A question about sum of probability... if $P(X\ge n)-P(X=n)=P(X>n)$

Question

If we know that $P(X\ge n)=(1-p)^{n-2}$
(This is not the main subject of the question, so I wont explain about it, hope this OK, but in sort: we get it because $P(X\ge n)=\sum_{k=n}^\infty (1-p)^{k-2}\cdot p$ ).

So,

1. It true that: $P(X\ge n)-P(X=n)=P(X>n)$?
2. $P(X>n)=(1-p)^{n-1}$ if we use what was given at the beginning (because instead of $k=n$ we put $k=n+1$) ?

I add one more Q:
So: $P(X=n)=(1-p)^{n-2}-(1-p)^{n-1}$???

If I'm wrong please fix my mistake, anyway, I have to find a way to calculate $P(X>n)$...

Thank you!

P.S. - The question is about discrete probability...

Answer 1

If your distibution is discrete, that is not continuous, we have that $$ P(X = n) = P(X\ge n) - P(X\ge n+1) =P(X\ge n) - P(X>n), $$ so indeed the first equation is true. Actually it is also true for continuous distributions, but also useless as $P(X=x)=0$ for all $x$.

Answer 2

If you are doing discrete probability, then those statements are valid. Otherwise, they are technically correct, but do not really mean anything as continuous probability makes no distinction between "less than" and "less than or equal to."

Answer 3

If $A$ and $B$ are disjoint events, i.e. $A\cap B=\emptyset$, then $P(A\cup B)=P(A)+P(B)$. Applying that on $A=\{X>n\}$ and $B=\{X=n\}$ gives: $$P\{X\ge n\}=P\{X>n\}+P\{X= n\}$$

By the same reasoning $P\{X> n\}$ will equalize $P\{X\ge n+1\}$ if and only if $P\{X\in (n,n+1)\}=0$. So if that condition is satisfied then:$$P\{X\ge n\}=P\{X\ge n+1\}+P\{X= n\}$$