Show that Chebyshev's inequality becomes an equality for this particular k For the discrete random variable X satisfying 
\begin{align}
p(X= 0) & = 1-\frac{1}{k^2}\\
p(X=-k) & = \frac{1}{2k^2} \\
p(X=+k) & = \frac{1}{2k^2} \\
\end{align}
with $k\geq1$, how can I show that Chebyshev's inequality becomes an equality for this particular value of k?
 A: So the proof of Chebyschev's inequality is:
$\begin{equation}
\begin{split}
P[|X-\mu|\geq k\sigma]  & = E[I_{|X=\mu|\geq k\sigma}]\\
& = E[I_{(\frac{X-\mu}{k\sigma})^2 \geq 1}]\\
& \leq E[(\frac{X-\mu}{k\sigma})^2]\\
& = \frac{1}{k^2} \frac{E[(X-\mu)^2]}{\sigma^2}\\
& = \frac{1}{k^2}
\end{split}
\end{equation}$
So you can see that the inequality comes in at the transition from line 2 to line 3, and is because if the indicator is satisfied then the argument of the indicator is greater than or equal to 1, which would indicate that it is greater than or equal to the maximum value of the indicator (1), thus it's expectation must be at least as large as the expectation of the indicator over these values.
For the values in the support for which the indicator isn't satisfied, we have that the indicator outputs a value of 0, but $(\frac{X-\mu}{k\sigma})^2 \geq 0$, so again the expression inside the expectation on line 3 must be greater than the expression inside line 2. So again, it's expectation must be at least as large as the expectation of the indicator over these values.
If you examine the distribution in the question, you can see it has mean 0 and variance 1. Plugging these into the second line you get:
$\begin{equation}
\begin{split}
P[|X-\mu|\geq k\sigma] & = E[I_{(\frac{X-\mu}{k\sigma})^2 \geq 1}]\\
& =  E[I_{(\frac{X}{k})^2 \geq 1}]
\end{split}
\end{equation}$
Now the key part, is that on the support of the given distribution the indicator and the argument of the indicator are equal. Specifically:
If $x = -k \rightarrow (\frac{x}{k})^2 =(\frac{-k}{k})^2 = 1 = I_{(\frac{-k}{k})^2 \geq 1 } = I_{(\frac{x}{k})^2 \geq 1 } $ 
if $x = k \rightarrow (\frac{x}{k})^2 =(\frac{k}{k})^2 = 1 = I_{(\frac{k}{k})^2 \geq 1 } = I_{(\frac{x}{k})^2 \geq 1 }$ 
and if $x = 0 \rightarrow (\frac{x}{k})^2 = (\frac{0}{k})^2 = 0 = I_{(\frac{0}{k})^2 \geq 1 } = I_{(\frac{0}{k})^2 \geq 1 }$
So for this specific distribution you can just swap the indicator with it's argument because they are equal on the support. (To be kind of pedantic here, they are not equal for all numbers, but they are equal on the values of X which have non-zero probability - Allowing you to exchange them in the expectation.)
This allows us to transition from line 2 to 3 as:
$\begin{equation}
\begin{split}
P[|X-\mu|\geq k\sigma]  & = E[I_{|X=\mu|\geq k\sigma}]\\
& = E[I_{(\frac{X}{k})^2 \geq 1}]\\
& = E[(\frac{X}{k})^2]\\
& = \frac{1}{k^2} E[X^2]\\
& = \frac{1}{k^2}
\end{split}
\end{equation}$
EDIT
So if you look at the Wikipedia page it states that equality is true in Chebyschev's inequality only for linear transformations of this distribution. On their page they have the support of $X$ being $\{-1,0,1\}$. Your distribution is just a scaling of that distribution by $k$, and hence a linear transformation. I tried to do a write up to show why only this set of distributions has this property, but got bogged down and didn't have time.
