1
$\begingroup$

Let ${\left( {{X_n}} \right)_{n \in \mathbb{N}}}$ be a sequence of identically distributed random variables such that the series $\sum\limits_{i = 1}^\infty {{X_i}} $ converges in probability. Show that ${X_n} = 0$ almost surely for all $n \in \mathbb{N}$.

My attempt: The sequence $\sum\limits_{i = 1}^n {{X_i}} $ converges in probability to $X$ if for all $\varepsilon > 0$, $\mathop {\lim }\limits_{n \to \infty } \mathbb{P}\left( {\left| {\sum\limits_{i = 1}^n {{X_i}} - X} \right| \geqslant \varepsilon } \right) = 0$.

We know that there exists a sub-sequence ${\left( {\sum\limits_{i = 1}^{{n_k}} {{X_i}} } \right)_{k \in \mathbb{N}}}$ such that $\mathop {\lim }\limits_{k \to \infty } \mathbb{P}\left( {\sum\limits_{i = 1}^{{n_k}} {{X_i}} = X} \right) = 1$.

We also know that $\mathop {\lim }\limits_{n \to \infty } {F_{\sum\limits_{i = 1}^{{n_k}} {{X_i}} }}\left( x \right) = {F_X}\left( x \right),\forall x \in C\left( {{F_X}} \right)$, where $C\left( {{F_X}} \right)$ is a set of points of continuity for ${{F_X}}$.

Since $X$ might be a generalized random variable, I'm not sure if I can assume that $\sum\limits_{i = 1}^n {{X_i}\left( \omega \right)} $ converges (as a sequence of real numbers) almost surely.

I think that just a hint in the right direction should be enough for me to show this. Using the central limit theorem in any of its forms is not allowed.

$\endgroup$
1
$\begingroup$

Hints: Set $S_n := \sum_{i=1}^n X_i$.

  1. Fix $\epsilon>0$. It follows from the convergence in probability that $$\lim_{n \to \infty} \sup_{m \geq n} \mathbb{P}(|S_n-S_m|>\epsilon) = 0.$$
  2. Deduce from the first step and $$\mathbb{P}(|X_1|>\epsilon) = \mathbb{P}(|S_{n+1}-S_n|>\epsilon)$$ that $$\mathbb{P}(|X_1|>\epsilon)=0.$$
  3. Conclude.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.