Intuitively, this inequality states that the probability that a random variable's value is much greater than its mean is low and that the larger this value, the smaller its probability. It states that \[\mathbf{P}(X\geq a)\leq\frac{\mu}{a},\hspace{10 mm}\forall a>0\] where $\mu$ is the mean, a.k.a. the expected value $\mathbf{E}(X)$ of the random variable. Note that this inequality needs that the random variable $X$ take non-negative values.

Here are some examples

- $\mathbf{P}(X\geq 10)\leq\mu/10$, which doesn't say much since $\mu$ could be greater than $10$. But the next one is more interesting.
- $\mathbf{P}(X\geq10\mu)\leq1/10$, which means that the probability that a random variable is $10$ times its mean is less than or equal to $0.1$. Now that wasn't obvious before!

Chebyshev inequality

Unlike Markov inequality that involves just the mean, Chebyshev inequality involves the mean and the variance. Intuitively, it states that the probability of a random variable taking values far outside of its variance is low and the farther this value from its mean, the lower its probability. More formally, it states that \[\mathbf{P}(|X-\mu|\geq c)\leq\frac{\sigma^2}{c^2},\hspace{10 mm}\forall c>0\] This implies \[\mathbf{P}({X-\mu|\geq k\sigma)\leq\frac{1}{k^2}\] This means that the probability that a value falls beyond $k$ times the standard deviation is at most $1/k^2$. This inequality tends to give tighter bounds than Markov inequality.

More concretely

- $\mathbf{P}(|X-\mu|\geq2\sigma)\leq1/4$, which means that the probability that a random variable takes values outside of 2 times its standard deviation is less than $1/4$.
- $\mathbf{P}(|X-\mu|\geq3\sigma)\leq1/9$, which means that the probability that a random variable takes values outside of 3 times its standard deviation is less than $1/9$.

From Markov inequality to Chebyshev inequality

Derivation of Chebyshev inequality from Markov inequality is relatively straight forward. Let $Y=|X-\mu|^2$. According to Markov inequality \[\mathbf{P}(Y\geq c^2)\leq \mathbf{E}(Y)/c^2\] Since $\mathbf{E}(Y)=\sigma^2$ and $\mathbf{P}(Y\geq c^2)$ is equivalent to $\mathbf{P}(|X-\mu|\geq c)$, by substituting them, we get Chebyshev inequality \[\mathbf{P}(|X-\mu|\geq c)\leq\frac{\sigma^2}{c^2}\]

Really useful short article. thanks for the pains taken to clarify these inequalities.

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