Let $K_{0}=1$, and in every round $i =1,...,n$ a player puts all capital $K_{i-1}$ in. A fair coin is thrown. If it falls on Heads he receives $\frac{5}{3}$ of his capital outlay, if it falls on Tails then $\frac{1}{3}$ of his capital outlay is returned to him.
$1.$ Show $\mathbb E[K_{n}]=1$
$2. K_{n} \to 0$ a.s., where $n \to \infty$
Ideas:
$1.$ I am open to corrections, but I think this is fairly straightforward, perhaps even inductive?
$\mathbb E[K_{1}]=\frac{1}{2}\frac{5}{3}K_{0}+\frac{1}{2}\frac{1}{3}K_{0}=1$
and it goes on $\mathbb E[K_{i}]=\frac{1}{2}\frac{5}{3}E[K_{i-1}]+\frac{1}{2}\frac{1}{3}E[K_{i-1}]=\mathbb E[X_{i-1}]$
Does this suffice?
$2.$
Because we need to show $K_{n} \to 0$ a.s., $n \to \infty$ I assume we need to use the law of large numbers
The only problem is:
$\sum_{i=1}^{n}\frac{K_{n}}{n}\to \mathbb E[X_{1}]$, a.s. but as proven in $1.$ $\mathbb E[X_{1}]=1$
Oh yes but of course, it is plain to see that $(K_{n})_{n}$ are not independent random variables, and therefore I need to find some smart way to "create" random variables that are independent yet involve $K_{n}$ in order to apply the law of large numbers.
Any ideas/assistance?
In this game, what is independent is the rate of the return $\frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $\frac{K_{n}}{K_{n-1}} = \left(\frac{5}{3}\right)^{X_n}\left(\frac{1}{3}\right)^{1-X_n}$ and $$ K_n = 1\cdot\left(\frac{5}{3}\right)^{X_1}\left(\frac{1}{3}\right)^{1-X_1}\cdots\left(\frac{5}{3}\right)^{X_n}\left(\frac{1}{3}\right)^{1-X_n}=\left(\frac{5}{3}\right)^{S_n}\left(\frac{1}{3}\right)^{n-S_n} $$ where $S_n= \sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have $$ \frac{1}{n}\log K_n = \frac{S_n}{n}\log\left(\frac{5}{3}\right)+\left(1-\frac{S_n}{n}\right)\log\left(\frac{1}{3}\right). $$ Strong law of large numbers implies $\frac{S_n}{n}\to E[X_1] =\frac{1}{2}$ almost surely. Thus, it holds $$ \frac{1}{n}\log K_n \to \frac{1}{2}\log\left(\frac{5}{9}\right)=c<0. $$ It says that $K_n^{\frac{1}{n}}\to e^c <1 $ and hence $K_n \to 0$ almost surely.