Given $x_0, y_0, \beta \in \mathbb{R}$, define a random process as: On turn $i$, probability of success is $x_i/y_i$. On success, $x_{i+1} = x_i + \beta$, else, $x_{i+1} = x_i$. And $y_{i+1} = y_i + \beta$ regardless. What is $E[x_n]$ ?
Intuitively, I would like to say something like: $E[x_{i+1}]= E[x_i] + \frac{E[x_i]}{y_i}\beta = E[x_i]\frac{y_{i+1}}{y_i}$ and thus that $E[x_n] = x_0\frac{y_n}{y_0}$.
However, I don't know if it's justified to use expectation like this. My thinking why linearity of expectation doesn't apply from the start is that $x_i$ and $x_{i+1}$ are from different sample spaces.
Best Answer
I will use $X_n$ instead of $x_n$, since capital letters are customary for random variables.
$X_1,X_2,X_3,\dots$ are all random variables defined on a common probability space $(\Omega,\mathcal F,P)$. Though it is not specified, the implication in this problem is that $(X_i)_{i=1}^\infty$ is a Markov chain. The probability measure $P$ is defined so that $$P(X_{n+1}=X_n+\beta|X_n)=\frac{X_n}{y_n}=1-P(X_{n+1}=X_n|X_n),$$ where $y_n=y_0+n\beta$. This uniquely determines the probability of $(X_1,\dots,X_n)$ taking any finite sequence.
With the setup rigorously specified, every step you performed was correct. Perhaps a step which can be filled in would be $$ E[X_{n+1}]=E[E[X_{n+1}|X_n]]=E[(X_n/y_n)\cdot (X_n+\beta)+(1-X_n/y_n)X_n]=\dots $$