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content/blog/expectations-are-linear.md

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+---
+title: "Expectations are Linear"
+date: 2026-04-26T09:03:28-04:00
+draft: false 
+tags: []
+math: true
+medium_enabled: false
+---
+
+> As an example, he asked me, in more words, what the expected rank when flipping over the top card of a deck of cards was (A=1, J=11, Q=12,  K=13). This is easy to compute directly as 7. Then he asked me the expectation of the *sum of the top two cards*.
+>
+> \- [From "Expectation and Copysets" by Justin Jaffray](https://buttondown.com/jaffray/archive/expectation-and-copysets/)
+
+What does your intuition say the answer is? Justin continues by stating that computing this expectation is as easy as summing their individual expectations.
+$$
+E[X + Y] = E[X] + E[Y]
+$$
+In other words, **expectations are linear**. I recommend reading his entire blog post. It's great and also talks about how this property is used in databases today. After a high-level explanation, he says: 
+
+> The fact that expectation is linear is easy to show if you just look at the definition, which we will not do here, but I trust you are capable of if you are interested and have not already seen it.
+
+In this episode of *Exercise for the Reader* ([last episode](/blog/implications-prenex-normal-form/)), we'll look at the definition and show why this property holds. This is true regardless of the underlying probability distribution and whether or not we're sampling with replacement.
+
+As Justin stated, let's start with the definition of expectation and then split the sum:
+$$
+\begin{align*}
+E[X + Y] &= \sum_{x \in X} \sum_{y \in Y} (x + y) \cdot P(X = x, Y=y) \\\\
+&= (\sum_{x \in X} \sum_{y \in Y} x \cdot P(X = x, Y = y)) + (\sum_{x \in X} \sum_{y \in Y} y \cdot P(X = x, Y = y))
+\end{align*}
+$$
+Notice that the left-hand-side of the multiplication does not depend on both variables anymore. Also it doesn't matter whether we do the summation over $X$ first or $Y$. Therefore, we can bring that variable out of the inner sum and simplify this to:
+$$
+E[X + Y] = (\sum_{x \in X} x \sum_{y \in Y} P(X = x, Y = y)) + (\sum_{y \in Y} y \sum_{x \in X} P(X = x, Y = y))
+$$
+We can then perform *marginalization* to substitute $\sum_{y \in Y} P(X = x, Y = y)$ with $P(X = x)$ and do the same for the right hand side of the sum.
+$$
+\begin{align*}
+E[X + Y] &= (\sum_{x \in X}xP(X = x)) + (\sum_{y \in y}yP(Y=y))) \\\\
+         &= E[x] + E[Y]
+\end{align*}
+$$
+
+---
+
+Why can we marginalize? For me to show why, we need to peel back the curtain on the notation.
+
+The set $\Omega$ contains the outcomes of all the events that we're concerned about. So, if we are considering events $X$ and $Y$ with outcomes $x_i$ and $y_i$, respectively. Then, our event space $\Omega$ is equal to $\\{ (x_i, y_i) \mid x_i \in X, y_i \in Y\\}$.
+
+Therefore when we say $X = x$, what we really mean is the set of outcomes where that is true. In mathematical terms, $\\{\omega \in \Omega \mid X(\omega) = x\\}$.
+
+Now, let's show why $\sum_{y \in Y} P(X = x, Y = y) = P(X = x)$.
+$$
+\sum_{y \in Y}P(X = x, Y = y) = \sum_{y \in Y} P(\\{\omega \in \Omega \mid X(\omega) = x\\} \cap \\{\omega \in \Omega \mid Y(\omega) = y\\})
+$$
+One of the three Kolmogorov axioms of probability is **countable additivity**. This is defined as:
+$$
+\sum_{x \in A}P(X = x) = P(\bigcup_{x \in A}X =x )
+$$
+Substituting that in and simplifying, we get:
+$$
+\begin{align*}
+\sum_{y \in Y}P(X = x, Y = y) &= P(\bigcup_{y \in Y}(\\{\omega \in \Omega \mid X(\omega) = x\\} \cap \\{\omega \in \Omega \mid Y(\omega) = y\\})) \\\\
+&= P(\\{\omega \in \Omega \mid X(\omega) = x\\} \cap \bigcup_{y \in Y}\\{\omega \in \Omega \mid Y(\omega) = y\\})
+\end{align*}
+$$
+Notice that the right hand side of the term is just $\Omega$. We can then simplify to,
+$$
+\begin{align*}
+\sum_{y \in Y}P(X = x, Y = y) &= P(\\{\omega \in \Omega \mid X(\omega) = x\\} \cap \Omega) \\\\
+&= P(\\{\omega \in \Omega \mid X(\omega) = x\\}) \\\\
+&= P(X = x)
+\end{align*}
+$$
+Since countable additivity is an axiom, we'll stop our derivations there.  See you next time.