# Memoryless property: how is $F^c(2) = F^c(1)F^c(1) = (F^c(1))^2$ and $F^c(1/2) = (F^c(1))^{1/2}$ implied?

I am currently studying the textbook Modeling and analysis of stochastic systems, third edition, by Kulkarni. Chapter 5.1.1 Memoryless Property says the following:

We begin with the definition of the memoryless property.

Definition 5.2 A non-negative random variable $$X$$ is said to have the memoryless property if $$P(X > s + t \mid X > s) = P(X > t), \ \ \ \ \ s, t \ge 0. \tag{5.3}$$

Theorem 5.1 Memoryless Property. A continuous non-negative random variable has memoryless property if and only if it is an $$\exp(\lambda)$$ random variable for some $$\lambda > 0$$.

Proof: We first show the "if" part. So suppose $$X \sim \exp(\lambda)$$ for some $$\lambda > 0$$. Then, \begin{align} P(X > s + t \mid X > s) &= \dfrac{P(X > s + t, X > s)}{P(X > s)} \\ &= \dfrac{P(X > s + t)}{P(X > s)} = \dfrac{e^{-\lambda(s + t)}}{e^{-\lambda s}} \\ &= e^{-\lambda t} = P(X > t). \end{align} Hence, by definition, $$X$$ has memoryless property. Next we show the "only if" part. So, let $$X$$ be a non-negative random variable with complementary cdf $$F^c(x) = P(X > x), \ \ \ \ x \ge 0.$$ Then, from Equation 5.3, we must have $$F^c(s + t) = F^c(s) F^c(t), \ \ \ \ s, t \ge 0.$$ This implies $$F^c(2) = F^c(1)F^c(1) = (F^c(1))^2,$$ and $$F^c(1/2) = (F^c(1))^{1/2}.$$ In general, for all positive rational $$a$$ we get $$F^c(a) = (F^c(1))^a.$$

I don't understand this part:

This implies $$F^c(2) = F^c(1)F^c(1) = (F^c(1))^2,$$ and $$F^c(1/2) = (F^c(1))^{1/2}.$$ In general, for all positive rational $$a$$ we get $$F^c(a) = (F^c(1))^a.$$

Since it was said that $$s, t \ge 0$$, why is it implies that we have $$s = t = 1$$ for $$F^c(2) = F^c(1)F^c(1) = (F^c(1))^2$$? It seems to me that we could just as validly have $$s = 0$$ and $$t = 2$$, or vice-versa, so that $$F^c(2) = F^c(0)F^c(2)$$. Furthermore, how is $$F^c(1/2) = (F^c(1))^{1/2}$$ implied?

• The argument on the rational numbers is to justify that the solutions of the so-called functional equation: $F^c(s+t) = F^{c}(s) F^{c}(t)$ for $s \geq 0$, $t \geq 0$ take the form $t \mapsto \rho^t$ for $\rho: = F^{c}(1)$. The relation holds when $t$ is a postive rational number $t := p /q$ and must hold for any positive number because $F^{c}(t)$ is continuous.
– Yves
May 4, 2021 at 12:24

The following relation holds for any $$s,t\geq 0$$ pair: $$F^c(s + t) = F^c(s) F^c(t)$$
So, you can safely substitute $$s=1, t=1$$ and derive a relation. You could also substitute $$s=0,t=2$$ as you mentioned if that helps you prove the only if part.
For the last equation, if you substitute $$s=t=1/2$$, you'll have $$F^c(1)=F^c(1/2)F^c(1/2)\rightarrow F^c(1/2)=\sqrt{F^c(1)}$$
• Oh, I see. So it wasn't saying that it is implied that we must have $s = t = 2$? May 4, 2021 at 13:19
• No, you can choose any pair satisfying the conditions, but $s=t=1$ is particularly useful for the proof. May 4, 2021 at 13:25