Clarify difference of average vs expected value

Resolves #1535

Author: hollasch

CHANGELOG.md

@@ -24,6 +24,7 @@ Change Log / Ray Tracing in One Weekend
 ### The Rest of Your Life
   - Fix    -- Fix typo of "arbitrary" (#1589)
   - New    -- Added a bit more explanation of Buffon's needle problem (#1529)
+  - Fix    -- Clarify the distinction between average and expected value (#1535)
 
 
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books/RayTracingTheRestOfYourLife.html

@@ -522,14 +522,28 @@
 If we assume all of the above, then we could solve for the arithmetic mean--the average--of the list
 $Y$ with the following:
 
-    $$ \operatorname{average}(Y) = E[Y] = \frac{1}{N} \sum_{i=0}^{N-1} y_i $$
+    $$ \operatorname{average}(Y_i) = E[Y] = \frac{1}{N} \sum_{i=0}^{N-1} y_i $$
     $$ = \frac{1}{N} \sum_{i=0}^{N-1} f(x_i) $$
     $$ = E[F(X)] $$
 
-Where $E[Y]$ is referred to as the _expected value of_ $Y$. If the values of $x_i$ are chosen
-randomly from a continuous interval $[a,b]$ such that $ a \leq x_i \leq b $ for all values of $i$,
-then $E[F(X)]$ will approximate the average of the continuous function $f(x')$ over the the same
-interval $ a \leq x' \leq b $.
+Where $E[Y]$ is referred to as the _expected value of_ $Y$.
+
+Note the subtle difference between _average value_ and _expected value_ here:
+
+- A set may have many different subsets of selections from that set. Each subset has an _average
+  value_, which is the sum of all selections divided by the count of selections. Note that a given
+  item may occur zero times, one times, or multiple times in a subset.
+
+- A set has only one _expected value_: the sum of _all_ members of a set, divided by the total
+  number of items in that set. Put another way, the _expected value_ is the _average value_ of _all_
+  members of a set.
+
+It is important to note that as the number of random samples from a set increases, the average value
+of a set will converge to the expected value.
+
+If the values of $x_i$ are chosen randomly from a continuous interval $[a,b]$ such that $ a \leq x_i
+\leq b $ for all values of $i$, then $E[F(X)]$ will approximate the average of the continuous
+function $f(x')$ over the the same interval $ a \leq x' \leq b $.
 
     $$ E[f(x') | a \leq x' \leq b] \approx E[F(X) | X =
         \{\small x_i | a \leq x_i \leq b \normalsize \} ] $$