Expected value of these random variables

Question

Say $X_0=1$ and $$ X_n= \begin{cases} 1-X_{n-1},&\mbox{with } p\\ X_{n-1},&\mbox{with } 1-p \end{cases} $$ Now, let $S_n=\sum_{i=1}^nX_i$. Is there any closed-form for $\text{E}S_n$? Like with binomial distribution we would simply have $np$.

Answer 1

You can use the Law of total probability and the fact that $EX_i = P(X_i=1)$: $$ \begin{aligned} EX_i&=P(X_i=1)=P(X_i=1|X_{i-1}=0)\,P(X_{i-1}=0)+P(X_i=1|X_{i-1}=1)\,P(X_{i-1}=1)\\ &=p\,P(X_{i-1}=0)+(1-p)\,P(X_{i-1}=1)\\ &=p\,(1-P(X_{i-1}=1))+(1-p)\,P(X_{i-1}=1)\\ &=p+(1-2p)\,P(X_{i-1}=1)\\ &=p+(1-2p)\,EX_{i-1} \end{aligned} $$

So you have a sequence with $e_0=1$, $e_i=p+(1-2p)e_{i-1}$, which is a more or less trivial problem in Calculus.