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