Update 1.md

_problems/unit-03/A-expectation-of-a-discrete-random-variable/1.md

@@ -5,3 +5,47 @@ statement: |
     A gambler plays roulette and bets \\$1 on black 19 times. They wins \\$1 with probability
     18/38 and loses \\$1 with probability 20/38. What are the expected winnings?
 ---
+The 19 plays yield outcomes which are independent of one another.
+
+Let $Y$ be a random variable representing a single play.
+
+$$
+Y \in \{B, B^C\} \text{, where we define the values as } B = \{\$ 1\} \text{ and } B^C = \{\$ -1\}
+$$
+
+The PMF for $Y$ is therefore
+
+$$
+P(Y=y) = \begin{cases} \frac{18}{38}, &y = \$ 1\\\\ \frac{20}{38}, &y = \$ -1 \\\\ 0, &otherwise\end{cases}
+$$
+
+The Expected Value notated as $E[Y]=$
+
+$$1*\frac{18}{38} + -1*\frac{20}{38} = \$ -0.053$$
+
+Let $Z$ be a random variable defined as 
+
+$$
+Z(n) = \sum_{i=1}^{n} Y(i)
+$$
+
+Let $n=19$ and 
+
+$$
+E[Z(19)] = E[\sum_{i=1}^{19} Y(i)]
+$$
+
+By the linear operator property of expected value, we have
+
+$$
+E[Z(19)]  = \sum_{i=1}^{19} E[Y(i)]
+$$
+
+By the given hypothesis of $IID$ or Identically and Independently Distributed property, the $E[Y(i)]$ is the same for each play.
+So substitute $Y$ for $Y(i)$ in 
+
+$$
+E[Z(19)]  = \sum_{i=1}^{19} E[Y] = \sum_{i=1}^{19} -0.053 = 19*(-0.053) = \$ -1
+$$
+
+Which is the desired expected value.