Question #252304

The joint probability distribution of X and Y is given by

f(x,y)=x+y, 0<x<1 and 0<y<1 zero otherwise.

Determine the joint distribution function of X and Y i.e F(x,y). Hence or otherwise calculate p(0<x<1/2,1/2<y<1)


1
Expert's answer
2021-12-24T08:18:35-0500

joint distribution function:

F(x,y)=P(Xx,Yy)F(x,y)=P(X\le x,Y\le y)


for x0x\le 0 or y0y\le 0 :

F(x,y)=0F(x,y)=0

for x1x\ge 1 and y1y\ge 1 :

F(x,y)=1F(x,y)=1


for 0<x<1,0<y<10<x<1 , 0<y<1 :


F(x,y)=0y0xf(u,v)dudv=0y0x(u+v)dudv=F(x,y)=\int^y_0\int^x_0f(u,v)dudv=\int^y_0\int^x_0(u+v)dudv=


=0y(x2/2+vx)dv=yx2/2+y2x/2=\int^y_0(x^2/2+vx)dv=yx^2/2+y^2x/2


for 0<x<10<x<1 and x1x\ge 1 :

F(x,y)=F(x,1)=x2/2+x/2F(x,y)=F(x,1)=x^2/2+x/2

for 0<y<10<y<1 and y1y\ge 1 :

F(x,y)=F(1,y)=y2/2+y/2F(x,y)=F(1,y)=y^2/2+y/2


P(0<x<1/2,1/2<y<1)=1/2101/2f(x,y)dxdy=P(0<x<1/2,1/2<y<1)=\int^1_{1/2}\int^{1/2}_0f(x,y)dxdy=


=yx2/2+y2x/20,1/21/2,1=1/8+1/2=5/8=yx^2/2+y^2x/2|^{1/2,1}_{0,1/2}=1/8+1/2=5/8


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