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The law of large numbers (WLLN) is a result that holds whatever be your interpretation of probability (i.e Bayesian, frequentist). You do not need to have an interpretation of probability to derive the result. In this question I want to clarify if the WLLN is the justification of the frequentist long run interpretation of probability.

If we let \begin{equation} X=\begin{cases} 0, \text{if event }A\text{ does not occur}\\ 1, \text{if event }A\text{ does occur}\\ \end{cases} \end{equation} and let $X_i,X_2,\ldots,X_n$ be iid $X$. We then have $E(X_i)=1\cdot\text{Pr}(A)+0\cdot \text{Pr}(\bar{A})=\text{Pr}(A)$ and $\text{Var}(X_i)=(1-\text{Pr}(A))^2\cdot \text{Pr}(A)+(0-\text{Pr}(A))^2(1-\text{Pr}(A))=(1-\text{Pr}(A))\text{Pr}(A)$, i.e the variance of a Bernoulli random variable (since we consider an indicator variable). We then have by the law of large numbers that

\begin{equation} \bar{X}\rightarrow \text{Pr}(A) \end{equation}

where the convergence is in probability by the weak law of large numbers and is almost surely by the strong law of large numbers.

So whatever be your definition of probability, the long run frequency the event $A$ occurs converges to the probability of $A$. Is this the reason the frequentists define probability as they do, i.e the long run frequency interpretation??

It seems like this answer agrees with me.

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    $\begingroup$ See these relevant posts on stats.stackexchange (by yours truly): stats.stackexchange.com/questions/232339/…, stats.stackexchange.com/questions/230415/…. As for the answer to your question, I am not certain, because I have never been able to force a frequentist to be explicit about what exactly they mean by "probability equals frequency", and since the meaning of the statement is too vague, it can't be confirmed or refuted. $\endgroup$ – Chill2Macht Oct 26 '16 at 20:04

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