91Ó°ÊÓ

Show that expected value is\(\mu \)using independence and the fact that

\(E\left( {{{\bar X}_i}} \right) = \mu ,\;\;\;i = 1,2, \ldots ,\)

as follows

\(\begin{array}{l}E\left( {{W_t}} \right) = E\left( {\alpha {{\bar X}_t} + \alpha (1 - \alpha ){{\bar X}_{t11}} + \ldots + \alpha {{(1 - \alpha )}^{t - 1}}{{\bar X}_1} + {{(1 - \alpha )}^t}\mu } \right)\\ = \alpha E\left( {{{\bar X}_t}} \right) + \alpha (1 - \alpha )E\left( {{{\bar X}_{t - 1}}} \right) + \ldots + \alpha {(1 - \alpha )^{t - 1}}E\left( {{{\bar X}_1}} \right)\\ + {(1 - \alpha )^t}\mu \\ = \mu \left( {\alpha + \alpha (1 - \alpha ) + \ldots + \alpha {{(1 - \alpha )}^{t - 1}} + {{(1 - \alpha )}^t}} \right)\\ = \mu \left( {\alpha \left( {1 + (1 - \alpha ) + \ldots + {{(1 - \alpha )}^{t - 1}}} \right) + {{(1 - \alpha )}^t}} \right)\\ = \mu \left( {\alpha \sum\limits_{i = 0}^\infty {{{(1 - \alpha )}^i}} - \alpha \sum\limits_{i = t}^\infty {{{(1 - \alpha )}^i}} + {{(1 - \alpha )}^t}} \right)\\ = \mu \left( {\frac{\alpha }{{1 - (1 - \alpha )}} - \alpha {{(1 - \alpha )}^t} \cdot \frac{1}{{1 - (1 - \alpha )}} + {{(1 - \alpha )}^t}} \right)\\ = \mu \end{array}\)

Where the geometric series were used.

Here, use mentioned independence to obtain the following

\(\begin{array}{l}V\left( {{W_t}} \right) = V\left( {\alpha {{\bar X}_t} + \alpha (1 - \alpha ){{\bar X}_{t1}} + \ldots + \alpha {{(1 - \alpha )}^{t - 1}}{{\bar X}_1} + {{(1 - \alpha )}^t}\mu } \right)\\ = {\alpha ^2}V\left( {{{\bar X}_t}} \right) + {\alpha ^2}{(1 - \alpha )^2}V\left( {{{\bar X}_t}1} \right) + \ldots + {\alpha ^2}{(1 - \alpha )^{2(t - 1)}}V\left( {{{\bar X}_1}} \right) + {(1 - \alpha )^t}\mu \\ = {\alpha ^2}V\left( {{{\bar X}_1}} \right) \times \left( {1 + {{(1 - \alpha )}^2} + \ldots + {{(1 - \alpha )}^{2(t - 1)}}} \right)\\ = {\alpha ^2}\frac{{1 - {{\left( {{{(1 - \alpha )}^2}} \right)}^t}}}{{1 - {{(1 - \alpha )}^2}}} \times \frac{{{\sigma ^2}}}{n} = \frac{{\alpha \left( {1 - {{(1 - \alpha )}^{2t}}} \right)}}{{2 - \alpha }} \times \frac{{{\sigma ^2}}}{n}\end{array}\)

The formula of how to compute\(w_i^\prime \)s is given in the exercise; however,\(\alpha \)is not specified. From the data in the mentioned example, one could use\(\bar s\)or, e.g.\(\sigma = 0.5\). The value\({\mu _0} = 40\)is given in the exercise. Assume that\(\alpha = 0.6\)(one could use any, it is not specified). Compute all control limits, and\(\sigma _i^2\)s using formula in\((b)\). Start with\(w_i^\prime s\)

\(\begin{array}{l}{w_0} = {\mu _0} = 40;\\{w_1} = 0.6{{\bar x}_1} + 0.4{w_0} = 0.6 \cdot 40.20 + 0.4 \cdot 40 = 40.12;\\{w_2} = 0.6{{\bar x}_2} + 0.4{w_1} = 0.6 \cdot 39.72 + 0.4 \cdot 40.12 = 39.88;\\ \vdots \\{w_{16}} = 40.29.\end{array}\)

The values of\({w_i}\)are, respectively,

\(40,39.88,40.20,40.07,40.06,39.88,39.74,40.14,40.25,40.00,40.29,40.36,40.51,40.19,40.21,40.29\)

Using formula

\(\sigma _i^2 = \frac{{\alpha \left( {1 - {{(1 - \alpha )}^{2t}}} \right)}}{{2 - \alpha }} \cdot \frac{{{\sigma ^2}}}{n}\)

the values of\({\sigma _i}\)'s are square root of the value above; thus

\(\begin{array}{l}\sigma _1^2 = \frac{{0.6 \cdot \left( {1 - {{(1 - 0.6)}^2}} \right)}}{{2 - 0.6}} \cdot \frac{{0.25}}{4} = 0.0225\\\sigma _2^2 = \frac{{0.6 \cdot \left( {1 - {{(1 - 0.6)}^4}} \right)}}{{2 - 0.6}} \cdot \frac{{0.25}}{4} = 0.0261\\ \vdots \\\sigma _5^2 = \sigma _6^2 = \ldots = \sigma _{16}^2\end{array}\)

which results in\({\sigma _i}\)'s, respectively

\(\begin{array}{l}{\sigma _1} = 0.1500\\{\sigma _2} = 0.1616\\{\sigma _3} = 0.1633\\{\sigma _4} = 0.1636\\{\sigma _5} = \sigma _6^2 = \ldots = \sigma _{16}^2 = 0.1636\end{array}\)

The control limits for\(t = 1\)are

\(\begin{array}{l}LCL = 40 - 3 \times 0.1500 = 39.55\\UCL = 40 + 3 \times 0.1500 = 40.45,\end{array}\)

The control limits for\(t = 2\)are

\(\begin{array}{l}LCL = 40 - 3 \times 0.1616 = 39.52\\UCL = 40 + 3 \times 0.1616 = 40.48\end{array}\)

The control limits for\(t = 3,4, \ldots ,16\)are

\(\begin{array}{l}LCL = 40 - 3 \times 0.1636 = 39.51\\UCL = 40 + 3 \times 0.1636 = 40.49\end{array}\)

Compare\({w_i}\)with corresponding\(LCL\)and\(UCL\)and see if the value is between those values. There is an out-of-control limit which is\({w_{13}}\)because

\(UCL = 40.49 < 40.51 = {w_{13}}\)

a. Use properties of expectation; b. Use properties of variance; c. An out-of-control limit \({w_{13}}\).