(a) Using Chebyshev’s Inequality, find upper bounds for the following probabilities: \(P(|X|\geq1)\), \(P(|X|\geq2)\), and \(P(|X|\geq3)\).
From Textbook: Chebyshev’s Inequality - Let \(X\) be a discrete random variable with expected value \(\mu = E(x)\), and let \(\in > 0\) be any positive real number. Then
\(P(|X - \mu| \geq \in) \leq \frac{V(x)}{\in^2}\).
Or, if \(\in = k\sigma\), then
\(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\).
So, in the case of \(P(|X|\geq1)\), given that \(\mu = 0\) and \(\sigma^2 = 1\):
\(P(|X - 0| \geq 1 \times 1) \leq \frac{1}{1^2}\) or \(\leq 1\)
And in the case of \(P(|X|\geq2)\), given that \(\mu = 0\) and \(\sigma^2 = 1\):
\(P(|X - 0| \geq 2 \times 1) \leq \frac{1}{2^2}\) or \(\leq \frac{1}{4}\) or \(\leq 0.25\)
Finally, in the case of \(P(|X|\geq3)\), given that \(\mu = 0\) and \(\sigma^2 = 1\):
\(P(|X - 0| \geq 3 \times 1) \leq \frac{1}{3^2}\) or \(\leq \frac{1}{9}\) or \(\leq 0.11\)
(b) The area under the normal curve between −1 and 1 is .6827, between −2 and 2 is .9545, and between −3 and 3 it is .9973 (see the table in Appendix A). Compare your bounds in (a) with these exact values. How good is Chebyshev’s Inequality in this case?
Chebyshev’s Inequality for \(P(|X|\geq1)\) is \(\leq 1\) in part (a) tells us that the probability of a deviation from the mean of one or more standard deviations is 1. The chances of being within 1 standard deviation of the mean is 0.6827, which means that the chances of being outise of 1 standard deviation is $1 - 0.6827 = 0.3173, which is clearly less than Chebyshev’s bound of 1.
Similarly, the chances of being outside of 2 standard deviations from the mean is \(1 - 0.9545 = 0.0455\), which again, is clearly less than Chebyshev’s bound 0.25 for \(P(|X - 0| \geq 2 \times 1)\).
Lastly, the chances of being outside of 3 standard deviations from the mean is \(1 - 0.9973 = 0.0027\), which again, is clearly less than Chebyshev’s bound 0.11 for \(P(|X - 0| \geq 3 \times 1)\).
How good is Chebyshev’s Inequality in this case? Well, the probabilities are definitely below the bounds, but the bounds are not very close to the actual probabilities. They certainly would not work well as substitutes for the probabilities.