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

Question

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 $P(X \ge x)=1-P(X \lt x)$?

Answer 1

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 $P(X=x)>0$, $$P(X> x)< 1- P(X< x) $$

Answer 2

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.