Section 10.5
Let X be any random variable with (finite) mean and (finite) variance2. We shall assume X is a continuous type random variable with p.d.f.f(x), but what follows applies to a discrete type random variable whereintegral signs are replaced by summation signs and the p.d.f. is replacedby a p.m.f. For any k 1, we observe that
2 = E[(X – )2] =
–
(x – )2f(x) dx =
{x : |x – | k}
(x – )2f(x) dx +
{x : |x – | < k}
(x – )2f(x) dx
{x : |x – | k}
(x – )2f(x) dx
{x : |x – | k}
k2 2 f(x) dx =
{x : |x – | k}
k2 2 f(x) dx
We now have that 2 k2 2 P(|X – | k) which implies that
1
P(|X – | k) —
k2
This is Chebyshev’s inequality and is stated in Theorem 10.5-1.
We may also write
1
P(|X – | < k) 1 – —
k2
1.
(a)
(b)
(c)
Let X be a random selection from one of the first 9 positive integers.
Find the mean and variance of X.
= E(X) =2 = Var(X) =
5
20
—
3
Find P(|X – 5| 4) .
If Y is any random variable with the same mean and variance as X,find the upper bound on P(|Y – 5| 4) that we get fromChebyshev’s inequality.
P(|X – 5| 4) =
P(X = 1 X = 9) =
2
—
9
P(|Y – 5| 4) =
P[|Y – 5| 4(3/20)1/2(20/3)1/2]
20/3 5
—— =—
1612
k
2.
(a)
(b)
Let X be a random variable with mean 100 and variance 75.
Find the lower bound on P(|X – 100| < 10) that we get fromChebyshev’s inequality.
P(|X – 100| < 10) =
P[|X – 100| < (2/3)(53)]
1
1 – ——– =
(2/3)2
Find what the value of P(|X – 100| < 10) would be, if X had aU(85 , 115) distribution.
1
—
4
P(|X – 100| < 10) =
20
— =
30
P(90 < X < 110) =
2
—
3
k
3.
(a)
(b)
Let Y have a b(n, 0.75) distribution.
Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get fromChebyshev’s inequality when n = 12.
Find the exact value of P(|Y / n – 0.75| < 0.05) when n = 12.
When n = 12, E(Y) = , and Var(Y) = , and
9
2.25
P(|Y / n – 0.75| < 0.05) =
P(|Y – 9| < 0.6) =
P[|Y – 9| < (0.4)(1.5)]
(A lower bound cannotbe found, since 0.4 < 1.)
When n = 12, P(|Y / n – 0.75| < 0.05) = P(|Y – 9| < 0.6) =
P(Y = 9) =
0.6488 – 0.3907 = 0.2581
(c)
(d)
Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get fromChebyshev’s inequality when n = 300.
By using the normal approximation, show that when n = 300, thenP(|Y / n – 0.75| < 0.05) = 0.9464.
When n = 300, E(Y) = , and Var(Y) = , and
225
56.25
P(|Y / n – 0.75| < 0.05) =
P(|Y – 225| < 15) =
P[|Y – 225| < (2)(7.5)]
1
1 – — = 0.75
22