# Memoryless property and geometric distribution

Suppose $X$ is a random variable taking values in $\mathbb N_0$ with the memoryless property,i.e., for each pair of number $s,t \in \mathbb N$, $$P(X\geq s+t\mid X>t)=P(X\geq s)$$

Show that a random variable $X$ with values in $\mathbb N_0$ has the memoryless property if and only if $X \sim \text{Geometric}(p)$ of parameter $p=P(X=1)$.

I could show that if $X \sim \text{Geometric}(p)$ (for an arbitrary parameter $p$), then $X$ has the memoryless property. I'll write that part of the answer:

$$P(X\geq s)=\sum_{i=s}^{\infty} P(X=i)$$$$=\sum_{i=s}^{\infty} (1-p)^{i-1}p$$$$=\sum_{i=0}^{\infty} (1-p)^{i-1}p-\sum_{i=0}^{s-1}(1-p)^{i-1}p$$$$=\frac{p}{1-p}(\frac{1}{1-(1-p)}-\frac{1-(1-p)^s}{1-(1-p)})$$$$=(1-p)^{s-1}.$$

Similarly, $$P(X\geq s+t\mid X>t)=\dfrac{P((X\geq s+t)\cap (X>t))}{P(X>t)}$$$$=\dfrac{P(X\geq s+t)}{P(X>t)}.$$ After calculating the probability of the numerator and the probability of the denominator, one can arrive to the same expression $$\dfrac{P(X\geq s+t)}{P(X>t)}=(1-p)^{s-1}.$$

So from here one deduces that the geometric random variable has the memoryless property.

I got stuck trying to show the other implication:

Let $s \in \mathbb N, t=1$, let $E=\{X>1\}$, then $$P(X\geq k+1)=P(X \geq k+1\mid X>1)P(X>1)+P(X\geq k+1 \mid X \leq 1)P(X \leq 1).$$

Since the probability $P(X\geq k+1 \mid X \leq 1)=0$, then $$P(X\geq k+1)=P(X \geq k+1\mid X>1)P(X>1)$$

Using the hypothesis, we have $$P(X \geq k+1)=P(X \geq k+1\mid X>1)P(X>1)=P(X \geq k)P(X>1)$$

I got stuck at that point, I want to conclude that $P(X=k)=(1-p)^{k-1}p$, where $p=P(X=1)$.

I would appreciate some help.

We have $$P(X\geq s+t\mid X>t)=P(X\geq s)$$ and $$P(X\geq s+t\mid X>t)=\dfrac{P(X\geq s+t)}{P(X>t)}$$ therefore$$P(X\geq s+t)=P(X\geq s)P(X>t).$$ Summing the last equality once over $s$ then once over $t$ we obtain $$I: \sum_{s=1}^{\infty}P(X\geq s+t)=\sum_{s=1}^{\infty}P(X\geq s)P(X>t)$$ $$II: \sum_{t=1}^{\infty}P(X\geq s+t)=\sum_{t=1}^{\infty}P(X\geq s)P(X>t)$$ Now in $I$ set $t=k$, and set $s=k$ in $II$ so that their left hand side be equal and consequently $$P(X\geq k)\sum_{t=2}^{\infty}P(X\geq t)=P(X\geq k+1)\sum_{t=1}^{\infty}P(X\geq t).$$ Now let $\alpha=\sum_{t=2}^{\infty}P(X\geq t)$ and evaluate for $k=1$ to obtain, $$\alpha=\frac{1-p}{p},$$where $p=P(X=1).$ Therefore $(1-p)P(X\geq k)=P(X\geq k+1)$. Since $P(X\geq 1)=1$ we have that $$P(X\geq k)=(1-p)^{k-1}; k=1,2,...$$ implying that $$P(X=k)=(1-p)^{k-1}-(1-p)^{k}=p(1-p)^{k-1}.$$Therefore $X\sim$ Geometric$(p)$.
• I know this is a pretty old question and answer, but how would I proceed if instead of $P(X\geq s+t)=P(X\geq s)P(x>t)$ I had $P(X>t)=P(X>s)P(X>t-s)$? Aug 3, 2016 at 19:05
• @implicati0n Since the random variable only takes values in $\mathbb N$ we have $P(X \ge s) = P(X > s - 1)$. May 8, 2020 at 22:31