How are the following two Chebychev's inequalities equivalent?

Question

I was looking at the following definition of Chebyshev's inequality

$$P(|X - E(X)| \geq r) \leq \frac{Var(X)}{r^2}$$

which includes the expected value and variance of $X$, and then I discovered there's another equivalent Chebyshev's inequality, which involves the standard deviation $\sigma$

$$P(|X - E(X)| \geq r\cdot \sigma) \leq \frac{1}{r^2}$$

but I am not understanding why are these formulas equivalent.

Could you please explain to me why this is the case?

Note that I know what is the standard deviation.

Answer 1

Let's replace $Var(X)$ with $\sigma^2$ in the first equation to give $$P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2}.$$

Now suppose $k= \dfrac{r}{\sigma}$, i.e. $r = k \sigma$, and substitute to give $$P(|X - E(X)| \geq k \sigma) \leq \frac{\sigma^2}{k^2\sigma^2} = \frac{1}{k^2}$$ which is your second equation using $k$ instead of $r$.

You can think of the $r$ in the first equation as having the same units as $X$ and $\sigma$, and the $r$ or $k$ in the second as being a unitless scalar multiple of the standard deviation, but ultimately they say the same thing.

Answer 2

It is exactly the same. If you use the first inequality you have $$P(|X−E(X)|\geq r⋅σ)\leq \frac{\operatorname{var}(X)}{(r\sigma)^2}=\frac{1}{r^2}.$$

Answer 3

Note that $Var(X) = \sigma^2$. \begin{align*} P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2} &\implies P((X - E(X))^2 \geq r^2) \leq \frac{\sigma^2}{r^2} \\ &\implies P((X - E(X))^2 \geq r^2\sigma^2) \leq \frac{1}{r^2} \\ &\implies P(|X - E(X)| \geq r\sigma) \leq \frac{1}{r^2} \end{align*}