Mean of compound sum of random variables with infinite expectations

Question

I know that if $\sum\limits_{i=1}^N X_i$ is a compound sum where $N$ is a random counting variable with $\mathbb{E}[N] < \infty$ which is independent of the i.i.d. sequence $X_1,X_2,X_3,...$ of non-negative random variables with $\mathbb{E}[X_1]<\infty$ then by a conditioning argument \begin{align} \mathbb{E} \left[\sum_{i=1}^N X_i \right] = \mathbb{E} \left[ \mathbb{E} \left[\sum_{i=1}^N X_i \mid N \right]\right] = \mathbb{E}[N] \mathbb{E}[X_1]. \end{align} But what if $\mathbb{E}[X_1] = \infty$? Do we then have $\mathbb{E} \left[\sum\limits_{i=1}^N X_i \right] = \infty$? And can we still use the conditioning argument when $\mathbb{E}[X_1] = \infty$ and things are no longer in $L^1$?

Answer 1

Let $m$ be a positive integer. Let $Y_k=\min \{X_k, m\}$. Then $(Y_i)$ is i.i.d. and $ \mathbb E (\sum\limits_{i=1}^{N} Y_i) =ENEY_1 \to \infty$ as $m \to \infty$. Since $X_k \geq Y_k$ (for any $m$) we get $ \mathbb E (\sum\limits_{i=1}^{N} X_i) =\infty$.