I've recently started reading through Stochastic Calculus and Financial Applications by J. Michael Steele and I'm not sure I understand how first step analysis works regarding hitting times. Exercise 1.1 of the book gives the following probability distribution for the up and down movements of a stock:
$$\mathbb{P}(Y_{n+1}=k+1|Y_n=k)=1/3\quad \mathrm{and}\quad \mathbb{P}(Y_{n+1}=k-1|Y_n=k)=2/3 \;\;\mathrm{for} \;k>20$$ $$\mathbb{P}(Y_{n+1}=k+1|Y_n=k)=2/3\quad \mathrm{and}\quad \mathbb{P}(Y_{n+1}=k-1|Y_n=k)=1/3 \;\;\mathrm{for} \;k<20$$ $$\mathbb{P}(Y_{n+1}=21|Y_n=20)=0.9\quad \mathrm{and}\quad \mathbb{P}(Y_{n+1}=19|Y_n=20)=0.1$$ The exercise asks for the expected value of the hitting time: $$\tau = \min (n:Y_n=18|Y_0=25)$$ The answer in: Finding the expected time for a stock to go from $\$25$ to $\$18$ given there is a support level at $20 with upward and downward biases. suggests that we can partition the expected time into three distinct parts. They derive an answer for the 21->20 case, but I would like some clarification.
Suppose that $\tau_{1} = \min(n:Y_{n+1}<Y_n)$, then by applying the Law of Total Expectation, conditioning on the first step, we have $$\mathbb{E}[\tau_{1}|Y_0=25]=\mathbb{P}(Y_1=26|Y_0=25)\cdot\mathbb{E}[\tau_{1}|Y_0=25,Y_1=26]+\mathbb{P}(Y_1=24|Y_0=25)\cdot\mathbb{E}[\tau_{1}|Y_0=25,Y_1=24]+1$$ The last '$+1$' accounts for the fact that we have 'moved forward' by adding the next $Y_1$ conditions. Then, applying the known probability distribution of $Y$, we have $$\mathbb{E}[\tau_{1}|Y_0=25]=\frac 13\cdot\mathbb{E}[\tau_{1}|Y_0=25,Y_1=26]+\frac 23 \cdot\mathbb{E}[\tau_{1}|Y_0=25,Y_1=24]+1$$ I know that $\{Y_0=25,Y_1=24\} \Rightarrow \tau_1=1$, which means $$\mathbb{E}[\tau_{1}|Y_0=25]=\frac 13\cdot\mathbb{E}[\tau_{1}|Y_0=25,Y_1=26]+\frac 23+1,$$ but this is where I get stuck. The book and the above answer imply that $$\mathbb{E}[\tau_{1}|Y_0=25,Y_1=26] = \mathbb{E}[\tau_1+1|Y_0=25].$$ Why exactly are we adding the '$+1$'? Is this true in general, i.e. $\mathbb{E}[\tau_{1}|Y_0=k,Y_1=k+1] = \mathbb{E}[\tau_1+1|Y_0=k]$? The book gives this rationale to a similar problem:
When we take our first step, two things happen. First, there is the passage of one unit of time; and, second, we will have moved from zero to either -1 or 1.
Doesn't the first '$+1$' in the beginning application of the Law of Total Expectation account for this passage of time? I realise that this question may be a little trivial, but any help is appreciated!