91Ó°ÊÓ

PMFs on the fringes The joint PMF contains all of the information about X and Y's distributions. This means that we can get the PMF of X from its joint PMF with Y, for example.

(a):

\({\rm{X}}\)has a marginal probability mass function.

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\sum\limits_{\rm{y}} {\rm{p}} {\rm{(x,y),}}\;\;\;{\rm{ for every x,}}\)

\({\rm{Y}}\)has a marginal probability mass function.

\({{\rm{p}}_{\rm{Y}}}{\rm{(y) = }}\sum\limits_{\rm{x}} {\rm{p}} {\rm{(x,y), for every y}}{\rm{.}}\)

Returning to the table from exercise \({\rm{3}}\), the following holds true for \({\rm{x = 0}}\).

\(\begin{aligned}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(0) = }}\sum\limits_{{{\rm{x}}_{\rm{2}}}} {\rm{p}} \left( {{\rm{0,}}{{\rm{x}}_{\rm{2}}}} \right){\rm{ = p(0,0) + p(0,1) + p(0,2) + p(0,3)}}\\{\rm{ = 0}}{\rm{.08 + 0}}{\rm{.07 + 0}}{\rm{.04 + 0}}{\rm{.00}}\\{\rm{ = 0}}{\rm{.19}}\end{aligned}\)

Similarly, for \({\rm{x}} \in \{ 1,2,3,4\} \), the following is true

\(\begin{aligned}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(1) = }}\sum\limits_{{{\rm{x}}_{\rm{2}}}} {\rm{p}} \left( {{\rm{1,}}{{\rm{x}}_{\rm{2}}}} \right){\rm{ = p(1,0) + p(1,1) + p(1,2) + p(1,3)}}\\{\rm{ = 0}}{\rm{.06 + 0}}{\rm{.15 + 0}}{\rm{.05 + 0}}{\rm{.04}}\\{\rm{ = 0}}{\rm{.3}}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(2) = }}\sum\limits_{{{\rm{x}}_{\rm{2}}}} {\rm{p}} \left( {{\rm{2,}}{{\rm{x}}_{\rm{2}}}} \right)\\{\rm{ = p(2,0) + p(2,1) + p(2,2) + p(2,3)}}\\{\rm{ = 0}}{\rm{.05 + 0}}{\rm{.04 + 0}}{\rm{.10 + 0}}{\rm{.06}}\end{aligned}\)

\( = {\bf{0}}.{\bf{25}}\),

\({{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(3) = }}\sum\limits_{{{\rm{x}}_{\rm{2}}}} {\rm{p}} \left( {{\rm{3,}}{{\rm{x}}_{\rm{2}}}} \right){\rm{ = p(3,0) + p(3,1) + p(3,2) + p(3,3)}}\)

\( = 0.00 + 0.03 + 0.04 + 0.07\)

\( = {\bf{0}}.{\bf{14}}\),

\({{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(4) = }}\sum\limits_{{{\rm{x}}_{\rm{2}}}} {\rm{p}} \left( {{\rm{4,}}{{\rm{x}}_{\rm{2}}}} \right){\rm{ = p(4,0) + p(4,1) + p(4,2) + p(4,3)}}\)

\( = 0.00 + 0.01 + 0.05 + 0.06\)

\( = {\bf{0}}.{\bf{12}}\),

\({{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}\left( {{{\rm{x}}_{\rm{1}}}} \right){\rm{ = 0,}}\;\;\;{{\rm{x}}_{\rm{1}}} \notin {\rm{\{ 0,1,2,3,4\} }}.\)

\({{\rm{X}}_{\rm{1}}}\) is determined by the variables above. We can also represent it as a table.

