Question

Consider the following joint distribution.

		X
	p(x,y)	2	4
Y	1	0.11	0.36
	6	0.33	0.2

Based on this distribution, fill in the blanks below.

		X
	p(x,y)	2	4
Y	1	0.11	0.36
	6	0.33	0.2

• E(X)=
• Sd(Y)= .
• Corr(X,Y)=

Answer 1

x	y	f(x,y)	x*f(x,y)	y*f(x,y)	x^2f(x,y)	y^2f(x,y)	xy*f(x,y)
2	1	0.1100	0.2200	0.1100	0.4400	0.1100	0.2200
4	1	0.3600	1.4400	0.3600	5.7600	0.3600	1.4400
2	6	0.3300	0.6600	1.9800	1.3200	11.8800	3.9600
4	6	0.2000	0.8000	1.2000	3.2000	7.2000	4.8000
	Total	1	3.1200	3.6500	10.7200	19.5500	10.4200
	E(X)=ΣxP(x,y)=					3.1200
	E(X2)=Σx2P(x,y)=					10.7200
	E(Y)=ΣyP(x,y)=					3.6500
	E(Y2)=Σy2P(x,y)=					19.5500
	Var(X)=E(X2)-(E(X))2=					0.9856
	Var(Y)=E(Y2)-(E(Y))2=					6.2275
	E(XY)=ΣxyP(x,y)=					10.4200
	Cov(X,Y)=E(XY)-E(X)*E(Y)=					-0.9680
	Correlation ρxy=Cov(X,Y)/sqrt(Var(X)*Var(Y))=					-0.3907

from above: E(x) =3.12

SD(Y)=sqrt(Var(y))=sqrt(6.2275) =2.4955

Corr(X,Y) =-0.3907