TITLE: Show the following pdf is memoryless
QUESTION [5 upvotes]: I've been thinking about this around 2 weeks for the midterm. but still can't prove it.
I used this
$$ P(X > r+s | X > s) = P(X > r) = \mathrm e^{−\lambda r}$$
$$P(X > r + s) / P(x >s ) = \mathrm e^{−\lambda(r+s)} / \mathrm e^{−\lambda s}$$
Could you please give me some hint? how can I prove ? Thanks.

REPLY [3 votes]: First we do it for random variable $X$ equipped with exponential distribution with parameter $\lambda=1$. 
$$P(X>r+s\mid X>s)=\frac{P(X>r+s\wedge X>s)}{X>s}=\frac{P(X>r+s)}{X>s}=$$$$\frac{e^{-r-s}}{e^{-s}}=e^{-r}=P(X>r)$$
Based on this result we can prove the same for $Y:=\lambda X$ (or if you wish $Y:=\lambda^{-1}X$, there is no essential difference).
This by:$$P\left(Y>r+s\mid Y>s\right)=P\left(X>\frac{r}{\lambda}+\frac{s}{\lambda}\mid X>\frac{s}{\lambda}\right)=P\left(X>\frac{r}{\lambda}\right)=P\left(Y>r\right)$$
Now realize that $Y$ has the pdf mentioned in your question.
For $x>0$ we have $F_X(x)=1-e^{-x}$ and consequently $F_Y(x)=1-e^{-\frac{x}{\lambda}}$ so that $f_Y(x)=\frac1{\lambda}e^{-\frac{x}{\lambda}}$.