Question #109037
the joint distribution function of X and Y is given by
F(x,y) = (1-e-x)(1-e-y) for x>0,y>0
= 0 otherwise
Discuss about the marginal densities of X and Y
Examine whether X and Y are independent (iii) Evaluate P(1<x<3,1<y<2)
1
Expert's answer
2020-04-13T18:44:25-0400

1) Marginal pdf of X:

fX(x)=fX,Y(x,y)dy.fX,Y(x,y) — joint pdf.fX,Y(x,y)=2xyF(x,y).F(x,y) — joint CDF.xF(x,y)=(1ey)ex.2xyF(x,y)=e(x+y).fX,Y(x,y)=e(x+y)(x>0,y>0);0 (elsewhere).fX(x)=0e(x+y)dy=ex.So marginal pdf of X fX(x)=ex(x>0);0 (elsewhere).Marginal pdf of Y (all the computations are symmetrical)fY(y)=ey(y>0);0 (elsewhere).2)X and Y are independent because fX,Y(x,y)=fX(x)fY(y).P{1<x<3,1<y<2}=1312fX,Y(x,y)dydx=1312fX(x)fY(y)dydx=1312exeydydx=(e2e1)(e3e1)0.074.f_X(x)=\int_{-\infty}^{\infty}f_{X, Y}(x,y)dy.\\ f_{X, Y}(x,y)\text{ --- joint pdf}.\\ f_{X, Y}(x,y)=\frac{\partial^2}{\partial x\partial y}F(x,y).\\ F(x,y)\text{ --- joint CDF}.\\ \frac{\partial}{\partial x}F(x,y)=(1-e^{-y})e^{-x}.\\ \frac{\partial^2}{\partial x\partial y}F(x,y)=e^{-(x+y)}.\\ f_{X, Y}(x,y)=e^{-(x+y)} (x>0, y>0); 0 \text{ (elsewhere)}.\\ f_X(x)=\int_{0}^{\infty}e^{-(x+y)}dy=e^{-x}.\\ \text{So marginal pdf of X }f_X(x)=e^{-x} (x>0); 0 \text{ (elsewhere)}.\\ \text{Marginal pdf of Y (all the computations are symmetrical)}\\ f_Y(y)=e^{-y} (y>0); 0 \text{ (elsewhere)}.\\ 2) X\text{ and } Y\text{ are independent because } f_{X,Y}(x,y)=f_X(x)f_Y(y).\\ P\{1<x<3, 1<y<2\}=\int_{1}^{3}\int_{1}^{2}f_{X,Y}(x,y)dydx=\int_{1}^{3}\int_{1}^{2}f_{X}(x)f_{Y}(y)dydx=\int_{1}^{3}\int_{1}^{2}e^{-x}e^{-y}dydx=(e^{-2}-e^{-1})(e^{-3}-e^{-1})\approx\\ \approx 0.074.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS