Chcę pokazać

Niech będzie zmienną losową w przestrzeni prawdopodobieństwa że$X:\Omega \to \mathbb N$$(\Omega,\mathcal B,P)$

$$E(X)=\sum_{n=1}^\infty P(X\ge n).$$

moja definicja z jest równa $E(X)$

$$E(X)=\int_\Omega X \, dP.$$

Dzięki.

Definicja $E(X)$ dla dyskretnego $X$ to $E(X) = \sum_i x_i \cdot P(X = x_i)$ .

$$P( X \ge i ) = P( X = i ) + P( X = i + 1 ) + \cdots$$

Więc

$$\begin{align} & \sum_i P( X \ge i ) = P( X \ge 1 ) + P( X \ge 2 ) + \cdots \\[8pt] = {} & P( X = 1 ) + P( X = 2 ) + P( X = 3 ) + \cdots + P(X = 2 ) + P( X = 3 ) + \cdots \end{align}$$

(zmieniamy terminy w ostatnim wyrażeniu)

$$\begin{align} & = 1 \cdot P( X = 1 ) + 2 \cdot P( X = 2 ) + 3 \cdot P( X = 3 ) + \cdots \\[8pt] & = \sum_i i \cdot P( X = i ) \end{align}$$

co było do okazania

Lubię styczniową odpowiedź. Czy mogę zasugerować sposób, aby zapisać serię, aby oko łatwiej uchwyciło zmianę układu (w ten sposób lubię pisać na tablicy)?

$$\begin{eqnarray} \sum_{k=1}^\infty P(X\geq k) &=& \quad P(X\geq 1) \quad=\quad P(X=1)&+&P(X=2)&+&P(X=3) &+& \;\dots\\ &+& \quad P(X\geq 2) &+& P(X=2) &+&P(X=3)&+& \;\dots \\ \\ &+& \quad P(X\geq 3) && &+& P(X=3)&+& \;\dots \\ \\ &+& \quad\quad\;\; \dots && &&&+& \;\dots\\ \end{eqnarray}$$ (The rearrangement is mathematically sound because this is a series of positive terms.)

I think the standard way of doing this is by writing

$$X =\sum_{n=1}^\infty \mathbf{1}(X\ge n)$$

$$E(X) =E\left(\sum_{n=1}^\infty \mathbf{1}(X\ge n)\right)$$

and then reverse order of expectation and sum (by Tonelli's theorem)

One of the other excellent answers here (from seanv507) has noted that this expectation rule actually follows from a stronger result that expresses the underlying random variable as an infinite sum of indicator variables. It is possible to prove a more general result, and this can be used to get the expectation rule in the question. If $X: \Omega \rightarrow \mathbb{N}$ (so its support is no wider than the natural numbers) then it can be shown (proof below) that:

$$X = \sum_{n=1}^{\max(X,m)} \mathbb{I}(X \geqslant n) \quad \quad \quad \text{for all } m \in \mathbb{N}.$$

Taking $m \rightarrow \infty$ then gives the useful result:

$$X = \sum_{n=1}^{\infty} \mathbb{I}(X \geqslant n).$$

It is worth noting that this result is stronger than the expectation rule in the question, since it gives a decomposition for the underlying random variable, and not just its moment. As noted in the other answer, taking expectations of both sides of this equation, and applying Tonelli's theorem (to swap the order of the sum and expectation operators), gives the expectation rule in the question. This is a standard expectation rule that is used when dealing with non-negative random variables.

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The above result can be proved fairly simply. Begin by observing that:

$$X = \underbrace{1+1+ \cdots +1}_{ X \text{ times}} + \underbrace{0+0+ \cdots +0}_{ \text{countable times}}.$$

For any $m \in \mathbb{N}$ we therefore have:

$$\begin{equation} \begin{aligned} X &= \underbrace{1+1+ \cdots +1}_{ X \text{ times}} + \underbrace{0+0+ \cdots +0}_{ \max(0,m-X) \text{ times}} \\[6pt] &= \sum_{n=1}^X \mathbb{I}(X \geqslant n) + \sum_{n=1}^{\max(0,m-X)} \mathbb{I}(X \geqslant X+ n) \\[6pt] &= \sum_{n=1}^X \mathbb{I}(X \geqslant n) + \sum_{n=X+1}^{\max(X,m)} \mathbb{I}(X \geqslant n) \\[6pt] &= \sum_{n=1}^{\max(X,m)} \mathbb{I}(X \geqslant n) . \\[6pt] \end{aligned} \end{equation}.$$