X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/1991bb9a2fa344a1bc72ce3222b43bec10ff2f34..23dcb45dd9df2aff12c4ca42d4d96ff2c039358e:/stopping.tex?ds=sidebyside

diff --git a/stopping.tex b/stopping.tex
index 9d7e74f..bb95663 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -366,7 +366,8 @@ Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
 With $t_n=32N^2+16N\ln (N+1)$, one obtains:  $\P_X(\tau > t_n)\leq \frac{1}{4}$. 
 Therefore, using the definition of $t_{\rm mix}$ and
 Theorem~\ref{thm-sst}, it follows that
-$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
+$t_{\rm mix}(\frac{1}{4})\leq 32N^2+16N\ln (N+1)=O(N^2)$ and that 
+
 
 
 Notice that the calculus of the stationary time upper bound is obtained
