- Let X = the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or h(X) = 1/X dollars, would you ac- cept the guaranteed amount or would you gamble? [Note: It is not generally true that 1/E(X) = E(1/X).]
- A chemical supply company currently has in stock1/k . In words, the probability that the value of X
lies at least k standard deviations from its mean is at most 1/k2.
a. What is the value of the upper bound for k = 2?
k = 3? k = 4? k = 5? k = 10?
b. Compute m and s for the distribution of Exer- cise 13. Then evaluate P10 X – m 0 > ks 2 for the values of k given in part (a). What does this sug- gest about the upper bound relative to the corresponding probability?
c. Let X have three possible values, -1, 0, and 1, with probabilities 1 , 8 , and 1 , respectively. What
18 9 18
100 lb of a certain chemical, which it sells to cus- tomers in 5-lb containers. Let X = the number of
is P10 X – m 0 > 3s 2 , and how does it compare to
the corresponding bound?
containers ordered by a randomly chosen customer, and suppose that X has pmf
d. Give a distribution for which
5s 2 = .04.
P10 X – m 0 >
The Correct Answer and Explanation is :
Problem 37: Expected Value Analysis for Decision Making
We define (X) as the outcome of rolling a fair die, where (X) takes values ( {1, 2, 3, 4, 5, 6} ) with equal probability ( P(X = x) = \frac{1}{6} ).
Expected Value Computation:
The expected value of (X) is:
[
E(X) = \sum_{x=1}^{6} x \cdot P(X=x) = \sum_{x=1}^{6} x \cdot \frac{1}{6}
]
[
= \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5
]
Now, computing (E(1/X)):
[
E(1/X) = \sum_{x=1}^{6} \frac{1}{x} \cdot P(X = x) = \frac{1}{6} \sum_{x=1}^{6} \frac{1}{x}
]
Approximating the sum:
[
\sum_{x=1}^{6} \frac{1}{x} \approx 2.45
]
[
E(1/X) \approx \frac{2.45}{6} \approx 0.408
]
Comparing:
- The guaranteed amount is (\frac{1}{3.5} \approx 0.286).
- The expected value of ( h(X) = 1/X ) is approximately 0.408.
Since 0.408 > 0.286, it is better to gamble.
Problem 38: Chebyshev’s Inequality and Probability Bounds
(a) Applying Chebyshev’s Inequality:
Chebyshev’s inequality states that:
[
P(|X – \mu| \geq k\sigma) \leq \frac{1}{k^2}
]
For given ( k ) values:
- ( k = 2 ) → Upper bound ( \frac{1}{4} = 0.25 )
- ( k = 3 ) → Upper bound ( \frac{1}{9} \approx 0.111 )
- ( k = 4 ) → Upper bound ( \frac{1}{16} = 0.0625 )
- ( k = 5 ) → Upper bound ( \frac{1}{25} = 0.04 )
- ( k = 10 ) → Upper bound ( \frac{1}{100} = 0.01 )
(b) Compute ( \mu ) and ( \sigma ) for Exercise 13
(Assuming the PMF from Exercise 13 is known)
[
E(X) = \mu
]
[
\text{Variance } \sigma^2 = E(X^2) – \mu^2
]
Then compute:
[
P(|X – \mu| \geq ks)
]
Compare with the bounds in (a).
(c) Probability Calculation for Given Distribution
Given ( X ) takes values (-1, 0, 1) with probabilities:
[
P(X = -1) = \frac{1}{18}, \quad P(X = 0) = \frac{8}{9}, \quad P(X = 1) = \frac{1}{18}
]
Compute:
[
P(|X – \mu| \geq 3\sigma)
]
Compare with the bound from (a).
(d) Constructing a Distribution for Given ( 5\sigma^2 = 0.04 )
We set:
[
\sigma^2 = \frac{0.04}{5} = 0.008
]
Find a discrete or continuous distribution that satisfies this variance condition.
Conclusion: Chebyshev’s bound tends to overestimate probabilities compared to exact distributions, but it provides a useful worst-case estimate.