Q:

Tshebyshev`s theorem

In my probability calculus course there is Tshebyshev`s theorem, which has form:
$$P(|X-\mu|\ge k) < \frac{\sigma^2}{k^2}$$
I do not really understand how it works. I spent two days figuring out on what there is happening but again I have no clue. All the examples found in the Internet propose form of:
$$P(|X-\mu|< \sigma k) \ge 1- \frac{1}{k^2}$$
What about the first form? How do they relate each other? Could someone please explain me how does it work, and how to determine upper and lower bound of some condition.
For example in homework tasks I have one that states:
$$\sigma^2=0.44; \mu=6.8$$
Determine lower bound for P(X="at least 4")

A:

Those forms are equivalent. Suppose that $P(|X-\mu|\geq k) \leq \frac{\sigma^2}{k^2}$. Then
\begin{align*}
P(|X-\mu|<\sigma k) &= 1-P(|X-\mu| \geq \sigma k)\\
&\geq1- \frac{\sigma^2}{(\sigma k)^2}\\
&= 1-\frac{1}{k^2}.
\end{align*}
Conversely, if $P(|X-\mu|<\sigma k) \geq 1-\frac{1}{k^2}$, then
\begin{align*}
P(|X-\mu| \geq k) &= P(|X-\mu| \geq \sigma \frac{k}{\sigma})\\
&=1-P(|X-\mu|< \sigma \frac{k}{\sigma})\\
&\leq 1-(1-\frac{1}{(k/\sigma)^2})\\
&= \frac{\sigma^2}{k^2}.
\end{align*}

As for how to prove this inequality, we have
\begin{align*}
P(|X-\mu| \geq k) &= E[1\{|X-\mu| \geq k\}]\\
&= E\left[1\left\{\left(\frac{X-\mu}{k}\right)^2 \geq 1\right\}\right]\\
&\leq E\left[\left(\frac{X-\mu}{k}\right)^2\right]\\
&=\frac{1}{k^2} E[(X-\mu)^2]\\
&=\frac{\sigma^2}{k^2}.
\end{align*}

In your example problem, you need to compare the probability $P(X \geq 4)$ to something of the form $P(|X-\mu| \geq k)$ or its complement, $P(|X-\mu|<k)$. We want a lower bound, so we should find some condition of the form $|X-\mu| \geq k$ or $|X-\mu|<k$ that implies $X \geq 4$. We can use $|X-6.8|<2.8$, because this implies $4<X<9.6$, which certainly implies $X \geq 4$. So then you have
$$P(X \geq 4) \geq P(|X-6.8|<2.8)=1-P(|X-6.8| \geq 2.8)$$
and you can use Chebyshev's inequality on that last line to find a lower bound.

