1

Let $(X_i)$ be sequence of independent random variables with $|X_i| < 1$ almost-surely and define $M_n :=\sum_{i=1}^{n}X_i$. Suppose it is known that $M_n$ converges in probability to some random variable $M$, where $M < \infty$ almost-surely. Does it follow that $\sum_{i=1}^{\infty}\mathbb{E}[X_i] < \infty$?

jII
  • 3,050
  • Yes, the series converges... the proof which i know is somewhat technical; it uses an symmetrization argument, i.e. $Y_i := X_i-\tilde{X}_i$ where $\tilde{X}_i$ is an independent copy of $X_i$ – saz Dec 12 '18 at 06:36
  • @saz Any reference or further details would be appreciated. – jII Dec 12 '18 at 13:30
  • 1
    Do you know Kolmogorov's three series convergence theorem? You can deduce it immediately from this result. – saz Dec 12 '18 at 13:54
  • @saz I've come across the three series theorem, although that applies to convergence a.s. whereas in this case the series is assumed to converge in probability. Is there an immediate way to connect the two? – jII Dec 12 '18 at 15:56
  • 1
    A well-known result (by Lévy) states that for independent random variables $X_i$ the series $\sum_i X_i$ converges almost surely iff it converges in probability; see e.g. this question – saz Dec 12 '18 at 15:59

0 Answers0