91Ó°ÊÓ

A random variable\({\rm{X}}\)with\({\rm{pmf}}\)–

\({\rm{p(x;\mu ) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

For\({\rm{x = 0,1,}}..{\rm{,}}\)is said to have Poisson Distribution with parameter\({\rm{\mu > 0}}\).

Notice first that there are total of\({\rm{5}}\)copies of the photography magazine! Because of that, the number purchased magazines cannot be bigger than\({\rm{5}}\). This indicates that we should look at random variable –

\({\rm{Y = min(5,X)}}\)

This random variable represents the number of sold copies, therefore we need to find its expectation. Rando Variable\({\rm{Y}}\)can take values from\(0\)up to\({\rm{5}}\)and it is a discrete random variable. However, we are going to use the following two propositions.

The Expected Value (mean value) of a discrete random variable\({\rm{X}}\)with set of possible values\({\rm{S}}\)and\({\rm{pmf p(x)}}\)is –

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {\rm{x}} \cdot {\rm{p(x)}}\)

The Expected Value (mean value) of any function\({\rm{g(X)}}\), where\({\rm{X}}\)is a discrete random variable\({\rm{X}}\)with set of possible values\({\rm{S}}\)and\({\rm{pmf p(x)}}\), denoted as\({\rm{E(g(X))(}}{{\rm{\mu }}_{{\rm{g(X)}}}}{\rm{),}}\)is –

\({\rm{E(g(X)) = }}{{\rm{\mu }}_{{\rm{g(X)}}}}{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {\rm{g}} {\rm{(x)}} \cdot {\rm{p(x)}}\)

The \({\rm{pmf}}\) of \({\rm{X}}\) is \({\rm{p(x;\mu ) = p(x;4)}}\) and \({\rm{g(X) = min(5,X)}}\).

Hence, the following is true –

\(\begin{aligned} E(Y) &= E(min(5,X)) = \sum\limits_{{\rm{x = 0}}}^\infty {{\rm{min}}} {\rm{(5,x)}} \cdot {\rm{p(x;4)}}\\\ &= {\rm{0}} \cdot {\rm{p(0;4) + 1}} \cdot {\rm{p(1;4) + 2}} \cdot {\rm{p(2;4) + 3}} \cdot {\rm{p(3;4) + 4}} \cdot {\rm{p(4;4) + }}\sum\limits_{{\rm{x = 5}}}^\infty {\rm{5}} \cdot {\rm{p(x;4)}}\\\ &= {\rm{1}}{\rm{.735 + 5 \times }}\left( {{\rm{1 - }}\sum\limits_{{\rm{x = 0}}}^{\rm{4}} {\rm{p}} {\rm{(x;4)}}} \right)\\&= 1 {\rm{.735 + 5}} \cdot {\rm{(1 - F(4;4))}}\\\ &= 3.59\end{aligned}\)

\({\rm{(1)}}\): for\({\rm{x}} \in {\rm{\{ 0,1,2,3,4\} }}\), the minimum of numbers\({\rm{5}}\)and\({\rm{x}}\)is obviously\({\rm{x}}\), and for\({\rm{x}} \in {\rm{\{ 5,6,}}..{\rm{\} }}\)the minimum is number\({\rm{5}}\);

\({\rm{(2)}}\): remember that\({\rm{pmf}}\)of\({\rm{X}}\)is –

\({\rm{p(x;4) = }}{{\rm{e}}^{{\rm{ - 4}}}}\frac{{{{\rm{4}}^{\rm{x}}}}}{{{\rm{x!}}}}{\rm{, x = 0,1,}}...{\rm{,}}\)

Which was used to compute values for\({\rm{x}} \in {\rm{\{ 0,1,2,3,4\} }}\);

\({\rm{(3)}}\): Appendix Table\({\rm{A}}{\rm{.2}}\)contains the Poisson cdf\({\rm{F(x;4)}}\).

Therefore, the value obtained is\({\rm{3}}{\rm{.59}}\).