We toss an infinite number of times with a symmetrical coin. $A_{2n}$ denotes the event that in the first $2n$ tosses there were as many eagles as tails. Show that there are infinitely many $A_{2n}$ events with probability 1.
Solution:
$S_{2n}$ is a sum of "additional" eagles in $2n$ (if in $2n = 10$ tosses there are $6$ eagles and $4$ tails then $S_{2n} = 2$).
$J_{2n}$ is an indicator variable for the event that the number of eagles is equal to number of tails (so in some sense it returns to origin / balance = its starting point). The total number of visits to the origin is given by: $$V = \sum_{n=0}^{\infty} J_{2n}$$ By the linearity of the expected value: $$\mathbb{E}[V] = \sum_{n=0}^{\infty} \mathbb{E}[J_{2n}] = \sum_{n=0}^{\infty} \mathbb{P} [ S_{2n} = S_0 ]$$
From Bergnoulli: $$\mathbb{P} [ S_{2n} = S_0 ] = {{2n}\choose{n}} \left( \frac{1}{2} \right)^{n} \left( \frac{1}{2} \right)^n = {{2n}\choose{n}} \frac{1}{2^{2n}}$$
From Stirling's formula: $$\mathbb{P} [ S_{2n} = S_0 ] = {{2n}\choose{n}} \frac{1}{2^{2n}} = \frac{2n!}{n! \cdot n!} \frac{1}{2^{2n}} \approx \frac{\sqrt{4 \pi n} \cdot e^{-2n} (2n)^{2n}}{(\sqrt{2 \pi n} \cdot e^{-n} n^n)^2 2^{2n}} = \frac{1}{\sqrt{2 \pi n}}$$
We know that $\sum_{n=0}^{\infty} n^{-a}$ converges only if $a > 1$. Therefore we have:
$$\mathbb{E}[V] = \sum_{n=0}^{\infty} \mathbb{P} [ S_{2n} = S_0 ] = \sum_{n=0}^{\infty} \frac{1}{\sqrt{2 \pi n}} = \frac{1}{\sqrt{2 \pi}} \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} = \infty$$
Let $q$ be the probability that $S_{2n}$ returns to its starting point $S_{0}$. Assuming that number of returns is finite, its distribution is given by: $$\mathbb{P}[V = k | \text{ where } k \neq \infty] = q^{k-1}(1-q) \ \ \ \text{for: } k = 1, 2, 3...$$
Then we have that: $$\mathbb{E}[V] = \sum_{n=1}^{\infty} k \mathbb{P}[V = k] = \sum_{n=1}^{\infty} k q^{k-1}(1-q) = \frac{1-q}{q} \sum_{n=1}^{\infty} k q^{k} = \frac{1-q}{q} \cdot \frac{q}{(1-q)^2}= \frac{1}{1-q} < \infty$$
But we know that $\mathbb{E}[V] = \infty$, so by contradiction $q = 1$ and therefore we have that: $$\mathbb{P}[\text{ number of returns is finite }] = \sum_{n=1}^{\infty} \mathbb{P}[V = k | \text{ where } k \neq \infty] = \sum_{n=1}^{\infty} q^{k-1}(1-q) \sum_{n=1}^{\infty} 0 = 0$$
Therefore: $$\mathbb{P}[\text{ (number of returns is finite)' }] = 1 - \mathbb{P}[\text{ number of returns is infinite }] = 1 - 0 = 1$$
Is that correct?