I'm reading about Chebyshev's Theorem and I am seeing conflicting information. My book states:
Chebyshev's Theorem is a general result that applies to most discrete random variables (and most continuous probability distributions as well).
I was watching videos and other people talking about this theorem and they say this theorem applies to any data set or distribution. Is this conflicting with my book when it says it applies to most random variables or are these two different statements?
The theorem is stated below:
Chebyshev's Theorem:
Let $X$ be a discrete random variable with finite mean $\mu_x$ and standard deviation $\sigma_x$. Let $k$ be greater than $1$. Then the probability that $X$ is more than $k$ standard deviations from the mean, $\mu_X$, is less than or equal to $\frac{1}{k^2}$. That is:
$\Pr\left(X < \mu_X -k \cdot \sigma_X \text{ or } X > \mu_X + k \cdot \sigma_X\right) = \Pr( \mid X – \mu \mid > k \cdot \sigma_X) \le \cfrac{1}{k^2} \tag{1}$
Equivalently,
$\Pr\left( \mu_X -k \cdot \sigma_X \le X \le \mu_X + k \cdot \sigma_X\right) = \Pr( \mid X – \mu \mid \le k \cdot \sigma_X) \ge 1-\cfrac{1}{k^2} \tag{2}$
My second question is the difference between equations (1) and (2). Is it correct to say that equation (1) is just setting an upper bound and equation (2) is setting a lower bound? So basically equation (2) is just applying the First Axiom of Probability? If so, why do the inequalitie in equation (2) become $ (\le X \le)$ from simply ">"?
Thank you in advance.
Best Answer
Chebyshev's inequality applies to any discrete or continuous distribution which meets the conditions. It is not particularly meaningful for some which do not, for example those which do not have a mean.
On your second question, what you really have in (1) is a statement of the form $\Pr(X \in A) \le c$; this clearly implies $\Pr(X \not\in A) \ge 1 - c$, which is the corresponding form of (2). So you need to use the opposite of "$\gt$", which is "$\le$": if you had had "$\ge$" inside the $\Pr(\cdots)$ in (1), the (2) would have needed "$\lt$" inside.