# A Quick Note: Some «Trivia» about Joint Expectations

## Intro

The subject of this post is rather trivial and, for this reason, its understanding is often assumed and taken for granted. We will discuss a property of expectation over a joint distribution of random variables that is implicitly used in many derivations, but rarely explicitly stated and I invite the reader to take a minute to convince themselves that it actually holds. Please, join in.

Let us finally formulate this artfully surrounded by an air of mysticism (with the intention to lure in unsuspecting readers, of course) property.

An expectation over a joint distribution of multiple variables of an expression that depends on a proper subset of these variables can be reduced to that computed over a marginal distribution of the variables involved or, in mathematical notation, \(E_{X_1,\dots,X_n}[f(X_{i_1},\dots,X_{i_k})] = E_{X_{i_1},\dots,X_{i_k}}[f(X_{i_1},\dots,X_{i_k})], \quad i_j \in \{1,2,\dots,n\} \mbox{ for } j = \overline{1,k}\)

Before moving forward, take a note of the notation. It is considered a sound practice to list the variables relative to which the expectation is computed in subscript, next to the āEā sign. For example, taking the expectation \(E_{X}[X^3 \cdot Y^2]\), provided \(X\) and \(Y\) are continuous, will result in the following integral: \(Y^2 \cdot \int X^3 \cdot p(X)\,dX\) , whereas \(E_{XY}[3\cdot X \cdot Y^2]\) is computed as \(\iint X^3 \cdot Y^2 \cdot p(X, Y) \,dX \,dY\).

## The Proof

For the sake of simplicity and brevity, we will limit ourselves to \(E_{XY}[f(Y)] = E_{Y}[f(Y)]\); naturally, the proof can be easily generalized to the case of an arbitrary number of variables.

Let \(X\) and \(Y\) be random variables with domains \(D_X\) and \(D_Y\) respectively. Assuming both variables to be *continuous*, the derivation steps below constitute the desired proof.
\(\begin{align*}
E_{XY}[f(Y)] & = \int_{D_X} \int_{D_Y} f(Y) \cdot p(Y, X) \,dY\,dX = \int_{D_X} \int_{D_Y} f(Y) \cdot p(Y) \cdot p(X | Y) \,dX\,dY = \\
& = \int_{D_Y} f(Y) \cdot p(Y) \cdot \underbrace{\left(\int_{D_X} p(X|Y) \,dX\right)}_{= \;1} \,dY = \int_{D_Y} f(Y) \cdot p(Y) \,dY = E_Y[f(Y)]
\end{align*}\)

In the case of *discrete* variables, additional assumptions are necessary:

and

\[D_Y = \{y_1, y_2, \dots, y_{d_Y}\};\quad (\mid D_Y\mid = d_Y)\]then the following holds:

\[\begin{align*} E_{XY}[f(Y)] &= \sum_{i = 1}^{d_X} \sum_{j = 1}^{d_Y} f(y_j) \cdot p(X = x_i, Y = y_j) = \sum_{i = 1}^{d_X} \sum_{j = 1}^{d_Y} f(y_j) \cdot p(Y = y_j) \cdot p(X = x_i | Y = y_j) = \\ & = \sum_{j = 1}^{d_Y} f(y_j) \cdot p(Y = y_j) \cdot \underbrace{\left(\sum_{i = 1}^{d_X} p(X = x_i| Y = y_j)\right)}_{=\;1} = \sum_{j = 1}^{d_Y} f(y_j) \cdot p(Y = y_j) = E_Y[f(Y)] \end{align*}\]ā Ry Auscitte