# When does the inequality change in probability problems and why does it?

From what I've learned in my stats class when you have $P(X \gt x)$ you make it $P(X \gt x) = 1-P(X \le x)$. I think I understand that you have to do this inequality change because you can't calculate a probability of any possible number being greater than $x$, but I don't understand why greater than changes to less than or equal to. Why is this the case?

Also, does it work that way in reverse? Is $P(X \ge x)=1-P(X \lt x)$?

Yes in does work in the reverse way... This and the fact that it is less or equal in your formula are both a straightforward consequence of the definition of a probability ; in particular if $A\cap B=\emptyset$, $$P(A\cup B)= P(A) + P(B)$$ and $$P(\Omega)=1$$

Indeed with $A=\{X>x\}$ (the event where $X$ is strictly bigger than $x$) and with $B=\{X\leq x\}$ (the event where $X$ is less or equal than $x$), you have $A\cap B=\emptyset$ and $A\cup B = \Omega$ (it is indeed the whole probability space, since this event $X$ is strictly bigger than $x$ or $X$ is less or equal than $x$ will always happen), so using the preceding rules, $$1= P(\Omega)= P(A\cup B)= P(A) + P(B)= P(X>x)+P(X\leq x)$$ so $$P(X>x)= 1- P(X\leq x).$$ But with $A'=\{X\geq x\}$ and $B'=\{X< x\}$ you will obtain $$1= P(\Omega)= P(A'\cup B')= P(A') + P(B')= P(X\geq x)+P(X< x)$$ so, again, $$P(X\geq x)= 1- P(X< x).$$

Now, if you just take $A''=\{X> x\}$ and $B''=\{X< x\}$ you will miss the event $C''=\{X=x\}$, since you will have $$1= P(\Omega)= P(A''\cup B''\cup C'')= P(A') + P(B')=P(C'')= P(X> x)+P(X< x)+P(X=x)$$ and if $P(X=x)>0$, $$P(X> x)< 1- P(X< x)$$

• I don't think I've ever seen $P(\Omega)=1$, what does this mean? Mar 3, 2018 at 21:54
• @Grace it is the "Universe" or all the events : a random variable $X$ is a function from $\Omega$ into to some space (usually $\mathbb{R}$, $\mathbb{N}$, etc..) Mar 3, 2018 at 21:57
• I know this wasn't part of my original question, but reading all of this made me realize I don't really understand why we we subtract the inverse probability from 1. I know that we subtract probabilities if we have an event A and we subtract that from 1 if we want to know the probability it will not occur, is this the same concept? I can kind of see how that would be true but it's hard for me to grasp. Mar 3, 2018 at 22:15
• @Grace yes it is exactly the same concept and the same proof : you take $A$ an event and $A^c$ the complementary event then (since, by definition $A\cap A^c=\emptyset$),$$1= P(\Omega)= P(A\cup A^c)= P(A) + P(A^c)= P(A)+P(A^c)$$ then $$P(A)=1-P(A^c)$$ which is 1 minus the probability of that $A$ will not occur... Mar 3, 2018 at 22:29
• @Grace Yes it is the same concept. Every event $A$ corresponds with an event $A^{\complement}$. This is the so-called complement of $A$. If $A$ is the event that tomorrow it will rain then $A^{\complement}$ is the event that tomorrow it will not rain. Note that exactly one of the events will occur. $A\cup A^{\complement}=\Omega$ corresponds with the fact that at least one will occur, and $A\cap A^{\complement}=\varnothing$ with the fact that not both events can occur. That gives $P(A)+P(A^{\complement})=1$ as explained in this answer. Mar 3, 2018 at 22:29

This equation is based on the fact that the sum of the probabilities for all possible values of $x$ is $1$ and that values of $x$ have order imposed on them. Then it is easy to see that $$\sum P(x)=1$$ and that given any given any $x$, you can divide the set of all possible values of $x$ into $3$ disjoint sets, the set which has $S_1 = \{ X | X < x \}$, $S_2 = \{x\}$ and $S_3 = \{X | X > x\}.$ So $P(S_1)+P(S_2)+P(S_3) = 1$.

Consequently $P(S_3) = P(X > x) = 1-P(S_1)-P(S_2) = 1- P(X<x)-P(X=x) = 1-P(X\leq x)$ From, this, it is also easy to see that if the same assumptions are met, the reverse can also be done.