Conditional Expectation with random walk

Question

For $(\xi_k^t)$ i.i.d. random variables, where $P(\xi_k^t=i) = P_i$, for $i=0,1,2$, and $P_0+P_1+P_2 = 1$ and $\mu = E(\xi_k^t) = P_1+2P_2.$

Consider the following where $Y_0=1$, $$ Y_{t+1} = \begin{cases} \xi_1^{t+1}+...+\xi_{Y_t}^{t+1}&\quad\text{if} \quad Y_t>0\\ 0 &\quad\text{if} \quad Y_t=0 \\ \end{cases}$$

If $Y$ is a markov chain, how can i prove $\mathbb {E} \bigl(\frac{Y_{t+1}}{\mu^{t+1}}|Y_t\bigr)= \frac{Y_t}{\mu^t}$? I'm not sure how to proceed, is this not just the definition of a MC? Could i just write it as : $$ \mathbb E \Bigl(\frac{Y_{t+1}}{\mu^{t+1}}|Y_t\Bigr) = \frac{\mathbb{E}(Y_{t+1}|Y_t)}{\mu^{t+1}} = \frac{\mathbb{E}(\sum_i\xi_i|Y_t)}{\mu^{t+1}}$$ and go from here?

Answer 1

If $Y_t=0$, then $Y_{t+1}=0$ so it is trivially true, otherwise $\frac{\mathbb E[Y_{t+1}|Y_t]}{Y_t}=\mathbb E\left[\frac{\xi_1^{t+1}+...+\xi_{Y_t}^{t+1}}{Y_t}\middle|Y_t\right]=\mathbb E[\xi_1^{t+1}|Y_t]=\mu$ by symmetry and since they are i.i.d. and independent of $Y_t$. So you have \begin{align*} &&\mathbb E[Y_{t+1}|Y_{t}] &= \mu Y_t\\ \iff&&\mathbb E\left[\frac{Y_{t+1}}{\mu^{t+1}}\middle|Y_{t}\right] &= \frac{Y_t}{\mu^t} \end{align*}