Download Beautiful Homework No. 4

Beautiful Homework No. 4
1 of 4
Engineering 323
Problem 3-43
Bridges
Definitions:
The expected value or mean of X is
E(X) = µ =
x ⋅ p( x)
x∈D
The Variance of X is
V(X) = σ =
2
(x - µ )2 ⋅ p( x) = E[( X − µ )2 ]
The Standard Deviation of X is
σ =
σ
2
Chebyshev’s inequality states that for any probability distribution of a random variable
X, and any number k, greater than 1, the probability that the value of X lies at least k
standard deviations from its mean is at most 1/k2 for each given k.
P( X - µ ) ≥ kσ ) ≤
1
k
2
Questions:
a. What is the value of the upper bound for the given k?
This is a simple problem of evaluating
k=2
k =3
k=4
k =5
k = 10
= .25
1
2
= .11
1
2
3
= .0625
1
4
2
1
2
5
1
10
2
= .04
= .01
1
k
2
for each given k.
Beautiful Homework No. 4
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Engineering 323
Problem 3-43
Bridges
b. Compute µ and σ for exercise 13. Then evaluate P( X - µ ≥ kσ ) for the values of
k given in part (a). What does this suggest about the upper bound relative to the
corresponding probability?
The pmf from 3-13 is:
P(X = 0) = .10
P(X = 1) = .15
P(X = 2) = .20
P(X = 3) = .25
P(X = 4) = .20
P(X = 5) = .06
P(X = 6) = .04
To find the expected value we use the formula given on page one.
µ = (0)(.1) + (1)(.15) + (2)(.20) + (3)(.25) + (4)(.20) + (5)(.06) + (6)(.04) = 2.64
To find the standard deviation first find the variance.
σ2 = (0 – 2.64)2 (.10) + (1 – 2.64)2 (.15) + (2 – 2.64)2 (.20) + (3 – 2.64)2 (.25) +
(4 – 2.64)2 (.20) + (5 – 2.64)2 (.06) + (6 - 2.64)2 (.04) = 2.3704
The deviation is then
σ
2
=σ
σ = 2.3704 = 1.54
Now to find P( | X - µ | ≥ kσ ) for the given k’s in part (a). What we are trying to
find is the range of values for X for which the equation | X - µ | ≥ kσ is true. This
equivalent to − kµσ ≥ X ≥ kµσ
For k = 2
P( | X – 2.64 | ≥ (2)(1.539) ) = P( | X – 2.64 | ≥ 3.078)
X ≥ 5.72) = P(-.44 ≥ X ≥ 5.72)= P(X = 6) = .04
We used P(X = 6) since 6 is the only value greater than 5.72 or less than - .44 in our
domain of the pmf.
For k=3
P( | X – 2.64 | ≥ (3)(1.539) ) = P( | X – 2.64 | ≥ 4.617)
= P(-1.98 ≥ X ≥ 7.26) = 0
Beautiful Homework No. 4
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Engineering 323
Problem 3-43
Bridges
Similarly there are no X values in this range, so the probability is 0.
The rest of the k’s follow.
For k=4
P( | X – 2.64 | ≥ (4)(1.539) ) = P( | X – 2.64 | ≥ 6.156)
= P(-3.52 ≥ X ≥ 8.8) = 0
For k = 5
P( | X – 2.64 | ≥ (5)(1.539) ) = P( | X – 2.64 | ≥ 7.695)
= P(-5.06 ≥ X ≥ 10.34) = 0
For k = 10
P( | X – 2.64 | ≥ (10)(1.539) ) = P( | X – 2.64 | ≥ 15.39)
= P(-12.76 ≥ X ≥ 18.04) = 0
a. Let X have three possible values, -1, 0, 1, with probabilities 181 , 89 , and 181 ,
respectively. What is P( | X - µ | ≥ 3σ ), and how does it compare to the
corresponding bound?
x
-1
0
1
P(x)
1
18
8
9
1
18
Again these are straightforward calculations.
µ = (-1)( 181 ) + (0)( 89 ) + (1)( 181 ) = 0
σ2 = (-1 – 0)2 ( 181 ) + (0 – 0)2 ( 89 ) + (1 – 0)2 ( 181 ) =
σ=
1
9
1
3
P( | X - µ | ≥ 3σ ) = P( | X – 0 | ≥ (3)( 13 ) ) = P( -1 ≥ X ≥ 1 ) = P(X=1) + P(X = -1)
=
1
18
+ 181 =
1
9
The bound for P( | X - µ | ≥ 3σ ) =
1
k
2
= 19 which is the same as the probability.
Beautiful Homework No. 4
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Engineering 323
Problem 3-43
Bridges
c.
Give a distribution for which P( | X - µ | ≥ 5σ ) = .04
We want to find p(-1), p(1), and p(0) for which P( | X - µ | ≥ 5σ ) = .04 holds.
This is like part (c) except instead being given the p(x) values and finding the
probability, we are given the probability and must find the individual p(x) values.
Using part (c) as a model we can intuitively find the appropriate p(x) values. First
notice p(-1) = p(1) in part (c). We will do the same here. It is easy to check that
this makes µ = 0. µ = (-1)p(-1) + (0)p(0) + (1)p(1) = p(1) – p(-1) = 0. The
variance is then attained by formula σ2 = (-1 – 0)2 p(-1) + (0 – 0)2 p(0) + (1 – 0)2
p(1) = p(1) + p(-1) = 2p(1). And finally the deviation is σ = 2a if we let
p(1) = a.
| X - µ | ≥ 5σ implies | X – 0 | ≥ 5 2a which implies X ≥ 5 2a .
Now let X = 1, we find our a = 1/50. Therefore our c = 1/50 and because p(-1)
+ p(0) + p(1) = 1 we can find b = 24/25.
Therefore
p(-1) = 1/50
P(0) = 24/25
P(1) = 1/50
This is a distribution where P( | X - µ | ≥ 5σ ) = .04
What was the point?
We constructed a PMF where the probability that the random variable is 5 standard
deviations away from the mean is exactly the Chebyshev limit of .04. Finding the
distribution in part (d) was nothing more then a mathematical exercise. The important
thing to remember is the Chebyshev limit is usually much greater than the actual
probability.