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There are many threads and answers on the Proof of the Strong Law of Large numbers. But my question is a specific example in these notes by Craig A. Tracy. The relevant parts are (page 3):

let $X_1, X_2, X_3, \dots$ denote an infinite sequence of independent random variables with common distribution. Set $S_n = X_1 + \dots + X_n$. […] (Take, for instance, in coining tossing the elementary event $\omega = HHHH\ldots$ for which $S_n(\omega) = 1$ for every $n$ and hence $\lim_{n\to\infty} S_n(\omega)/n = 1$.)

My question is: Is $S_n(\omega) = 1$ wrong?

Analysis: I guess $\omega$ denotes the event of constant, consecutive heads (as opposed to tails).

$$ S_n(\omega) = X_1(\text{HH}\ldots) + X_2(\text{HH}\ldots) + \ldots + X_n(\text{HH}\ldots) $$

The probability of a single heads result is $\frac12$. The probability of $k$ heads results is $\frac1{2^k}$. $S_n(\omega)$ should therefore be $n \cdot \frac1{2^k}$. This is different from his claim $S_n(\omega) = 1$.

If we assume $S_n(\omega) = 1$, I think the limes is also wrong.

$$ \lim_{n \to \infty} \frac{S_n(\omega)}{n} = \lim_{n\to\infty} \frac{1}{n} = 0 \neq 1 $$

But my claim $S_n(\omega) = n \cdot \frac{1}{2^k}$, is not correct either:

$$ \lim_{n \to \infty} \frac{S_n(\omega)}{n} = \lim_{n\to\infty} \frac{n \cdot \frac{1}{2^k}}{n} = \frac{1}{2^k} \neq 1 $$

If we assume $k = n$, we get $S_n(\omega) = 0$ as well. $k$ must be $0$ to hold true, which would be pointless. So where am I wrong here?

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    $\begingroup$ It should be $X_n(\omega)=1$. $\endgroup$ – Lord Shark the Unknown Aug 9 '17 at 4:30
  • $\begingroup$ @LordSharktheUnknown Still makes no sense to me. The probability of $k$ heads is $1/2^k$. Hence $X_n(\omega) = 1/2^k$. Please elaborate. $\endgroup$ – meisterluk Aug 9 '17 at 4:38
  • $\begingroup$ Tossing a coin for the $n$-th time gives a random variable $X_n$ which takes the value $0$ with probability $1/2$ and $1$ with probability $1/2$; so $X_n$ is never $1/2^n$. $\endgroup$ – Lord Shark the Unknown Aug 9 '17 at 4:42
  • $\begingroup$ @LordSharktheUnknown Ah, so $X_i$ represents "$i$-th toss is head". So $S_n(\omega) = X_1(H) + X_2(H) + \ldots + X_n(H)$. Thanks. Can you make an answer out of it? Then I can give you some credit. $\endgroup$ – meisterluk Aug 9 '17 at 4:46
  • $\begingroup$ @meisterluk Sorry but your last comment re-confuses everything with everything. No, $X_i$ does not represent "$i$-th toss is head". In fact, $X_i$ represents the result of the $i$-th toss, thus $X_i(\omega)=1$ for every $\omega$ such that the $i$-th toss produces heads and $X_i(\omega)=0$ for every $\omega$ such that the $i$-th toss produces tails. Doesn't your textbook expand on this? $\endgroup$ – Did Aug 9 '17 at 7:17

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