I am trying to teach myself statistics and am reading DeGroot's book. Anyways, I'm in Chebychev's Inequality and am confused by this example:

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I understand the second part about if we know the distribution beforehand, we get a smaller required equal to 15. However, I am unable to replicate the first part based on the definition of Chebychev's inequality.

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I tried the following. Pr(0.4 <= Xbar <= 0.6) should be the same thing as Pr(abs(Xbar - μ) <= 0.1) and so:

Pr(abs(Xbar - μ) >= 0.1) <= σ^2 / (n * 0.1^2)

We know the variance of the distribution is n * (0.5) * (1 - 0.5) since its binomial. However, when I plug that variance in, obviously the ns cancel out and we are unable to find the necessary sample size. I am not sure what I am doing wrong. I also don't understand his explanation and why he approaches the problem the way he does. I understand the left hand side of his answer and realize he plugs in T/n as the sample mean. I don't get how he got the right hand side.


  • $\begingroup$ The $n$'s don't cancel completely. See my Answer details. $\endgroup$ – BruceET Sep 8 '19 at 21:00

First issue. Evaluating the bound. In Chebychev's Inequality,

$$P\left(\left|T -\frac n2\right| \le \frac{n}{10}\right) \le 1 - \frac{V(T)}{(n/10)^2} = 1 - \frac{n/4}{n^2/100}\\ = 1 - \frac{100n}{4n^2} = 1 - \frac{25}{n}.$$

Second issue: probability and bound change with sample size. For $n = 84,$ Chebyshev's Inequality says

$$P(.4n \le T \le .6n) = P(33.6 \le T \le 50.6)\\ = P(34 \le T \le 50) \le 1 - 25/84 = 0.7024.$$

The exact probability is $P(34 \le T \le 50) = 0.9370,$ as computed in R.

sum(dbinom(34:50, 84, .5))
[1] 0.9370277

For $n = 15,$ the Chebyshev bound is useless; we already know the probability is positive: $P(6 \le T \le 9) \le 1 - 25/15 = -0.6667.\,$ The exact probability is $0.6982 \approx 0.7,$ as claimed. This is just the author's way of saying that the Chebychev bound isn't very useful in a particular practical application.

sum(dbinom(6:9, 15, .5))
[1] 0.6982422

Because the mean and variance of $T$ depend on $n,$ you have to use one value of $n$ at a time.

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