0
$\begingroup$

Let $X$ be a real-valued random variable with positive expectation (wlog, $\mathbf{E}[X] = 1$, say).

For $N \in \mathbf{N}$, let $X_1, \cdots, X_N$ be independent, identically-distributed copies of $X$, and let $\bar{X}_N = \frac{1}{N} \sum_{i =1}^N X_i$ be their sample mean. Now, consider the quantity

$$p_N \triangleq \mathbf{P} ( \bar{X}_N > 0 ) \in [0, 1].$$

My question is: Is it known whether $p_N$ is increasing with $N$?

Intuitively, it seems like it ought to be (edit, added after answer: I meant to say `eventually' here). If it can be proved with some moment assumption on $X$, I would also be happy with that, though it would be nice to do so without this assumption. A counter-example would also be interesting.

$\endgroup$
2
$\begingroup$

The answer is, that $p_N$ is not necessarily increasing. Not that $\mathbb{P}(\bar X_N > 0) = \mathbb{P}(X_1 + \ldots + X_N > 0)$. Put $\mathbb{P}(X=1) = 0.99$ and $\mathbb{P}(X=-98) = 0.01$. Then $\mathbb{E}X_1 = 0.01$, but $p_2 < p_1$.

$\endgroup$
  • $\begingroup$ Thank you, that makes sense. Do you get the sense that there would be any nontrivial assumptions one could make on $X$ which would cause it to be increasing? $\endgroup$ – πr8 Feb 12 at 17:10
  • 1
    $\begingroup$ Only an idea: Assume that $X$ has symmetric distribution, i.e. $X \sim -X$ and $X_i \sim X + \mu$ with $\mu > 0$. $\endgroup$ – Dieter Kadelka Feb 12 at 17:16

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.