Answer :
The marginal density functions of the joint probability density functions, X and Y (a) [tex]f(x) = \frac{2x-1}{6}, f(y) = \frac{2}{9} (6-y)[/tex]
(b) P(X>2, Y>1.5 ) = 0.3194
(c) X and Y are not independent
The given function is :
[tex]f(x) =\frac{3x-y}{9} , 1 < x < 3 , 1 < y < 2 and 0 , eleswhere[/tex]
(a) Marginal Density Functions of X and Y are
[tex]f_{x}(x) =\int\limits^2_1{f(x,y)} \, dy =\int\limits^2_1{\frac{3x-y}{9} } \, dy =\frac{1}{9} [2xy -\frac{y^{2} }{2} ]^{2} _{1} = \frac{1}{9}(3x-\frac{3}{2}=\frac{2x-1}{6} , 1 < x < 3\\ f(x) = \frac{2x-1}{6}, 1 < x < 3[/tex]
[tex]f_{y}(y) = \int\limits^3_1{f(x,y)} \, dx = \int\limits^3_1{\frac{3x-y}{9} } \, dx =\frac{2}{9}(6-y)\\ f_{y}(y) = \frac{2}{9}(6-y) , 1 < y < 2[/tex]
(b) P(X>2, Y>1.5)
[tex]=\int\limits^2_\frac{3}{2} \int\limits^3_2 {f(x,y)} \, dxdy \\ =\int\limits^2_\frac{3}{2} {[\int\limits^3_2 {\frac{3x-y}{9} } \, dx }] \, dy \\=0.3194[/tex]
(c) To check whether X and Y are independent, we have:
[tex]f_{x}(x) =\frac{2x-1}{6} ,f_{y} (y) =\frac{2(6-y)}{9} \\ X,Y are independent if \\f_{xy} =f_{x}*f_{y} = \frac{2x-1}{6}*\frac{2(6-y)}{9} \\=\frac{1}{27}(12x+y-2xy-6)\\ \neq f(x,y)[/tex]
Therefore, X and Y are not independent.
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