Consider a geometric distributed random variable X with parameter p, i.e. $p(X = n) = (1 -p)^{n-1} . p$ for n $\geq$ 1.
Show that for all natural numbers n, m $\geq 1$ applies: $$p(X \geq n + m \text{| } X \geq n) = p(X \geq m-1)$$
On the left side I get $(1-p)^{m}$, which is $p(X \gt m)$ and not $p(X \geq m-1)$. What am I doing wrong?
I think the task is incorrect, in the last line of my post, i proved the right side, but the left side is not the same as in the task.
I thought at first that $P(X \gt m)$ and $P(X \geq m-1)$ are equivalent. but that's obviously not the case, is it?
My attempts to correct the task:
$P(X\geq n+m|X\geq n)=\frac{P(X\geq (n+m) \cap X\geq n)}{P(X\geq n)}=\frac{P(X \geq n+m)}{P(X\geq n)}=P(X \gt m)$
$\ P(X \geq n + m \mid X \gt n) = \frac{\ P(X \geq n + m)}{\ P(X \gt n)} =\frac{(1 - p)^{m+n-1}}{(1 - p)^{n}} = (1 - p)^{m-1} = \ P(X \gt m-1)$
$\ P(X \geq n+m-1 \mid X \gt n) = \frac{\ P(X \geq n+m-1 )}{\ P(X \gt n)} =\frac{(1 - p)^{m-1+n-1}}{(1 - p)^{n}} = (1 - p)^{m-1-1} = \ P(X \geq m-1)$
EDIT: This task contains a typo. $P(X \geq m-1)$ should be $P(X \geq m+1)$ so there is no solution to this problem. You can simply come up with the solution for the corrected task, I could provide my solution here later.
If you define distribution of Geometric rv as $P(X=n) = (1-p)^{n-1}p$, the probability $P(X \geq n+m| X\geq n)$ is $P(X \geq m-1) = (1-p)^m$. Keep in mind that partial geometric sum is $\sum_{k=0}^{n} x^k = \frac{1-x^{n+1}}{1-x}$
EDIT: $P(X > n) = 1 - P(X \leq n) = 1- \sum_{k=1}^{n}p (1-p)^{k-1}$ because the definition of Geometric rv I used counts all events, including the last one (success), so $P(X \leq k)$ means probability that there were $k-1$ failures and 1 success. Therefore, $P(X>n)$ means there were at least $n+1$ events, implying there were $n$ failure before it.