memoryless property of exponential distributions with random variables

It is true that $P(X>t+s|X>t)=P(X>s)$ for certain values $t$ and $s$.

However, how can I show that this still holds if:

• $T$ is a continuous random variable. That is $P(X>T+s|X>T)=P(X>s)$
• Both $T$ and $S$ are continuous RVs: $P(X>T+S|X>T)=P(X>S)$

All the random variables are independent.

I am assuming both $T$ and $S$ are independent of $X \sim exp(\lambda)$! If not, you must have some joint distribution.

Rough sketch: Suppose $f_T$ is the pdf of $T$. Then,

$\begin{eqnarray} P(X >T+s \mid X>T) &=& \frac{ P(X >T+s) }{P(X >T)} \\ &=& \frac{\int_o^\infty P(X > t+s \mid T=t ) f_T(t)\,dt}{\int_o^\infty P(X > t \mid T=t) f_T(t)\,dt} \\ &=& \frac{\int_o^\infty e^{-\lambda (t+s)} f_T(t)\,dt}{\int_o^\infty e^{-\lambda t} f_T(t)\,dt} \\ &=& e^{-\lambda s } \frac{\int_o^\infty e^{-\lambda t} f_T(t)\,dt}{\int_o^\infty e^{-\lambda t} f_T(t)\,dt} \\ &=& P(X>s) \end{eqnarray}$

For the second problem, assume $S$ and $T$ are independent. If not, you have to work with the joint distribution. Suppose $f_S$ is the pdf of $S$. Then,

$\begin{eqnarray} P(X >T+S \mid X>T) &=& \frac{ P(X >T+S) }{P(X >T)} \\ &=& \frac{\int_o^\infty \int_o^\infty P(X > t+s \mid T=t, S=s ) f_S(s) f_T(t)\,ds \,dt}{\int_o^\infty P(X > t \mid T=t) f_T(t)\,dt} \\ &=& \frac{\int_o^\infty \int_o^\infty e^{-\lambda (t+s)} f_S(s) f_T(t)\,dt}{\int_o^\infty e^{-\lambda t} f_T(t)\,dt} \\ &=& \int_o^\infty e^{-\lambda s } f_S(s) \,ds \frac{\int_o^\infty e^{-\lambda t} f_T(t)\,dt}{\int_o^\infty e^{-\lambda t} f_T(t)\,dt} \\ &=& \int_o^\infty e^{-\lambda s } f_S(s) \,ds \\ &=& P(X>S) \end{eqnarray}$