# $P(X > \mu + \sigma) > 0$?

I'm doing some data analysis. Let $X$ follow an unknown distribution, with a certain average $E[X]$ and a standard deviation $\sigma$. Both values are calculated using the (corrected) point estimators.

The data is not normal, and let's assume that the underlying distribution is continuous. Now I want to know if we can say anything about $P(X > E[X] + \sigma)$. My intuition says that this is always strictly positive. But I don't know if this is true. The name standard deviation and the formula make me think that this should be possible. So, what I actually want to know is: the fact that I found a standard deviation $\sigma$, does that imply I have a value $d \ge E[X] + \sigma$ in my dataset?

More generally, is it possible to say anything about $P(X > E[X] + k\sigma)$, with $k > 0$, being greater than some value?

(Wikipedia has a list of about 50 inequalities about probability... I checked Chebyshev, but the sign in the inequality is in the wrong way for my question.)

Suppose you have a Bernoulli random variable where $X=1$ with probability $p$ and $X=0$ with probability $1-p$. Then $E[X]=p$ and $\sigma_X= \sqrt{p(1-p)}$.
For any $k \gt 0$, if $p=\frac{1}{k^2+1}$ then $E[X]=\frac{1}{k^2+1}$ and $\sigma_X= \frac{k}{k^2+1}$ so $E[X]+k\sigma_X=1$ meaning that in this example $P(X\gt E[X]+k\sigma_X)=0$.
Strictly speaking this is a discrete random variable, but any discrete distribution can be approximated arbitrarily closely by the distribution of a continuous random variable. So the answer is no, unless you have more information about the distribution of your $X$.
• It does not matter: give me a $k$ and I will then choose a $p$ less than $\frac{1}{k^2+1}$ and then adjust my discrete distribution into a continuous distribution with very narrow uniform intervals around $0$ and $1$. On the cumulative distribution graph, I can turn discontinuous jumps into continuous almost-vertical lines with arbitrarily small impacts on $E[X]$ or $\sigma_X$. Feb 14, 2015 at 12:46