Question

Need clarification on these distribution problems. Thank you.

Q3 Grinding Distributions Let the joint probability density function of X and Y be fxy(x, y) {2004, 24cy, x,y> 0 and 0 < x+y = 1 otherwise Q3.1 What is P[X > 0.75 Y > 0.5]? Enter your answer here Save Answer Q3.2 Are X and Y independent? O No O Yes Save Answer Q3.3 Are X and Y identically distributed? Ο Νο O Yes Save Answer Q3.5 What is P[Y > X? Enter your answer here Save Answer Q3.6 What is P[Y > X]? Enter your answer here Save Answer

Answer 1

Q3.1) Using definition of conditional Probabiity we get

$P[X>0.75| Y>0.5]={P(X>0.75, Y>0.5)\over P(Y>0.5)}~~~~~~~~~~~(1)$

Since $0\leq x+y\leq 1$ , so $P(X>0.75,Y>0.5)=0$ Since the sum 0.75+0.5=1.25

Thus by (1)

Q3-2) We find the marginal pdf's o X and Y

Marginal PDF of X is

f«(a) = $fxr(x, y)ds 24 /** rydy = 24 1-1 = 24.0 2 0 = 12c[1 – ] 12[2° +1 2 2.c

Marginal PDF of Y is

$f_{Y}(y)=\int_{0}^{1-y} f_{XY}(x,y) dx\\ ~~~~~~~~~~~~~=24 \int_{0 }^{1-y} xydx\\ ~~~~~~~~~~~~~=24y \left ( {x^2\over 2} \right )_{0}^{1-y} \\ ~~~~~~~~~~~~=12y [1-y]^2 \\ ~~~~~~~~~~~~~= 12[y^3+y-2y^2]$

Since $f_{Y}(y)\times f_{X}(x)\not=f_{XY}(x,y)$

So X and Y are not independent.

Q-3-3) Since the marginal PDF's are similar

So X and Y are indentically distributed

Q-3-4) WE have

$P(Y>X)=\int_{0}^{0.5 }\int_{x}^{1-x} f_{XY}(x,y) dydx\\ ~~~ ~~~~~~~~~~~~~=24\int_{0}^{0.5 }\int_{x}^{1-x} xydydx\\ ~~~ ~~~~~~~~~~~~~=24 \int_{0}^{0.5}x \left ( {y^2\over 2} \right )_{x}^{1-x}dx \\ ~~~~~~~~~~~~~~~~=12\int_{0}^{0.5 }x(1-2x)dx \\ ~~~~~~~~~~~~~~~~~= 12 \left ( {x^2\over 2}-2{x^3\over 3} \right )_{0}^{0.5 }\\ ~~~~~~~~~~~~~~~~~=0.5$

Q-3-5) As X and Y are continuous random variables so

$P(Y\geq X)=P(Y>X)= 0.5$