Our tutor told us that using Chebycheff inequality (CI), i.e. $P(|X-\mathbb{E}(X)|>\epsilon)\leq \frac{\mathbb{V}(X)}{\epsilon^2}$, where $\epsilon >0$, allows us to find upper and lower bounds in all situations if we use "clever" manipulations of inequalities. She came up with some examples to explain her statement and I didn't question it...
Now, I constructed a few exapmples on my own which lead to some confusion.
Let's consider a random variable $X$, $0<a<\mathbb{E}(X)<b$ and $\mathbb{V}(X)\geq0$.
1.) Find an $L\geq 0$ such that $P(X>a)\geq L$. This is equivalent to $P(X\leq a)\leq 1-L$, $$ \begin{align*} &P(X\leq a)=P(X-\mathbb{E}(X)\leq a -\mathbb{E}(X))=P(X-\mathbb{E}(X)\leq -|a -\mathbb{E}(X)|)\\ &=P(\mathbb{E}(X)-X\geq |a -\mathbb{E}(X)|)\leq \frac{\mathbb{V}(X)}{(a -\mathbb{E}(X))^2}=:1-L\\ &\implies L= 1-\frac{\mathbb{V}(X)}{(a -\mathbb{E}(X))^2}. \end{align*}$$ O.k. no problem, in this case it might be that $L$ is better than $0$.
2.) Find an $U\leq 1$ such that $P(X>a)\leq U$, $$ \begin{align*} &P(X>a)=P(X-\mathbb{E}(X)> a -\mathbb{E}(X))=P(X-\mathbb{E}(X)> -|a -\mathbb{E}(X)|)\\ &=P(\mathbb{E}(X)-X< |a -\mathbb{E}(X)|)\leq??? \end{align*}$$ In this case it seems that we can't find a better bound using Chebycheff, right?
3.) Find an $U\leq 1$ such that $P(X<a)\leq U$, $$ \begin{align*} &P(X<a)=P(X-\mathbb{E}(X)< a -\mathbb{E}(X))=P(X-\mathbb{E}(X)< -|a -\mathbb{E}(X)|)\\ &=P(\mathbb{E}(X)-X> |a -\mathbb{E}(X)|)\leq\frac{\mathbb{V}(X)}{(a -\mathbb{E}(X))^2}=:U. \end{align*}$$ In this case we find an upper bound that can be better than $1$.
4.) Find an $L\geq 0$ such that $P(X<a)\geq U$. This is equivalent to $P(X\geq a)\leq 1-U$, $$ \begin{align*} &P(X\geq a)=P(X-\mathbb{E}(X)\geq a -\mathbb{E}(X))=P(X-\mathbb{E}(X)\geq -|a -\mathbb{E}(X)|)\\ &=P(\mathbb{E}(X)-X\leq |a -\mathbb{E}(X)|)\leq ??? \end{align*}$$ In this case it seems that we can't find a better bound using Chebycheff, right?
If I look at $P(X>b)$ and $P(X<b)$ I run into the same problems. So I doubt that Chebycheff inequality helps us to find a better bound in each case. Am I right or are there some magic manipulations that produce bounds in case 2.) and 4.)?
I guess your tutor meant that inequality for the event $|X-E[x]|\geq\varepsilon$ $$P(|X-E[x]|\geq\varepsilon)\leq \frac{\sigma^2}{\varepsilon^2}$$ can be rewritten as inequality for the complement event $|X-E[x]|<\varepsilon$ $$P(|X-E[x]|<\varepsilon)\geq 1-\frac{\sigma^2}{\varepsilon^2}$$