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When I look up the strong law of large numbers it says that (Looking at Discrete Random Variables) $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)$$ That got me wondering are the following equivalent $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)=\lim_{n\rightarrow\infty}P\left(\bar{X}_{n}=\mu\right)$$

I tried to think of this question in terms or translate this question in terms of a Measure on some object (to think of it somewhat pictorially) but I didn't know how to translate the summing of more random variables to this that type of thinking.

If this is not necessarily true are there conditions that would allow it to be true?

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marked as duplicate by Did Aug 24 at 20:09

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If $\overline{X}_n$ is continuous, then $P(\overline{X}_n=\mu)=0$. –  angryavian Aug 24 at 18:32
    
you are write I had discrete random variables in mind when writing this question but it seems that this still may not equivalent –  Kamster Aug 24 at 18:36

2 Answers 2

up vote 2 down vote accepted

As noted already, if $\overline{X}_n$ is continuous, then $P(\overline{X}_n = \mu)=0$.

If the $X_n$ are discrete, then the averages $\overline{X}_n$ will also be discrete, but the space of values that they can take become arbitrarily large as $n \to \infty$. So, $\lim_{n \to \infty}P(\overline{X}_n=\mu)=0$, even though $P(\overline{X}_n = \mu)$ may be nonzero.

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yea I guess Im having a hard time wrapping my head around between the difference of the the two different mean –  Kamster Aug 24 at 18:49

They are not.

If $\bar{X}_n$ is continuous, then $P(\bar{X}_n=c)=0$ for any constant $c$ (including the specific case $c=\mu$). So the limit on the RHS is just $0$ and the probability on the LHS is 1 (in the case when the SLLN applies.).

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Thanks, you are totally right, is it possible this could be true for discrete random variables? –  Kamster Aug 24 at 18:35
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To OP: maybe you can get some intuition for the discrete case by working through a simple case in which each $X_i$ is Bernoulli. –  Kim Jong Un Aug 24 at 18:59

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