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\({\rm{\mu > 0}}\).

The two Poisson distributions are given, first with mean values\({{\rm{\mu }}_{\rm{1}}}\)and second with mean value\({{\rm{\mu }}_{\rm{2}}}\). Each of the CPA prepares the same number of forms.

Notice first that the given\({\rm{pmf}}\)is actually sum of the two given\({\rm{pmfs}}\)(for the Poisson Distributions) multiplied with\({\rm{0}}{\rm{.5}}\). The following is true –

\(\begin{array}{c}{\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{,}}{{\rm{\mu }}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.5}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{1}}}}}\frac{{{\rm{\mu }}_{\rm{1}}^{\rm{x}}}}{{{\rm{x!}}}}{\rm{ + 0}}{\rm{.5}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}}}\frac{{{\rm{\mu }}_{\rm{2}}^{\rm{x}}}}{{{\rm{x!}}}}\\{\rm{ = 0}}{\rm{.5}} \cdot {\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{1}}}} \right){\rm{ + 0}}{\rm{.5}} \cdot {\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{2}}}} \right)\end{array}\)

First check if\({\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{,}}{{\rm{\mu }}_{\rm{2}}}} \right) \ge {\rm{0}}\)holds for every\({\rm{x}} \in {{\rm{N}}_{\rm{0}}}\). Obviously, since\({\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{1}}}} \right)\)and\({\rm{p}}\left( {{\rm{x;}}{{\rm{\mu }}_{\rm{2}}}} \right)\)are two\({\rm{pmfs}}\), they are both non-negative, and the following is true –

Second, the sum of the probabilities needs to be equal to\({\rm{1}}\). Again, since there are given two\({\rm{pmfs}}\), the following holds –

Therefore, the given function is in fact a legitimate\({\rm{pmf}}\).

(b)

Proposition: For a random variable \({\rm{X}}\) with Poisson Distribution with parameter \({\rm{\mu > 0}}\), the following is true –

\({\rm{E(X) = V(X) = \mu }}\)

The following holds –

\(\begin{aligned} E(X) &= \sum\limits_{x = 0}^\infty x \cdot p\left( {x;{\mu _1},{\mu _2}} \right) = \sum\limits_{x = 0}^\infty x \left[ {0.5 \cdot p\left( {x;{\mu _1}} \right) + 0.5 \cdot p\left( {x;{\mu _2}} \right)} \right] \\ &= 0.5\sum\limits_{x = 0}^\infty x p\left( {x;{\mu _1}} \right) + 0.5\sum\limits_{x = 0}^\infty x p\left( {x;{\mu _2}} \right) \\ &= 0.5 \cdot E\left( {{Y_1}} \right) + 0.5 \cdot E\left( {{Y_2}} \right) \\ &= 0.5{\mu _1} + 0.5{\mu _2} \\ \end{aligned} \)

Because random variables \({{\rm{Y}}_{\rm{1}}}\), and \({{\rm{Y}}_{\rm{2}}}\) follow given Poisson Distributions.

Therefore, the expression is obtained as \({\rm{0}}{\rm{.5}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + 0}}{\rm{.5}}{{\rm{\mu }}_{\rm{2}}}\).

(c)

Compute the variance using –

\({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\)

Where\({\rm{E(X)}}\)is known. The following holds –

\({{\rm{(E(X))}}^{\rm{2}}}{\rm{ = }}{\left( {{\rm{0}}{\rm{.5}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + 0}}{\rm{.5}}{{\rm{\mu }}_{\rm{2}}}} \right)^{\rm{2}}}\)

Also, similarly to (b) it is obtained –

\(\begin{aligned}E\left( {{X^2}} \right) &= \sum\limits_{x = 0}^\infty {{x^2}} \cdot p\left( {x;{\mu _1},{\mu _2}} \right) = \sum\limits_{x = 0}^\infty {{x^2}} \left[ {0.5 \cdot p\left( {x;{\mu _1}} \right) + 0.5 \cdot p\left( {x;{\mu _2}} \right)} \right] \\ &= 0.5\sum\limits_{x = 0}^\infty {{x^2}} p\left( {x;{\mu _1}} \right) + 0.5\sum\limits_{x = 0}^\infty {{x^2}} p\left( {x;{\mu _2}} \right) \\ &= 0.5 \cdot E\left( {Y_1^2} \right) + 0.5 \cdot E\left( {Y_2^2} \right) \\ \end{aligned} \)

For any random variable\({\rm{Z}}\)the following holds –

\({\rm{E}}\left( {{{\rm{Z}}^{\rm{2}}}} \right){\rm{ = V(Z) + (E(Z)}}{{\rm{)}}^{\rm{2}}}\)

If the random variable \({\rm{Z}}\) follows Poisson distribution with parameter \({\rm{p}}\), using the proposition given above, it is obtained –

\({\rm{E}}\left( {{{\rm{Z}}^{\rm{2}}}} \right){\rm{ = V(Z) + (E(Z)}}{{\rm{)}}^{\rm{2}}}{\rm{ = \mu + }}{{\rm{\mu }}^{\rm{2}}}\)

Using this, it is obtained –

\(\begin{aligned} E\left( {{X^2}} \right) &= 0.5 \cdot E\left( {Y_1^2} \right) + 0.5 \cdot E\left( {Y_2^2} \right) \\ &= 0.5 \cdot \left[ {{\mu _1} + \mu _1^2} \right] + 0.5 \cdot \left[ {{\mu _2} + \mu _2^2} \right] \\ \end{aligned} \)

Finally, it is obtained that –

\(\begin{aligned}V(X) &= E\left( {{X^2}} \right) - {[E(X)]^2} \\ &= 0.5 \cdot [{\mu _1} + \mu _1^2] + 0.5 \cdot [{\mu _2} + \mu _2^2] - {[0.5{\mu _1} + 0.5{\mu _2}]^2} \\ \end{aligned} \)

Therefore, the expression is obtained as\({\rm{0}}{\rm{.5}} \cdot {\rm{(}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + \mu }}_{\rm{1}}^{\rm{2}}{\rm{) + 0}}{\rm{.5}} \cdot {\rm{(}}{{\rm{\mu }}_{\rm{2}}}{\rm{ + \mu }}_{\rm{2}}^{\rm{2}}{\rm{) - (0}}{\rm{.5}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + 0}}{\rm{.5}}{{\rm{\mu }}_{\rm{2}}}{{\rm{)}}^{\rm{2}}}\).

(d)

The change in the\({\rm{pmf}}\)is obtained as –

\({\rm{p(x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{,}}{{\rm{\mu }}_{\rm{2}}}{\rm{) = 0}}{\rm{.6}} \cdot {\rm{p(x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{) + 0}}{\rm{.4}} \cdot {\rm{p(x;}}{{\rm{\mu }}_{\rm{2}}}{\rm{)}} \ge {\rm{0, x}} \in {{\rm{N}}_{\rm{0}}}\)

Therefore, the expression is obtained as\({\rm{p(x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{,}}{{\rm{\mu }}_{\rm{2}}}{\rm{) = 0}}{\rm{.6}} \cdot {\rm{p(x;}}{{\rm{\mu }}_{\rm{1}}}{\rm{) + 0}}{\rm{.4}} \cdot {\rm{p(x;}}{{\rm{\mu }}_{\rm{2}}}{\rm{)}} \ge {\rm{0, x}} \in {{\rm{N}}_{\rm{0}}}\).