Deriving the geometric distribution from memorylessness

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?

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• From glancing at Wikipedia's definition of discrete memorylessness it seems that you need to be careful with which inequalities are strict ($>$) and which are non strict ($\ge$). May 14 at 1:29
• Is there no "general" notion of 'memorylessness' then? May 14 at 2:43
• I think I solved the issue. The point is, according to 'my' definition of memorylessness (where the condition holds for all $t,s\geq0$), the geometric distribution is not a memoryless random variable. The reason is that if $T$ is geometrically distributed with parameter $p>0$, then $P(T>t+s)=P(T>t)P(T>s)$ holds for $s,t\geq0$ iff the sum of the fractional parts of $s$ and $t$ is strictly less than $1$. (In particular, it holds for integers.) So we must 'relax' the definition of memorylessness for discrete rvs. Following 'my' definition of memorylessness, the exponential is indeed unique. May 14 at 4:31