \alpha }}{\beta }\\\alpha = {\beta ^2}\end{array}\)

鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹��. (2)

Substitute equation (1) in equation (2) and solve for\(\alpha \)and\(\beta \)

\(\begin{array}{c}0.2\beta = {\beta ^2}\\\beta = 0.2\end{array}\)

Substitute in equation (1)

\(\begin{array}{c}0.2 = \frac{\alpha }{{0.2}}\\\alpha = 0.04\end{array}\)

Now, the average time required to serve a random sample of\(n = 20\)customers is\({\overline x _n} = 3.8\)minutes.

So,

\({\overline x _n} = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\)

\(\begin{array}{c}\sum\limits_{i = 1}^n {{x_i} = {{\overline x }_n}} \times n\\ = 20 \times 3.8\\ = 76\end{array}\)

Due to conjugate pairs of posterior and prior distributions, the posterior distribution of\(\theta \),\(\xi \left( {\theta |{x_1},{x_2},...{x_n}} \right)\), is also gamma with parameters\(\alpha + n\)and\(\beta + \sum\limits_{i = 1}^n {{x_i}} \).

So, mean of posterior distribution is:

\({\mu _1} = \frac{{\alpha + n}}{{\beta + \sum\limits_{i = 1}^n {{x_i}} }}\)

\(\begin{array}{l} = \frac{{20 + 0.04}}{{0.2 + 76}}\\ = \frac{{20.04}}{{76.20}}\\ = 0.263\end{array}\)

The Bayes estimator of\(\theta \)when squared error loss function is used is 0.263.