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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.

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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 $p_2 < p_1$.

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  • $\begingroup$ 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
    Commented Feb 12, 2020 at 17:10
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    $\begingroup$ 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
    Commented Feb 12, 2020 at 17:16

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