diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex
index 4195844..dca129c 100644
--- a/inputs/lecture_16.tex
+++ b/inputs/lecture_16.tex
@@ -197,6 +197,7 @@ Typically $\cF_n = \sigma(X_1, \ldots, X_n)$ for a sequence of random variables.
     \footnote{In this form it means, that there is some filtration,
     that we don't explicitly specify}.
     Then $(f(X_n))_n$ is a sub-martingale.
+    Likewise, if $f$ is concave, then $((f(X_n))_n$ is a super-martingale.
 \end{corollary}
 \begin{proof}
     Apply \autoref{cjensen}.
diff --git a/inputs/prerequisites.tex b/inputs/prerequisites.tex
index b78ea23..cdad15c 100644
--- a/inputs/prerequisites.tex
+++ b/inputs/prerequisites.tex
@@ -280,7 +280,7 @@ This is taken from section 6.1 of the notes on Stochastik.
     Let $X$ be a random variable and $a > 0$.
     Then
     \[
-        \bP[|X - \bE(X)| \ge  a] \le \frac{\Var(X)}{a}.
+        \bP[|X - \bE(X)| \ge  a] \le \frac{\Var(X)}{a^2}.
     \]
 \end{theorem}
 \begin{proof}
@@ -288,7 +288,7 @@ This is taken from section 6.1 of the notes on Stochastik.
     \begin{IEEEeqnarray*}{rCl}
         \bP[|X-\bE(X)| \ge a]
         &=& \bP[|X - \bE(X)|^2 \ge a^2]\\
-        &\overset{\text{Markov}}{\ge}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
+        &\overset{\text{Markov}}{\le}& \frac{\bE[|X - \bE(X)|^2]}{a^2}.
     \end{IEEEeqnarray*}
 \end{proof}