Question

1) Suppose that three random variables, X, Y, and Z have a continuous joint probability density function f(x, y. z) elsewhere a) Determine the value of the constant b) Find the marginal joint p. d. fof X and Y, namely f(x, y) (3 Points) c) Using part b), compute the conditional probability of Z given X and Y. That is, find f (Z I x y) d) Using the result from part c), compute P(Z<0.5 x - 3 Points) 2 Points) 0.25 and y0.75) 3 Points)

Answer 1

Given the joint PDF $f_{X,Y,Z}\left ( x,y,z \right )=c\left ( x+2y+3z \right );0\leqslant x\leqslant 1,0\leqslant y\leqslant 1,0\leqslant z\leqslant 1$ .

a) The condition for PDF is,

$\int \int \int f_{X,Y,Z}\left ( x,y,z \right )dzdydx=1\\ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}c\left ( x+2y+3z \right )dzdydx=1\\ c\int_{0}^{1}\int_{0}^{1}\left ( xz+2yz+3z^2/2 \right ]_{0}^{1}dydx=1\\ c\int_{0}^{1}\int_{0}^{1}\left ( x+2y+3/2 \right ]dydx=1\\ c\int_{0}^{1}\left ( xy+y^2+3/2y \right ]_{0}^{1}dx=1\\$

$c\int_{0}^{1}\left ( x+1+3/2 \right ]dx=1\\ c\left ( x^2+5/2x \right ]_{0}^{1}=1\\ c\left ( 7/2 \right )=1\\ {\color{Blue} c=2/7}$

b) The marginal PDF of   is

$f_{X,Y}\left ( x,y \right )=\int f_{X,Y,Z}\left ( x,y,z \right )dz\\ f_{X,Y}\left ( x,y \right )=\int_{0}^{1}c\left ( x+2y+3z \right )dz\\ f_{X,Y}\left ( x,y \right )=c\left ( xz+2yz+3z^2/2 \right ]_{0}^{1}\\ {\color{Blue} f_{X,Y}\left ( x,y \right )=\frac{2x+4y+3}{7} ;0\leqslant x\leqslant 1,0\leqslant y\leqslant 1}$

b) The conditional PDF,

$f_{Z|X,Y}\left (z| x,y \right )=\frac{f_{X,Y,Z}\left ( x,y,z \right )}{f_{X,Y}\left ( x,y \right )}\\ f_{Z|X,Y}\left (z| x,y \right )=\frac{c\left ( x+2y+3z \right )}{c\left (x+2y+3/2 \right )}\\ {\color{Blue} f_{Z|X,Y}\left (z| x,y \right )=\frac{x+2y+3z }{x+2y+3/2 };0\leqslant x\leqslant 1,0\leqslant y\leqslant 1,0\leqslant z\leqslant 1}$

d) The conditional probability,

$P\left ( Z<0.5|x=0.25,y=0.75 \right )=\int_{0}^{0.5}f_{Z|X=0.25,Y=0.75}\left (z| x,y \right )dz\\ P\left ( Z<0.5|x=0.25,y=0.75 \right )=\int_{0}^{0.5}\frac{0.25+2\times 0.75+3z }{0.25+2\times 0.75+3/2 }dz\\ P\left ( Z<0.5|x=0.25,y=0.75 \right )=\int_{0}^{0.5}\frac{1.75+3z }{3.25}dz\\ P\left ( Z<0.5|x=0.25,y=0.75 \right )=\left [\frac{1.75z+3z^2/2 }{3.25} \right ]_{0}^{0.5}\\ P\left ( Z<0.5|x=0.25,y=0.75 \right )=\frac{1.75\times 0.5+3\left ( 0.5 \right )^2/2 }{3.25} \\ {\color{Blue} P\left ( Z<0.5|x=0.25,y=0.75 \right )=0.3846}$

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