sequence of discrete random variables - exchange expectation and countable summation

Question

For a sequence of non-negative random variables $(X_n)_{n\geq1}$ holds $$\mathbb E\left[\sum_{n\geq 1}X_n\right]=\sum_{n\geq 1}\mathbb E[X_n].$$

Is this because one can define an increasing sequence of r.v.s $Y_n:=\sum_{k=1}^nX_k$? What would be the argument, if i do not know, whether the sequece of $X_n$ converges or not?

Answer 1

The sequence $ Y_n$ as you defined it increases monotonously to $ Y := \sum_{n=1}^\infty X_n $, hence by the monotone convergence theorem (MCT), $$ \mathbb{E}[Y] = \mathbb{E}[\lim_{n \rightarrow \infty} Y_n] \overset{\text{MCT}}{=} \lim_{n \rightarrow \infty} \mathbb{E}[Y_n] = \sum_{n=1}^\infty \mathbb{E}[X_n]. $$

Some details:

• The MCT is applied to integration against the underlying probability measure $\mathbb{P}$ using that $\mathbb{E}[X] = \int X \mathrm{d}\mathbb{P}$ by definition.
• Convergence of $ Y_n $ to $Y$ is in the sense that we admit for the possibility that $ Y_n(\omega) \rightarrow \infty $ for some $ \omega \in \Omega$ (the underlying probability space). In this case, $Y(\omega) = \infty$, which is poses no obstacle for the MCT.