Let $X:\Omega \to \mathbb N$ be a random variable on probability space $(\Omega,\mathcal B,P)$ .show that $$E(X)=\sum_{n=1}^{\infty}P(X\ge n).$$
my definition from $E(X)$ is equal $$E(X)=\int_{\Omega}XdP.$$ $\mathcal B$ is Borel $\sigma $-algebra
Thanks.
I think we can write $E(X) = \sum_i x_i \cdot P(X = x_i)$.
$P( X \ge i ) = P( X = i ) + P( X = i + 1 ) + \dots$ and get summation from both side
but this idea for discrete but in question do not mention X is discrete or Continuous.
Assume $X$ is absolutely continuous since you've seen the proof in the discrete case.
Let $H=\{(e,t)\in\Omega \times \mathbb{R} : 0\leq t \leq X(e)\}$ i.e. the area under the graph.
Then $E[X]=\int_0^\infty P(X\geq t) dt$. (your equality is in this case an inequality as will be seen later). Consider the projection functions $p_1: \Omega \times \mathbb{R} \mapsto X$ and $p_2: \Omega \times \mathbb{R}\mapsto \mathbb{R}$ then we can write $$H=p_2^{-1}([0,\infty))\cap (f\circ p_1-p_2)^{-1}([0,\infty))$$
To see H is a $\mathcal{B}\times\mathcal{B}(\mathbb{R})$ measurable set. Define now $$H_x=\{t\in \mathbb{R} : (e,t) \in \mathbb{R} \}=\{t\in \mathbb{R} : 0\leq t \leq X(e) \}=[0,X(e)]$$ and $$ H^t=\{e\in \Omega : (e,t)\in H \}=\{f\geq t\}$$ if $t\geq 0$ and the empty set otherwise (remember this to when we insert it in a second).
Now ($\lambda$ is the Lebesgue measure) using this rule $$P\otimes\lambda(H)=\int_\Omega \lambda(H_x) dP=\int_\Omega \lambda([0,X(e)]) dP=\int_\Omega X(e) dP=E[X]$$ Duh, but doing it the other way around yields $$P\otimes\lambda(H)=\int_R P(H^t) dt=\int_0^\infty P(X\geq t) dt$$ So $$\int_0^\infty P(X\geq t) dt = E[X]$$ Since $t\mapsto P(X\geq t)$ is decreasing it is measurable and the mentioned inequality $$E[X]\geq \sum_{k=1}^\infty P(X\geq k )$$ follows directly from the observation that for $t\in(0,\infty)$ $$P(X\geq t)\geq \sum_{k=1}^\infty P(X\geq k)1_{(k-1,k]}(t)$$ integrating it and interchanging sum and integration.
