Existence of Random variables when $X_n \to -\infty$ a.s. and $EX_n\to \infty$

Question

Do there exist random variables $\{X_n\}_{n\ge 1}$ $$$$such that $X_n \to -\infty$ a.s. and $EX_n\to \infty$?

How would you prove or disprove this analytically?

Answer 1

To get a simple example, fix the probability space $[0,1]$ equipped with the Borel $\sigma$-algebra and Lebesgue measure. Define random variables $X_n:[0,1] \to \mathbb{R}$ by setting $$X_n(\omega) = \begin{cases} -n & \text{if } \omega > \frac{1}{n} \\ n^3 & \text{if } \omega \leq \frac{1}{n} \end{cases}$$

Then $X_n(\omega) \to - \infty$ for all $\omega > 0$ and hence $X_n \to - \infty$ a.e. However you can calculate $\mathbb{E}[X_n] = \frac{1}{n} n^3 - (1- \frac{1}{n})n = n^2 - n + 1 \to \infty$ as $n \to \infty$.

Answer 2

Example: $P(X_n=-n)=1-\frac{1}{n}$ and $P(X_n=n^3)=\frac{1}{n}$. So $E(X_n)=n^2-n+1\to \infty$. However $X_n\to -\infty$ a.e.

Answer 3

Firstly note that when $\mu,\nu$ are probability measures and $p \in [0,1]$ then $\lambda = p\mu + (1-p)\nu$ is probability measure (that is: a convex combination of pr.measures is a pr. measure).

Let $\mu$ be probability measure (distribution) of positive part of Cauchy random variable, that is for any $B \in \mathcal B(\mathbb R)$ we have $\mu(B) = \int_{B \cap \mathbb R_+} \frac{2\cdot dx}{\pi(1+x^2)}$

Let $\nu_n = \delta_{-n}$ be probability measure such that $\delta_{-n}(B) = 1 $ iff $-n \in B$ (that is one-point distribution).

Let $p_n = \frac{1}{n^2} \in [0,1]$, and finally let $\lambda_n = (1-p_n)\nu_n + p_n \mu$.

Define $X_n$ to have distribution $\lambda_n$, then we have: $$\mathbb E[X_n] = \int_{\mathbb R} x d\lambda_n(x) = \int_{\mathbb R} \frac{x}{n^2} d\mu(x) - n(1-\frac{1}{n^2}) = + \infty$$ since $\int_{\mathbb R} xd\mu(x) = \int_{\mathbb R_+} \frac{2x}{\pi(1+x^2)}dx = +\infty $.

So obviously sequence of $+\infty$ tends to $+\infty$, so $\mathbb E[X_n] \to +\infty$.

Looking at $X_n$, take any $M \in \mathbb N$. We have: $$\sum_{n=1}^\infty \mathbb P(X_n > -M) \le M + \sum_{n=M+1}^\infty \mathbb P(X_n > -M) = M + \sum_{n=M+1}^\infty \mathbb P(X_n \neq -n) = M + \sum_{n=M+1}^\infty \frac{1}{n^2}$$ which is finite. This means that (Borel Cantelli) on the set of measure $1$ we have only finitelly many $X_n(\omega)$ being greater than $-M$, so $\limsup_{n \to \infty} X_n(\omega) \le -M$ for any $\omega$ in that set of measure $1$.

Since $M \in \mathbb N$ was arbitrary, we deduce that $\limsup X_n = -\infty$ almost surely, but obviously $\liminf X_n \le \limsup X_n$, so $\liminf X_n = \limsup X_n = -\infty$ almost surely, and that means that almost surely we have the limit which is $-\infty$