Question

Suppose X is a random variable with p(X = 0) = 4/5, p(X = 1) = 1/10, p(X = 9) = 1/10. Then

(a) Compute Var [X] and E [X].

(b) What is the upper bound on the probability that X is at least 20 obained by applying Markov’s inequality?

(c) What is the upper bound on the probability that X is at least 20 obained by applying Chebychev’s inequality?

Answer 1

x	P(x)	xP(x)	x2P(x)
0	4/5	0.000	0.000
1	1/10	0.100	0.100
9	1/10	0.900	8.100
	total	1.000	8.200
	E(x) =μ=	ΣxP(x) =	1.0000
	E(x2) =	Σx2P(x) =	8.2000
	Var(x)=σ2 =	E(x2)-(E(x))2=	7.200

a)

Var(X)=7.200

E(X)=1.00

b)

upper bound on the probability that X is at least 20 obained by applying Markov’s inequality

P(X>a) <=E(X)/a

P(X>20)<=1/20

c)

upper bound on the probability that X is at least 20 obained by applying Chebychev’s inequality

=P(X>20)=P(X-1>19)>=Var(X)/k²

P(X>20)<=7.2/19

P(X>20)<=36/95