fixed error

inputs/lecture_20.tex

@@ -208,7 +208,7 @@ we need the following theorem, which we won't prove here:
     then $X^T$ is also a supermartingale,
     and we have $\bE[X_{T \wedge n}] \le  \bE[X_0]$ for all $n$.
     If $(X_n)_n$ is a martingale, then so is $X^T$
-    and $\bE[X_{T \wedge n}] \le  \bE[X_0]$.
+    and $\bE[X_{T \wedge n}] = \bE[X_0]$.
 \end{theorem}
 \begin{proof}
     First, we need to show that $X^T$ is adapted.