91Ó°ÊÓ

X and Y have a discrete joint distribution and the joint p.f is,

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}\frac{1}{{30}}\left( {x + y} \right)\;for\;x = 0,1,2\;and\;y = 0,1,2,3\\0\;otherwise\end{array} \right.\)

a.

The marginal p.f of X is-

\(\begin{array}{c}{\bf{Pr}}\left( {{\bf{X = x}}} \right){\bf{ = }}{{\bf{f}}_{\bf{X}}}\left( {\bf{x}} \right)\\{\bf{ = }}\sum\limits_{{\bf{y = 0}}}^{\bf{3}} {{{\bf{f}}_{{\bf{X,Y}}}}\left( {{\bf{x,y}}} \right)} \\{\bf{ = }}\sum\limits_{{\bf{y = 0}}}^{\bf{3}} {\frac{{\bf{1}}}{{{\bf{30}}}}\left( {{\bf{x + y}}} \right)} \end{array}\)

So,

\(\begin{array}{c}{f_X}\left( x \right) = \frac{1}{{30}}\left( {\left( {x + 0} \right) + \left( {x + 1} \right) + \left( {x + 2} \right) + \left( {x + 3} \right)} \right)\\ = \frac{1}{{30}}\left( {4x + 6} \right)\\ = \frac{{2x + 3}}{{15}}\end{array}\)

Thus, the marginal p.f of X is\(\frac{{2x + 3}}{{15}}\)for x=0,1,2.

Now, the marginal p.f for Y is -

\(\begin{array}{c}{\bf{Pr}}\left( {{\bf{Y = y}}} \right){\bf{ = }}{{\bf{f}}_{\bf{Y}}}\left( {\bf{y}} \right)\\{\bf{ = }}\sum\limits_{{\bf{x = 0}}}^{\bf{2}} {{{\bf{f}}_{{\bf{X,Y}}}}\left( {{\bf{x,y}}} \right)} \\{\bf{ = }}\sum\limits_{{\bf{x = 0}}}^{\bf{2}} {\frac{{\bf{1}}}{{{\bf{30}}}}\left( {{\bf{x + y}}} \right)} \end{array}\)

So,

\(\begin{array}{c}{f_Y}\left( y \right) = \frac{1}{{30}}\left( {\left( {0 + y} \right) + \left( {1 + y} \right) + \left( {2 + y} \right)} \right)\\ = \frac{1}{{30}}\left( {3y + 3} \right)\\ = \frac{{y + 1}}{{10}}\end{array}\)

Thus, the marginal p.d.f of Y is\(\frac{{y + 1}}{{10}}\)for y=0,1,2,3.

To check if Xand Yare independent, we have to show that,\({\bf{Pr}}\left( {{\bf{X = x,Y = y}}} \right){\bf{ = Pr}}\left( {{\bf{X = x}}} \right){\bf{Pr}}\left( {{\bf{Y = y}}} \right){\bf{,}}\forall \left( {{\bf{x,y}}} \right)\)

Now clearly we can see that,

\(\Pr \left( {X = x,Y = y} \right) = \frac{1}{{30}}\left( {x + y} \right)\)and\(\Pr \left( {X = x} \right)\Pr \left( {Y = y} \right) = \frac{1}{{150}}\left( {2x + 3} \right)\left( {y + 1} \right)\;,\forall \left( {x,y} \right)\)

So,

\(\Pr \left( {X = x,Y = y} \right) \ne \Pr \left( {X = x} \right)\Pr \left( {Y = y} \right)\;\forall \left( {x,y} \right)\)

Therefore, we can conclude that Xand Y are not independent.