In general when $X$ is an exponential random varaible, the memoryless property is stated as $$\mathbb{P}( X> s+t | X> s ) = \mathbb{P}(X > t).$$
But a direct computation shows if $S$ is also a exponential random variable, then $$\mathbb{P}( X> S+t | X> S ) = \mathbb{P}(X > t)$$is true as well. If $X\sim Exp(\lambda)$, $S\sim Exp(\mu)$, then $$\begin{align} \mathbb{P}( X> S+t, X> S ) &= \mathbb{P}( X> S+t) \\ & = \int_{0}^\infty \int_{s+t}^\infty \lambda e^{-\lambda x} \mu e^{-\mu s}dxds\\ &= \int_{0}^\infty \mu e^{-\mu s} e^{-\lambda (s+t)} ds\\ &= \frac{\mu}{\mu+\lambda} \cdot e^{-\lambda t} \bigg[\int_{0}^\infty (\mu+ \lambda) e^{-(\mu+\lambda) s} ds\bigg]\\ &= \mathbb{P}(X>S) \mathbb{P}( X>t) \end{align}$$So is there a generalized version of this property? And why don't we introduce it as part of the standard definition.
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$\begingroup$There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+t\mid X> S) \\= \frac{P(X>S+t)}{P(X>S)}\\=\frac{\int P(X>s+t)f_S(s)ds}{\int P(X>s)f_S(s)ds} \\=\frac{\int P(X>t)P(X>s)f_S(s)ds}{\int P(X>s)f_S(s)ds} \\=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
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