A discrete random variable \({\rm{X}}\) with a set of possible values \({\rm{S}}\) and \({\rm{pmfp(x)}}\) has an Expected Value (mean value) of

\[\text{E(X)=}{{\text{ }\!\!\mu\!\!\text{ }}_{\text{X}}}\text{=}\sum\limits_{\text{x }\!\!\hat{\mathrm{I}}\!\!\text{ S}}{\text{x}}\text{ }\!\!\times\!\!\text{ p(x)}\text{.}\]

As a result, the anticipated value is

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ = 0 \times }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(0) + 1 \times }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(1) + 2 \times }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(2) + 3 \times }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(3) + 4 \times }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(4)}}\\{\rm{ = 0 \times 0}}{\rm{.19 + 1 \times 0}}{\rm{.3 + 2 \times 0}}{\rm{.25 + 3 \times 0}}{\rm{.14 + 4 \times 0}}{\rm{.12}}\\{\rm{ = 0 + 0}}{\rm{.3 + 0}}{\rm{.5 + 0}}{\rm{.42 + 0}}{\rm{.48}}\\{\rm{ = 1}}{\rm{.7}}\end{aligned}\)

(b):

We get the following marginal pmf of \({{\rm{X}}_{\rm{2}}}\), same like in \({\rm{(a)}}\).

\(\begin{aligned}{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}{\rm{(0) = }}\sum\limits_{{{\rm{x}}_{\rm{1}}}} {\rm{p}} \left( {{{\rm{x}}_{\rm{1}}}{\rm{,0}}} \right){\rm{ = p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0)}}\\{\rm{ = 0}}{\rm{.08 + 0}}{\rm{.06 + 0}}{\rm{.05 + 0}}{\rm{.00 + 00 = 0}}{\rm{.19}}\\{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}{\rm{(1) = }}\sum\limits_{{{\rm{x}}_{\rm{1}}}} {\rm{p}} \left( {{{\rm{x}}_{\rm{1}}}{\rm{,0}}} \right){\rm{ = p(0,1) + p(1,1) + p(2,1) + p(3,1) + p(4,1) = 0}}{\rm{.3}}\\{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}{\rm{(2) = }}\sum\limits_{{{\rm{x}}_{\rm{1}}}} {\rm{p}} \left( {{{\rm{x}}_{\rm{1}}}{\rm{,2}}} \right){\rm{ = p(0,2) + p(1,2) + p(2,2) + p(3,2) + p(4,2) = 0}}{\rm{.28}}\\{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}{\rm{(3) = }}\sum\limits_{{{\rm{x}}_{\rm{1}}}} {\rm{p}} \left( {{{\rm{x}}_{\rm{1}}}{\rm{,3}}} \right){\rm{ = p(0,3) + p(1,3) + p(2,3) + p(3,3) + p(4,3) = 0}}{\rm{.23}}\\{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}\left( {{{\rm{x}}_{\rm{2}}}} \right){\rm{ = 0}}\;\;\;{{\rm{x}}_{\rm{2}}} \notin \{ 0,1,2,3,4\} \end{aligned}\)

We can also represent it as a table.

(c):

If and only if, two random variables \({\rm{X}}\) and \({\rm{Y}}\)are independent.

1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\),

whenever \({\rm{(x,y)}}\) occurs Discrete RVs \({\rm{X}}\) and \({\rm{Y}}\),

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\)and when \({\rm{X}}\)and \({\rm{Y}}\)continuous rv's, otherwise they are dependent.

By calculating probabilities

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 4}}} \right){\rm{ = }}{{\rm{p}}_{{{\rm{X}}_{\rm{1}}}}}{\rm{(4)}}\\{\rm{ = 0}}{\rm{.12}}\\{\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ = }}{{\rm{p}}_{{{\rm{X}}_{\rm{2}}}}}{\rm{(0)}}\\{\rm{ = 0}}{\rm{.19}}\end{aligned}\)

and probability

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 4,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0}}} \right){\rm{ = p(4,0)}}\\{\rm{ = 0}}\end{aligned}\)

we can notice that

We can deduce that the random variables are interdependent.