How do I prove that this discrete distribution satisfies Chebyshev’s inequality? I have been given a discrete distribution, which has variance $\sigma^2$ and mean $\mu$. There are three $X$ values in terms of $k$, $\mu$ and $\sigma$, while the probabilities are in those terms too.
I need to find out how this distribution satisfies Chebyshev? The $|X-\mu|$ of each $X$ is $k\sigma$. But I'm not sure how I'd go on to say how the distribution satisfies Chebyshev.
 A: I think that what you are asking for is a random variable $X$ with a specified
mean $\mu$ and variance $\sigma^2$ for which equality holds in the Chebyshev Inequality for one specified value of $k$: that is,

$$P\{|X-\mu| \geq k\sigma\} ~ \mathbf{ equals}~ \frac{1}{k^2}.$$

As Jack D'Aurizio has pointed out, $P\{|X-\mu| \geq k\sigma\}\leq \frac{1}{k^2}$ always holds for all positive real numbers $k$.
Consider a discrete random variable $Y$ that takes on values $0,k$, and $-k$ with probabilities 
$1-\frac{1}{k^2}, \frac{1}{2k^2}, \frac{1}{2k^2}$ respectively.
Obviously, $E[Y]= 0$ while 
$$E[Y^2] = (-k)^2\times \frac{1}{2k^2} + 0\times (1-\frac{1}{k^2})
+ (k)^2\times \frac{1}{2k^2} = 1$$
so that $\sigma_Y^2 = E[Y^2] -(E[Y])^2 = 1$.  Clearly, $P\{|Y| \geq k\} = P\{Y=-k\} + P\{Y=k\} = \frac{1}{k^2}$, that is,

$$P\{|Y-\mu_Y| \geq k\sigma)\} ~ \mathbf{ equals}~ \frac{1}{k^2}$$

and thus equality holds in Chebyshev's inequality for this particular
zero-mean unit-variance random variable for the specified value of $k$.
I will leave it to you to prove that if you set $X = \sigma Y + \mu$
where $\mu$ and $\sigma^2$ are the mean and variance that you
need $X$ to have,  then $X$ is a
random variable with mean $\mu$ and variance $\sigma^2$ and it is
also true that
$$P\{|X-\mu| \geq k\sigma\} = \frac{1}{k^2}.$$
A: Chebyshev's inequality holds for every distribution with finite $\mu$ and $\sigma$.
