Memoryless property of the exponential distribution 
If $X\sim \exp(\lambda)$, then $E[X\mid X>20]=20+E[X]$

Can anyone give an intuitive proof of this property? In the context that  $X$ is exponential and thereby memoryless. I am having trouble in visualizing continuous memoryless distributions. Any intuitive example for the same is also appreciated.  
 A: An exponential distributed random variable is the measure of waiting time until the arrival of some event, the likes of which occur independently and at some constant average rate (ie: a Poisson Process).
This means that at every point in time the chance of such an event happening is the same as every other point in time, regardless of whether another such event happened recently or not.
So the expected waiting time until the next event is not determined by when you start waiting.
So if you wait for twenty units with no event occurring, you are facing the exact same situation you were twenty units ago; only now with the added knowledge that 20 units have passed without an event.   That 20 units doesn't affect the expected wait time until the next event there after; but if you are given knowledge of it ahead of time (or in retrospect), then...

Or by calculus.
$$\begin{align}
\mathsf E(X\mid X>20) & = \dfrac{\int_{20}^\infty \lambda t e^{-\lambda t}\operatorname d t}{\int_{20}^\infty \lambda e^{-\lambda t} \operatorname d t}
\\[1ex] & = \dfrac{e^{-20 \lambda}(20\lambda +1 )}{\lambda}\dfrac{1}{e^{-20\lambda}}
\\[1ex] & = 20+\frac 1 \lambda
\\[1ex] & = 20+\mathsf E(X)\end{align}$$
A: Memorylessness says the conditional distribution of $X-20$ given $X>20$ is the same as the distribution of $X$.
Therefore the conditional expected value of $X-20$ given $X>20$ is the same as the expected value of $X$:
$$
\operatorname{E} (X - 20 \mid X>20) = \operatorname{E}(X).
$$
But
$$
\operatorname{E} (X - 20 \mid X>20) = \operatorname{E} (X \mid X>20) - \operatorname{E} (20 \mid X>20) = \operatorname{E} (X \mid X>20) - 20.
$$
Hence
$$
\operatorname{E} (X) = \operatorname{E} (X-20 \mid X>20) = \operatorname{E} (X \mid X>20) - 20.
$$
So
$$
\operatorname{E} (X) + 20 = \operatorname{E} (X \mid X>20).
$$
A: The intuitive answer is that memoryless means that if you are waiting for an event that has not yet happened, it doesn't matter how long you have been waiting.  If you get to $x=20$ and it hasn't happened, the expected time until it happens is the same from now as it was at the start.  Since you have already waited $20$ so far, the expected total time is $20+E(X)$.  The fact that we are now at $20$ and it hasn't happened is the definition of the $|X\gt 20]$ in your expected value, so $E[X|X \gt 20]=20+E[X]$  
To do it mathematically, the pdf of the exponential distribution is $\lambda e^{-\lambda x}$.  The chance it doesn't happen by $x=20$ is $1-\int_0^{20}\lambda e^{-\lambda x}dx=e^{-20 \lambda}$  Plug this into your formula and observe that $E[X|X \gt 20]=20+E[X]$
