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. This problem is adapted from Stochastic Calculus and
          Financial Applicationsby J. Michael Steele,
          Springer, New York, 2001, Chapter 1, Section 1.6, page 9.
Consider a naive model for a stock that has a support
      level of \$20/share because of a corporate buy-back
      program. Suppose also that the stock price moves randomly
      with a downward bias when the price is above \$20, and
      randomly with an upward bias when the price is below \$20.
      To make the problem concrete, we let $Y_n$ denote the stock
      price at time $n$, and we express our stock support
      hypothesis by the assumptions that
\begin{eqnarray*} \Pr[ Y_{n+1} = 21 | Y_{n} = 20] &=& 9/10 \\ \Pr[ Y_{n+1} = 19 | Y_{n} = 20] &=& 1/10 \end{eqnarray*} 
We then reflect the downward bias at price levels above \$20 by requiring that for $k > 20$: 
\begin{eqnarray*} \Pr[ Y_{n+1} = k+1 | Y_{n} = k ] &=& 1/3 \\ \Pr[ Y_{n+1} = k-1 | Y_{n} = k ] &=& 2/3. \end{eqnarray*} 
We then reflect the upward bias at price levels below \$20 by requiring that for $k < 20$: 
\begin{eqnarray*} \Pr[ Y_{n+1} = k+1 | Y_{n} = k ] &=& 2/3 \\ \Pr[ Y_{n+1} = k-1 | Y_{n} = k ] &=& 1/3 \end{eqnarray*} 
I would like to calculate the expected time, $T_{25,18}$ for the stock to fall from $\$25$ through the support level all the way down to $\$18$. 
My first step is to use first-step analysis (no pun intended). This will give me a recursive set of 9 equations. However, one of the hints given is to show that the expected time to go from $\$25$ to $\$20$ is $T_{25,20} = 15$ steps and that $T_{21,20} = 3$. Is it claimed that these are trivial to find. However, I am really not sure how to do this part. There appears to be no upper boundary above $\$25$ -- Does anyone have any hints as to how I can find this? Thanks. 

Let $\tau=\inf\{n>0: Y_{n+1}<Y_n\mid Y_0=25\}$. Then
$$\mathbb E[\tau] = 1 + \frac23 + \frac13\mathbb E[\tau+1]\implies \mathbb E[\tau] =3.$$
It follows that $T_{21,20}=\tau=3$ and $T_{25,20}=5\tau=15$. The quantities $T_{20,19}$ and $T_{19,18}$ may be computed by a similar argument, and $$T_{25,18} = T_{25,20}+T_{20,19}+T_{19,18}. $$
