# Explanation about proof of strong law of large numbers in Tracy's notes

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?

• It should be $X_n(\omega)=1$. – Lord Shark the Unknown Aug 9 '17 at 4:30
• @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. – meisterluk Aug 9 '17 at 4:38
• 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 $1/2^n$. – Lord Shark the Unknown Aug 9 '17 at 4:42
• @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. – meisterluk Aug 9 '17 at 4:46
• @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? – Did Aug 9 '17 at 7:17