# I want to show $E(X)=\sum_{n=1}^{\infty}P(X\ge n)$ [closed]

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.

• Posted more than 5 hours earlier on stats.SE where it has already received an answer. Commented Nov 13, 2013 at 19:15
• @DilipSarwate. it answer not complete and I am Confused.if it possible for you help me to solve this question .thanks Commented Nov 13, 2013 at 19:48
• Explained $n$ times on this very site.
– Did
Commented Nov 13, 2013 at 20:02
• @Did.ok if it possible for you give me a link about this question.I am sure you are Excellent in probabilty. Commented Nov 13, 2013 at 20:10

## 1 Answer

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.