A sequence of days which stay outside of the market will consist
of segments of days in one collar alternating with segments of
days in the other collar.

Consider now the random variable $B^u$ which gives the number of
days we can stay within the upper collar. Note that $B^u_k =
\bigcap_{j=1}^k B_j$. Using the previous lemma,
\begin{eqnarray*}
   E(B^u) &<& \sum_{k=0}^\infty \Prob(B_k) \\
         &<& C_N + \sum_{k=N}^\infty { e^{-k\lambda^2/4} \over
      \lambda\sqrt{\pi k}}
    < \infty,
\end{eqnarray*}
where $\lambda = (\ln\kappa_u-\mu)/\sigma$. An similar bound
exists for expectation time in the lower collar $E(B^l)$.

Now consider the random variable $C$ which gives the number of
passages from upper to lower collars before the next consecutive
day at-market. The probability to pass from one collar to the
other must be at least,
\[
     p_c = \Prob(\, Z_j \ge (\ln\kappa_u-\ln\kappa_l) \,).
\]
Considering the various cases by which a random walk might stay
outside of either collar, we arrive at, essentially, a sum over
$i$ changes from upper to lower of walks of approximate length
weighted by the probability of $i$ changes of collars,
\begin{eqnarray*}
   E(T_1) &\approx& \Prob(\, \ln\kappa_l < Z_1 < \ln\kappa_u\,)
   + \sum_{i=1}^\infty (p_c^{i-1}) (i/2)(E(B^u)+E(B^l)+i) \\
          &<& K_c + K \sum_{i=1}^\infty i^2 p_c^i < \infty.
\end{eqnarray*}
The proof for $E(T_1^2)$ is similar.
