Interval Estimates

Section 10.2 Interval Estimates - Chebyshev

An interval centered on the mean in which at least a certain proportion of the actual data must lie.

Theorem 10.2.1. Chebyshev's Theorem.

Given a random variable X with given mean \(\mu\) and standard deviation \(\sigma\text{,}\) for \(a \in \mathbb{R}^+\) ,

\begin{equation*} P( \big | X - \mu \big | \lt a ) \gt 1 - \frac{\sigma^2}{a^2} \end{equation*}

Proof.

Notice that the variance of a continuous variable X is given by

\begin{align*} \sigma^2 & = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) dx\\ & \ge \int_{-\infty}^{\mu-a} (x - \mu)^2 f(x) dx + \int_{\mu + a}^{\infty} (x - \mu)^2 f(x) dx\\ & \ge \int_{-\infty}^{\mu-a} a^2 f(x) dx + \int_{\mu + a}^{\infty} a^2 f(x) dx\\ & = a^2 \big ( \int_{-\infty}^{\mu-a} f(x) dx + \int_{\mu + a}^{\infty} f(x) dx \big )\\ & = a^2 P( X \le \mu - a \text{ or } X \ge \mu + a )\\ & = a^2 P( \big | X - \mu \big | \ge a) \end{align*}

Dividing by \(a^2\) and taking the complement gives the result.

Corollary 10.2.2. Alternate Form for Chebyshev's Theorem.

For positive k,

\begin{equation*} P( \big | X - \mu \big | \lt k \sigma ) \gt 1 - \frac{1}{k^2} \end{equation*}

Corollary 10.2.3. Special Cases for Chebyshev's Theorem.

For any distribution, it is not possible for f(x)=0 within one standard deviation of the mean. Aslo, at least 75% of the data for any distribution must lie within two standard deviations of the mean and at least 88% must lie within three.

Proof.

Apply the Chebyshev Theorem with \(a = \sigma\) to get

\begin{equation*} P(\mu - \sigma \lt X \lt \mu + \sigma) \gt 1 - \frac{\sigma^2}{\sigma^2} = 0 \end{equation*}

Apply the Chebyshev Theorem with \(a = 2 \sigma\) to get \(1 - \frac{1}{2^2} = 0.75\) and with \(k = 3 \sigma\) to get \(1 - \frac{1}{3^2} = \frac{8}{9} > 0.8888\text{.}\)

Checkpoint 10.2.4. WebWork - Chebyshev.

Checkpoint 10.2.5. WebWork - More Chebyshev.

Example 10.2.6. - Comparing known distribution to Chebyshev.

TBA