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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 $$F^c(s + t) = F^c(s) F^c(t), \ \ \ \ s, t \ge 0.$$

How does Equation 5.3 imply that $F^c(s + t) = F^c(s) F^c(t), s, t \ge 0$?

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    $\begingroup$ Have you tried to apply any definition of conditional probability? $\endgroup$ – whuber May 3 at 12:17
  • $\begingroup$ @whuber en.wikipedia.org/wiki/Conditional_probability which property are you referring to? It is not clear to me that there are any here that work for this. $\endgroup$ – The Pointer May 3 at 12:20
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    $\begingroup$ @whuber Also, I do not understand why this was closed. It seems to me that this is a perfectly good and valid question... $\endgroup$ – The Pointer May 3 at 12:21
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    $\begingroup$ @whuber. There are few things that puzzle me concerning why this question was closed, and I was hoping you can supply me with some clarity. On what basis has it been determined that this question is a self-study question? $\endgroup$ – microhaus May 3 at 15:37
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    $\begingroup$ @ThePointer Here's a a nudge in the right direction. Use the definition P(A|B) = P(A and B)/P(B), and P(A and B) = P(A) if A is a subset of B. $\endgroup$ – Double Ought Not May 3 at 17:32