Explanation of $\mathbb P(X>x)$ I'm studying random variables right now and trying to wrap my head around the idea of $\mathbb P(X>x)$. I understand the idea that $\mathbb P(X=x)$ is the probability of our random variable equalling some outcome, but I'm having a little trouble grasping the inequality version. If someone could maybe provide an example/help me with some intuition that would be fantastic.
Thanks!
 A: $\mathbb P(X > x)$ is simply the probability of the random variable $X$ being larger than some value $x$. For example, suppose a random variable $X$ can take the values $1$, $2$, $3$, or $4$ with the following probabilities:
\begin{matrix}
\text{Outcome}&1&2&3&4\\
\text{Probability}&0.1&0.3&0.4&0.2
\end{matrix}
What is $\mathbb P(X > 2)$? It is simply the sum of probabilities of all values greater than $2$, i.e., $$\mathbb P(X > 2) = \mathbb P(X = 3) + \mathbb P(X = 4) = 0.4 + 0.2 = 0.6$$
Notice that an alternative way to find $\mathbb P(X > x)$ is
$$\mathbb P(X > x) = 1 - \mathbb P(X \leq x)$$
So, now, we can find $\mathbb P(X > 2)$ as
$$\mathbb P(X > 2) = 1 - \mathbb P(X \leq 2) = 1 - \left[\mathbb P(X = 1) + \mathbb P(X = 2)\right] = 1 - [0.1 + 0.3] = 1 - 0.4 = 0.6$$
Notice that in both methods, the answer is the same. However, the method you choose may vary depending on the type of problem you are trying to solve.
A: Actually, the probability is defined on sets from a sigma-algebra on the universe (or sample space) $\Omega$. $P(X=x)$ is a shortcut for $P(\{\omega\in\Omega|X(\omega)=x\})$.
Likewise, $P(X>x)$ is a shortcut for $P(\{\omega\in\Omega|X(\omega)>x\})$.
A event is an element of this sigma-algebra.
A real random variable is simply a (measurable) function from $\Omega$ to $\Bbb R$.
Now, for the intuitive part, throw a die, and denote by $X$ the random variable being the number that appears on the top of the die, modulo $3$. The universe if $\Omega=\{1,2,3,4,5,6\}$. The sigma-algebra is the powerset of $\Omega$ and the random variable is the function defined by


*

*$X(1)=1$

*$X(2)=2$

*$X(3)=0$

*$X(4)=1$

*$X(5)=2$

*$X(6)=0$


What is $P(X>0)$? Intuitively, that means the "probability that $X$ is greater than $0$", or if you prefer the "probability that $X$ is $1$ or $2$", since $X$ can only take the values $0,1,2$.
More formally, the event $X>0$ is $\{1,2,4,6\}$, so really,
$$P(X>0)=P(\{1,2,4,6\})$$
Notice that I can't tell you the exact value of this probability, because I have still not defined it: you have to define a function (more precisely, a probability measure $P$ on the sigma-algebra described above. In the finite case, you can define the probability of every elementary event, that is $P(\{1\}), P(\{2\}), \dots, P(\{6\})$. It's quite customary in the case of a die to define $P(\{\omega\})=\frac16$ for all $\omega\in\Omega$, but it's not mandatory. If you choose this probability, then
$$P(X>0)=P(\{1,2,4,6\})=\frac46=\frac23$$
Likewise,
$$P(X=0)=P(\{3,6\})=\frac26=\frac13$$
Of course, you can check that $P(X=0)+P(X>0)=1$ here, as the events are the complement of one another.
