Suppose that $X_1, X_2,\cdots, X_n$ are iid random variables from $N(\theta,1)$, $\theta$ is rational. Then we know that $\bar X \sim N(\theta,1/n)$. It is said that $\bar X$ is almost surely irrational. I am wondering why is it irrational a.s.? How can I interpret it?
why is this $\bar X$ a.s. irrational?
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statistics
probability-distributions
1 Answers
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Let me denote $\bar X$ by $Z$. Since $Z$ is a random variable with continuous density, the probability of the event $E_q$ that $Z = q$ is $0$ for any fixed real $q$. Now, the event $Z \in \mathbb Q$ can be written as $\bigcup_{q \in \mathbb Q} E_q$; hence by countable additivity, this event also has a zero probability.
We effectively ignored a lot of the details given in the question (e.g., $Z$ is normally distributed, that $\theta$ is rational). The conclusion does not change as long as $Z$ has a continuous distribution.
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0@Sri, +1 for ignoring the irrelevant aspects of the question. – 2011-11-12