Can somebody provide an explanation of memoryless property in this example? The following example is taken from the book "Introduction to Probability Models" of Sheldon M. Ross (Chapter 5, example 5.4).

The dollar amount of damage involved in an automobile accident is an
  exponential random variable with mean 1000. Of this, the insurance
  company only pays that amount exceeding (the deductible amount of)
  400. Find the expected value and the standard deviation of the amount the insurance company pays per accident."

In the solution, the author states that: 

By the lack of memory property of the exponential, it follows that if
  a damage amount exceeds 400, then the amount by which it exceeds it is
  exponential with mean 1000.

After reading several implications of this property, I easily map such statement to something like: if you have been waiting for 400s without seeing the bus, then the expected time until the next bus is always 1000s. (Please correct me if I'm wrong)
In case I've understood well, what makes me confuse is this next equation:
$$
E[Y|I=1] = 1000
$$
where:
$X$:  the dollar amount of damage resulting from an accident
$Y=(X-400)^+$: the amount paid by the insurance company (where $a^+$ is $a$ if $a>0$ and 0 if $a<=0$).
$I = 1*(X > 400) + 0*(X<=400)$
I don't get why that equality holds given the memoryless property. Straightforwardly, I think with respect to 400 subtraction, it should be something like: $E[Y|I] = 1000 - 400 = 600$ (or some other value). Can anyone give me an explanation about this?
In case you are not clear about my description, please refer to this link with example 5.4.
 A: Comment:  The link didn't work on my computer. Here is a simulation of the situation
as I understand it from your description. 
A million accidents are simulated using R statistical software.
Those costs below \$ 400 are set to $0.$ The sample mean and SD (including the
$0$ costs) are shown. The histogram shows the simulated distribution very accurately, except
that the bin $(0,100)$ contains only $0$'s. This a very long-tailed distribution; the maximum payout in a million simulated accidents is almost \$ 14,500.
x = rexp(10^6, 1/1000)  # R uses rate (reciprocal of mean)
x[x<=400] = 0           # adjust for deductible
mean(x)
## 936.641              # aprx avg payment/accident
sd(x)
## 1047.132             # aprx SD payment/accident
mean(x==0)              
## 0.329995             # aprx fraction with 0 payment
pexp(400, 1/1000)
## 0.32968              # exact theoretical fraction with 0 payment
max(x)
## 14474.76             # max payment in a million sim accidents

You did not give Ross's answer
or yours. If my interpretation is right, results should agree to at least two
significant digits.

Note: The bus example is technically correct. If it does not seem
reasonable, that is because waiting times for buses are not actually
exponentially distributed in real life. The exponential distribution is
so easy to use (because of the no-memory property and because of it's
simple mathematical form) that there is a temptation to use it in
circumstances where it is not appropriate. (You may find a lot ot
textbook examples of that type; perhaps fewer in Ross's book than in most.)
I would have to see extensive actuarial evidence to believe that payouts
on car accidents are truly exponential. Totaling a new Mercedes would
cost someone a lot more than \$14,500.
A: Indeed. We are looking at: $$\begin{align}\mathsf E(Y \mid I=1) ~=~& \mathsf E(Y\mid X>400) &:~& I=\big[ X>400\big] \text{ an indicator function}
\\ ~=~& \mathsf E(X-400\mid X>400) &:~& Y=(X-400)^+ = \max(X-400, 0)
\\ =~& \mathsf E(X\mid X>400) - 400 &:~& \text{Linearity of Expectation}
\\ =~& (400+\mathsf E(X))-400 &:~& \mathsf E(X\mid X>400)=400+\mathsf E(X) \tag{1}
\\ =~& \mathsf E(X) &:~& \text{algebraic cancelation} \end{align}$$
$(1)$ is a consequence of the memoryless property since $X$ has an exponential distribution.   The distribution of damage above 400 is the same as the distribution of the damage above $0$.   So the expected total damage, when given that it is more than 400, is 400 plus the expected damage above 400.

This should lead on to : $$\mathsf E(Y) ~=~ \mathsf E(X)~\mathsf P(X>400)$$
Via the Law of Total Expectation and that $\mathsf E(Y\mid I=0)~=~0$. 
