Question

The National Highway Traffic Safety Administration is interested in the effect of seat belt use on saving lives of children under 5. In this study, there were 7,060 accidents where there was at least one fatality in the years between 1985 to 1989 (3015 children were involved). Let X denote the type of seat belt that the child wore (if any) and let Y denote whether the child survived or not and. The joint pmf of (X, Y) is given on the next slide. a. What is E(XY)? b. What is E[max(X,Y)]? c. E(XY) = 0.12 What is Cov(X,Y)? d. Cov(X,Y) = -0.0576, E(X) = 0.74, E(Y) = 0.24. What is Cor(X, Y) = ρ(X, Y)?

		Y
	p(x,y)	Survivors (0)	Fatalities (1)
X	no belt (0)	0.37	0.17
	Adult belt (1)	0.14	0.02
	Child Seat (2)	0.24	0.05

Answer 1

a)

E(XY)=xyP(x,y)=0*0*0.37+0*1*0.17+1*0*0.14+1*1*0.02+2*0*0.24+2*1*0.05=0.12

b)

Emax(XY)=maxx,y)P(x,y)=0*0*0.37+1*0.17+1*0.14+1*0.02+2*0.24+2*0.05=0.91

c)

marginal pmf of X:

x	P(x)	xP(x)	x^2P(x)
0	0.5400	0.0000	0.0000
1	0.1600	0.1600	0.1600
2	0.2900	0.5800	1.1600
total	0.99	0.74	1.32
E(x)	=		0.7400
E(x^2)	=		1.3200
Var(x)	E(x^2)-(E(x))^2		0.7724

marginal pmf of Y:

y	P(y)	yP(y)	y^2P(y)
0	0.7500	0.0000	0.0000
1	0.2400	0.2400	0.2400
total	0.9900	0.2400	0.2400
E(y)	=		0.2400
E(y^2)	=		0.2400
Var(y)	E(y^2)-(E(y))^2		0.1824

Cov(X,Y)=E(XY)-E(X)*E(Y)=-0.0576

d)

Cor(X, Y)=Cov(X,Y)/sqrt(Var(X)*Var(Y))=-0.15346