Merge pull request #48 from mids-w203/unit2_A1

cleaning up probability space notation

_problems/unit-02/A-computing-the-distribution-induced-by-a-random-variable/1.md

@@ -2,43 +2,40 @@
 index: 1
 statement: |
     Suppose outcome space $\Omega = \{1,2,3,4,5,6\}$ represents a 6-sided die, and the probability function assigns probability $1/6$ to each outcome. Let random variable $X: \Omega \rightarrow \mathbb{R}$ be defined as, 
-    $$X(x) = \begin{cases} 1, &x \le 2\\\\ 2, &otherwise\end{cases}$$
+    $$X(\omega) = \begin{cases} 1, &\omega \le 2\\\\ 2, &otherwise\end{cases}$$
     Compute the pmf of $X$
 ---
-Let $Y$ be a random varable defined with positive probability over the outcome space $\\{1, 2, 3, 4, 5, 6\\}$
 
-Define the probability of $Y$  as follows:
+Let $p$ be the probability function associated with $\Omega$ (part of the probability space). The problem tell us that,
 
 $$
-P(Y = y) = 
+p(\\{\omega\\}) = 
 \begin{cases} 
-\frac{1}{6} & \text{for } y \in \\{1,2,3,4,5,6\\} \\\\
+\frac{1}{6} & \text{for } \omega \in \\{1,2,3,4,5,6\\} \\
 0, & \text{otherwise} 
 \end{cases}
 $$
 
-The implication of the defined transformation is 
+Random variable $X: \Omega \rightarrow \mathbb{R}$ is defined as, 
 
 $$
-\begin{aligned}
-Y = \\{1,2\\} & \Rightarrow X = 1 \\\\
-Y = \\{3,4,5,6\\} & \Rightarrow X = 2
-\end{aligned}
+X(\omega) = \begin{cases} 1, &\omega \le 2\\\\ 2, &otherwise\end{cases}
 $$
 
-In bracket notation, let random variable $X: \Omega \rightarrow \mathbb{R}$ be defined as, 
+We can rewrite this more clearly to enumerate all the values $\omega$ can take on:
 
 $$
-X(x) = \begin{cases} 1, &y \le 2\\\\ 2, &y \in \\{1,2,3,4,5,6\\}\end{cases}
+X(\omega) = \begin{cases} 1, &\omega \in \\{1,2\\} \\\\ 2, &\omega \in \\{3,4,5,6\\}\end{cases}
 $$
 
-Computing inducted probabilities, we have
+
+Computing induced probabilities, we have
 
 $$
-P(X = 1) = P(Y = 1 \cup Y = 2) = \text{  (Independence)  } P(Y = 1) + P(Y = 2)  = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}
+P(X = 1) = p(X^{-1}(1)) = p( \\{1,2\\} )  =  \text{ (countable additivity) } p( \\{1\\} ) + p( \\{2\\} )  = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}
 $$
 
-By law of complementary probability and the definition of the output space of $X$ we have
+Since $X$ has support $\\{1,2\\}$ we know that the events $X=1$ and $X=2$ are complements. Using the complement rule,
 
 $$ 
 P(X=2) = 1 - P(X=2) = \frac{2}{3}
@@ -47,7 +44,7 @@ $$
 or
 
 $$
-P(X = 2) = P(Y = 3 \cup Y = 4 \cup Y = 5 \cup Y = 6)  = \text{ (Independence) } P(Y = 3) + P(Y = 4) +  P(Y = 5) + P(Y = 6)
+P(X = 2) = p(X^{-1}(2)) = p( \\{3,4,5,6\\} )  = \text{ (countable additivity) } p( \\{3\\} ) + p(\\{4\\}) +  p(\\{ 5\\}) + p(\\{ 6\\})
 = 4 * \frac{1}{6} = \frac{2}{3}
 $$