Suppose you have a fair coin. You start with $1$ dollar, if toss H, your money doubles, if toss T, your money halves. What is the expected value of the money you have if you toss the coin infinitely?
Why do the following two arguments lead to different answers? Why is 2 incorrect?
Let $X$ denote a toss, then $\mathbb{E}(X)=\frac{1}{2}2+\frac{1}{2}\frac{1}{2}=\frac{5}{4}$. Then We have $\mathbb{E}\left(\prod_{i=1}^nX_i\right)=\prod_{i=1}^n\mathbb{E}(X_i)=\left(\frac{5}{4}\right)^n$.
Let $M_n$ be the amount of money you have at step $n$, then $M_n=2^{n_H-n_T}$, with $n_H+n_T=n$, where $n_H$ is the number of heads in the first $n$ tosses, and $n_T$ is the number of tails in the first $n$ tosses. Since the coin is fair, so for large $n$, $n_H=n_T$, and hence $M_n\to1$.