Exercise on almost sure convergence of random variables

probability theoryrandom variables

We define the independent random variables $(X_n)_{n\in \mathbb N}$, with density functions $$f_{X_n}=\frac 2 {\lambda_n^2}x\cdot\mathcal X_{(0,\lambda_n)}(x).$$
First off, I had to show that the events $A:=\{X_n\leq \frac 1 3\lambda_n\}$ and $B:=\{X_n\geq \frac 2 3\lambda_n\}$ occurr an infinite number of times, and here I had no problems. Then, I showed that the succession converges in law to a variable $X$ when $\lambda_n \rightarrow\lambda\lt\infty$, and finally I must show that the succession converges almost surely. The solutions of my text say that, definitely, we have that $X\geq \frac 2 3\lambda$ and $X\leq \frac 1 3\lambda$, and this is impossible so there's no almost sure convergence. However the $X_n$ don't converge to a constant random variable, so I don't see why the succession cannot assume infinite values both lower than $\frac 1 3\lambda$ and larger than $\frac 2 3 \lambda$; I mean, even if I take $n$ outcomes of the variable $X$, I should have infinite values lower than $\frac 1 3\lambda$ and larger than $\frac 2 3 \lambda$ as $n$ goes to infinity. Can you give me a clarification? Thanks a lot.

Best Answer

Suppose you have a non-random sequence $(x_n)$ such that $x_n \le \lambda / 3$ occurs for infinitely many $n$, and $x_n \ge 2\lambda_n / 3$ occurs for infinitely many $n$. You should then conclude that the sequence $(x_n)$ cannot converge.

For the same reason, each instance of the sequence $(X_n)$ of random variables cannot converge, so $(X_n)$ does not converge almost surely. (Remember that "$(X_n)$ converges almost surely" is "$\lim_{n \to \infty} X_n$ exists almost surely.") This is much stronger than convergence in distribution.

Response to comment:

  • I think your confusion is in not correctly understanding the notions of convergence in distribution and convergence almost surely. You should carefully check the definitions.

  • Specifically, when checking convergence of $(X_n)$ to $X$ in distribution, it suffices to check that the CDFs converge [at points of continuity of the limit CDF]. It does not matter how the $X_n$ depend on each other; you just consider the sequence of distributions separately. However when checking for convergence almost surely, you must consider each instance of the sequence $(X_n)$ and check if it is convergent as a non-random sequence. Here, the dependence between the random variables in the sequence matters very much.

  • In the example in your comment, you have not specified the sequence of random variables converging to $X \sim \text{Bernoulli}(1/2)$, so there is no way to say that "$0$ and $1$ will occur infinitely many times." Here are two silly examples where $(X_n)$ converges almost surely to $X$, but $0$ and $1$ do not appear infinitely many times.

    1. Let $X_n = X$ for all $n$. Then either the sequence is $1,1,1,\ldots$ (with probability $1/2$) or $0,0,0,\ldots$ (with probability $1/2$). In either event, we have a non-random sequence that converges. The limit is a $\text{Bernoulli}(1/2)$ random variable. This is what is meant by almost sure convergence. Note that it is not true that both $0$ and $1$ occur infinitely many times; in any instance, only one of the two will appear.
    2. Let $Y \sim \text{Bernoulli}(1/2)$ and let $X_n = X + Y/n$. Then four sequences $(X_n)$ can happen, each with probability $1/4$; they are $$1+1, 1+1/2, 1+1/3, \ldots$$ or $$1 - 1, 1-1/2, 1-1/3, \ldots$$ or $$1, 1/2, 1/3, \ldots$$ or $$-1, -1/2, -1/3, \ldots.$$ The first two sequences converge to $1$ while the second two sequences converge to $0$. The distribution of the limit is $\text{Bernoulli}(1/2)$. Note that in any instance of the sequence $(X_n)$, neither $1$ nor $0$ appears infinitely often!
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