2
$\begingroup$

We have encountered the following problem that we think that should be true. Let $\{X_n\}_{n\geq 0}$ a sequence of random variables which we know that $\mathbb{E}[X_n]$ tends to infinity.

The question is the following: can we assure that the sequence does NOT converge in distribution to a Poisson?

$\endgroup$

2 Answers 2

4
$\begingroup$

No. It's easy to construct a sequence $Y_n$ with $Y_n \to 0$ a.s. but $E Y_n \to +\infty$. (You can even have $E Y_n \equiv +\infty$ if you wish.) Now let $X$ be a fixed Poisson random variable and $X_n = X + Y_n$.

$\endgroup$
2
$\begingroup$

As Nate Eldridge noted, the answer is no. For a positive result, you need some extra condition. Suppose e.g. the variances $\text{Var}(X_n) < c (E X_n)^2$ for $n$ sufficiently large, with some constant $c \in (0,1)$. Then $X_n$ can't converge in distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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