The joint probability density function of two random variables is given by

$f_{xy}$ (X,Y)= 15$e^{-3x-5y} $ x≥0,y≥0

• Find the probability that x<2 and Y>0.2.
• Find the marginal densities of X and Y
• Are X and Y Independent?
• Find E(X/Y) and E(Y/X).

**Mumbai University > Electronics and Telecommunication Engineering > Sem 5 > Random Signal Analysis

Marks: 10M

Year: Dec 2015

P(X<2,Y>0.2)=15$∫_0^2 {e^{-3x}} dx$. $∫_{0.2}^∞ {e^{-5y}} dy$

=15$({e^{-3x}/(-3))_0^2}. {e^{-5y}/(-5))_{0.2}^∞}$ = 15 ${1-e^{-6} )}{e^{-1} }$/15 =0.367

$f_X$ (x)=$∫_0^∞ 15{e^{-3x-3y}} dy$

=15 ${e^{-3x}} ({e^{-5y}/(-5))_0^∞}$ = 3${e^{-3x}}$ x>0

$f_X$ (x) = 3$e^{-3x}$ x>0

=0 otherwise

$f_Y$ (y)=$∫_0^∞ 15{e^{-3x-3y}} dx$

=15$e^{-5y} ({e^{-3x}/(-3))_0^∞}$= 5${e^{-5y}}$ y>0

$f_Y$ (y) = 5$e^{-5y} y\gt0 =0 otherwise Now X, Y are independent if $f_{XY}$ (x,y)=$f_X$ (x).$f_Y$ (y) Since$ f_X$ (x).$f_Y$ (y)=$(3.e^{-3x} )(5{e^{-3y}} )$ = 15$e^{-3x-3y}$ =$f_{XY}$ (x,y) Hence X and Y are independent. Since (X,Y) are independent, the conditional probability density functions are equal to marginal probability density function ∴$f_{X/Y}$ (x/y) = $f_X$ (x) =3$e^{-3x}$, x\gt0 ∴$f_{Y/X}$ (y/x) = $f_Y$ (y) =3$e^{-3y}$, y\gt0 ∴E(X/Y=x)=$∫_0^∞ x. $$f_{X/Y} (x/y)dx$ =$∫_0^∞ x. {f_X} (x) dx$ =$∫_0^∞ x. 3{e^{-3x}} dx$ =$[3x(e^{-3x}/(-3))_0^∞ . -3({e^{-3x}/9)}_0^∞ ]$ =1/3 Similarly ∴$f_{Y/X} (y/x)$ = $f_Y$ (y) =5$e^{-3y}$, y\gt0 ∴E(Y/X=y)=$∫_0^∞ y. {f_{Y/X}} (y/x)dy$ =$∫_0^∞ y.{f_Y} (y) dy$ =$∫_0^∞ y.3{e^{-3y}} dy$ =$[3y(e^{-3y}/(-3))_0^∞ -3({e^{-3y}/9})_0^∞ ]$=1/3