Minor edits

Advanced/Stopping Problems/parts/0 probability.tex

@@ -60,6 +60,9 @@ As a function, $\mathcal{H}$ maps values in $\Omega$ to values in $\mathbb{Z}^+_
 	\item ...and so on.
 \end{itemize}
 
+Intuitively, a random variable assigns a \say{value} in $\mathbb{R}$ to every possible outcome.
+
+
 \definition{}
 We can compute the probability that a random variable takes a certain value by computing the probability of
 the set of outcomes that produce that value. \par
@@ -92,18 +95,18 @@ Find $\mathcal{P}(\mathcal{X} = x)$ for all $x$ in $\mathbb{Z}$.
 %
 
 
-\definition{}
+\definition{}<defexp>
 Say we have a random variable $\mathcal{X}$ that produces outputs in $\mathbb{R}$. \par
 The \textit{expected value} of $\mathcal{X}$ is then defined as
 \begin{equation*}
 	\mathcal{E}(\mathcal{X})
-	~\coloneqq~ \sum_{x \in A}\Bigl(x \times \mathcal{P}\bigl(\mathcal{X} = x\bigr)\Bigr)
+	~\coloneqq~ \sum_{x \in \mathbb{R}}\Bigl(x \times \mathcal{P}\bigl(\mathcal{X} = x\bigr)\Bigr)
 	~=~ \sum_{\omega \in \Omega}\Bigl(\mathcal{X}(\omega) \times \mathcal{P}(\omega)\Bigr)
 \end{equation*}
 That is, $\mathcal{E}(\mathcal{X})$ is the average of all possible outputs of $\mathcal{X}$ weighted by their probability.
 
 \problem{}
-Say we flip a coin with $\mathcal{P}(\texttt{H}) = \nicefrac{1}{3}$ three times. \par
+Say we flip a coin with $\mathcal{P}(\texttt{H}) = \nicefrac{1}{3}$ two times. \par
 Define $\mathcal{H}$ as the number of heads we see. \par
 Find $\mathcal{E}(\mathcal{H})$.
 
@@ -113,6 +116,14 @@ Find $\mathcal{E}(\mathcal{H})$.
 Let $\mathcal{A}$ and $\mathcal{B}$ be two random variables. \par
 Show that $\mathcal{E}(\mathcal{A} + \mathcal{B}) = \mathcal{E}(\mathcal{A}) + \mathcal{E}(\mathcal{B})$.
 
+\begin{solution}
+	Use the second definition of $\mathcal{E}$, $\sum_{\omega \in \Omega}\Bigl(\mathcal{X}(\omega) \times \mathcal{P}(\omega)\Bigr)$.
+
+	\vspace{2mm}
+
+	Make sure students understand all parts of \ref{defexp}, and are comfortable with the fact that a random variable \say{assigns values} to outcomes.
+\end{solution}
+
 \vfill
 
 \definition{}