In this document at page 20, one can read $M(t)$ as being defined as the expected value of $N_t$ where $N_t$ is the number of occurence of an event at time=t. As demonstration of the result, the author says:
$$E[N_t] = \sum_{n=0}^{\infty}{P(N_t>n)}$$
I don't understand where it comes from. The expected value should be
$$E[N_t] = \sum_{n=0}^{\infty}{n P(N_t=n)}$$
How do you obtain the author's expression?
Think of it like this. Let a = P(N = 1), b = P(N = 2), and c = P(N = 3) and the only values N can take on are (1,2,3). Thus, if we use the standard definition for the expected value, we get the following expression:
E[N] = (a) + (b + b) + (c + c + c)
We can represent this sum as a right triangle
a
b b
c c c
Under the standard way, we are adding these up "row-wise", but what if we added them "column-wise"?
That would be equivalent to E[N] = (a + b + c) + (b + c) + (c)
(a + b + c) = P(N > 0)
(b + c) = P(N > 1)
(c) = P(N > 2)
Which is the alternative expression for expectation we desired.
Suppose $X$ is a discrete random variable with range $\mathbb{N}$. Then $$ \begin{split} \mathbb{E}[X] &= \sum_{n=0}^\infty n \mathbb{P}[X=n] \\ &= \sum_{n=1}^\infty \sum_{k=1}^n \mathbb{P}[X=n] \\ &= \sum_{k=1}^\infty \sum_{n=k}^\infty \mathbb{P}[X=n] \\ &= \sum_{k=1}^\infty \mathbb{P}[X \ge k] \\ &= \sum_{k=0}^\infty \mathbb{P}[X > k] \end{split} $$
Because...
$$E[X] = \sum_{n = 0}^{\infty} n P(X = n) = \sum_{n = 0}^{\infty} P(X > n) = \sum_{n = 0}^{\infty} (1 - P(X \leq n)),$$
The continuous analog is
$$E[X] = \int_{0}^{\infty} (1 - F_{X}(x)) \, dx = \int_{0}^{\infty} R(x) = \int_{0}^{\infty} P(X>x) \, dx,$$
where $R(x)$ is the reliablity function and is defined by $R(x) = 1 - F(x)$.
