Intuition behind Chebyshev's inequality Is there any intuition behind Chebyshev's inequality or is that only pure mathematics? What strikes me is that any random variable (whatever distribution it has) applies to that.
$$
    \Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.
  $$
 A: Square integrable variables are not any random variables. They are in fact pretty regular !
Once you know that your variable has a variance, it's natural that the distance to the mean of your variable can be controlled in probability by this variance. Chebyshev's inequality is probably the simplest way to achieve that.
A: It's useful to view Chebyshev's inequality as more of an application of Markov's inequality which for a nonnegative random variable $X$ and $\alpha > 0$ is given by,
$$
\begin{align}
P(X \geq \alpha) \leq \frac{\text{E}(X)}{\alpha} .
\end{align}
$$
(Notice how we arrive at Chebyshev's inequality by applying Markov's inequality to the event $\{(X - \mu)^2 \geq k^2 \sigma^2 \}$ which is equivalent to $\{|X - \mu| \geq k \sigma \}$ and therefore has the same probability.)
Now the intuition behind Markov's inequality is that there is an implicit relationship between probability and expectation, and that for nonnegative random variables knowing the expected value places certain constraints on the behavior of the tail.  That is, if one already knows how large $X$ is on average, then the probability of large values must be controlled or $\text{E}(X)$ will itself be "pulled" towards a larger value.
To illustrate, suppose that $\text{E}(X) = 1$.  Is it possible that $P(X \geq 2) > 1/2$?  Obviously not, because then $\text{E}(X) > 1$ and we've contradicted ourselves.
A: To me it means:
The further away the random variable from the mean is, the more seldom it becomes. k gives you the number of standard deviations (if taken as a natural number) and the probabilty will automatically be limited by $1/k^2$.
My intuition why this is a meaningful statement for all random variables is the following: The measure of the whole space is limited, namely = 1. You cannot fill up the whole space in the reals with positive values (measure will be inf), so the distribution must vanish on the sides.
A: 
The intuition is that if $g(x) \geq h(x) ~\forall x \in \mathbb R$, then $E[g(X)] \geq E[h(X)]$ for any random variable $X$ (for which these expectations exist). This is what one would intuitively expect: since $g(X)$ is always at least as large as $h(X)$, the average value of $g(X)$ must be
  at least as large as the average value of $h(X)$. 

Now apply this intuition
to the functions
$$g(x) = (x-\mu)^2 ~ \text{and}~ 
h(x)= \begin{cases}a^2,& |x - \mu| \geq a,\\0, & |x-\mu|< a,\end{cases}$$
where $a > 0$ and where $X$ is a random variable with finite mean $\mu$ and finite variance $\sigma^2$. This gives
$$E[(X-\mu)^2] = \sigma^2 \geq E[h(X)] = a^2P\{|X-\mu|\geq a\}.$$
Finally, set $a = k\sigma$ to get the Chebyshev inequality.


Alternatively, consider the variance $\sigma^2$as representing the moment of
  inertia of the probability mass about the center of mass (a.k.a. mean 
  $\mu$). The total probability mass $M$ in the region $(-\infty, \mu-k\sigma] \cup [\mu+k\sigma, \infty)$ that is far far away from the mean $\mu$ contributes
  a total of at least $M\cdot (k\sigma)^2$ to the sum or integral
  for $\sigma^2 = E[(X-\mu)^2]$, and so, since everything else
  in that sum or integral is nonnegative, it must be that 
  $$\sigma^2 \geq M\cdot (k\sigma)^2 \implies
 M = P\{|X-\mu| \geq k\sigma\} \leq \frac{1}{k^2}.$$

Note that for a given value of $k$,
 equality will hold in the Chebyshev inequality when there
are equal point masses of $\frac{1}{2k^2}$ at
$\mu \pm k\sigma$ and a point mass of $1 - \frac{1}{k^2}$ at
$\mu$. The central mass contributes nothing to the 
variance/moment-of-inertia-about-center-of-mass calculation
while the far-away
masses each contribute 
$\left(\frac{1}{2k^2}\right)(k\sigma)^2 = \frac{\sigma^2}{2}$
to add up to the variance $\sigma^2$
A: Find the worst case distribution, and everything else must have a lesser probability.
Since all that matters is whether a point is inside or outside the ball of radius $k \sigma$ centered on $\mu$, we should make sure all of the probability mass inside the ball is at $\mu$, and everything outside the ball is exactly on the boundary $|x - \mu| = k \sigma$; doing so would minimize their contribution to the standard deviation, thus letting us place as much mass outside of the ball as we can.
i.e. we should consider the distribution
$$ P(X = x) = \begin{cases} 1 - \rho & x = \mu
\\ \rho/2 & x = \mu \pm k \sigma
 \\ 0 & \text{otherwise} \end{cases} $$
where $\rho = P(|X-\mu| \geq k \sigma)$.
This has mean $\mu$ and standard deviation $k \sigma \sqrt{\rho}$, and thus $\rho = 1/k^2$.
(Making this rigorous would require either proving that this is the worst case. Of course, once we know what the answer should be, it may be easier to prove it more directly)
A: First of all, putting $\mu$'s and $\sigma$'s all over is ridiculous -- like natural units in physics, let's set $\mu=0$ and $\sigma=1$. Then Chebyshev's inequality states that:
$$P(|X|>k)\leq1/k^2$$
I.e. for a distribution to have a unit standard deviation, there is a natural limit on how much of the distribution can be some given amount $k>1$ away from the mean. This isn't particularly surprising -- as you add stuff to the distribution outside standard deviation range, you inevitably increase the standard deviation. In order to keep the standard deviation at 1, you need to squeeze the things inside the standard deviation range and reduce their deviation, so the overall deviation stays at 1.
But there's got to be a limit on how much you can reduce the total deviation, right? You can't make the contribution of the things inside to the deviation negative -- it can only go down to zero. So what exactly is this limiting case?
Well, at the limiting case, you have a Dirac delta at $X=0$ and two Dirac deltas at $X=k$ and $X=-k$. What's the maximum height of these two Dirac deltas? Answer this, and you have Chebyshev's inequality.

The standard deviation of this distribution is --
$$\sqrt{\frac{p}2(-k)^2+\frac{p}2k^2}=k\sqrt{p}$$
We set this equal to 1, and we get $p=1/k^2$ as the maximum amount of stuff you can pile at $k$ and beyond even if you use the distribution with the least possible standard deviation. 
(A full proof would consider the possibility of an asymmetric distribution with a $p$ stick at $X=k$ and a $1-p$ stick at $X=-\frac{p}{1-p}k$, but it turns out the limit on $p$ is even stricter, at $p\leq\frac1{k^2+1}$.)
