It is "well-known" that there are basically only two types of memoryless distributions: The exponential distribution (in the continuous case) and the geometric distribution (in the discrete case). I wish to show this result. That is, given any real, positive random variable, I want to deduce that there are only two possibilities: Either the random variable has an exponential distribution, or it has a geometric distribution.
So let a positive random variable $T$ be given. Let us suppose that it satisfies the memorylessness property: $ P(T > t + s ) = P(T>t) P (T> s)$ for all $t,s \geq 0$. (Let's also assume that $P(T>t)\neq0$ for all $t\geq 0$, as this seems to be a natural assumption.)
If I assume that $P(T=0)=0$, then it's OK, because I am able to deduce from that that then $T$ should have an exponential distribution. So I assume that $P(T=0)>0$. The goal now is to prove that $T$ has a geometric distribution. Let's put $p:=P(T=0)$. Then $p \in (0,1]$.
I then argue as follows. Because $T$ satisfies memorylessness, we have that $T-t \;
| \; T>t$ has the same distribution as $T$, for every $t\geq 0$.
But then I get $p=P(T=0) = P(T-1 = 0 \; | \; T>1) = P(T =1 \; | \; T>1) = 0$, contradiction. Yet this doesn't seem to make much sense.
Where did I make a mistake in my reasoning?