# 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!

$\mathbb P(X > x)$ 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.

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.