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Conditional Expectation

Conditional Expectation

Conditional expectation is defined in many different ways and written in various notations. We summarize the definitions and nottations about conditional expectations.

Definition. conditional probability given sigma algebra

$Z$ is said to be conditional expectation given $\mathcal{G}$ if

\[A \in \mathcal{G}, \ \int_{A} X(\omega) \ dP(\omega) = \int_{A} Z(\omega) \ dP(\omega)\]

We write $Z$ by

\[\mathrm{E} \left[ \left. X \right| \mathcal{G} \right] (\omega) := Z(\omega) .\]
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Definition. conditional probability given r.v.

\(\mathrm{E} \left[ X \mid \sigma(Y) \right]\) is called be conditional expectation given $Y$. We write

\[\mathrm{E} \left[ \left. X \right| Y \right] (\omega) := \mathrm{E} \left[ \left. X \right| \sigma(Y) \right] (\omega) .\]
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Definition. conditional probability over Y=y

$g$ is said to be conditional expectation of $X$ given $Y = y$ if

\[\forall B \in \mathcal{B}(\mathbb{R}), \ \int_{Y^{-1}(B)} X(\omega) \ dP(\omega) = \int_{B} g(y) \ dP^{Y}(y) .\]

We denote $g$ by

\[\mathrm{E} \left[ \left. X \right| Y = y \right] := g(y) \quad (y \in \mathbb{R}) .\]
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Remark

Since \(\sigma(Y) = \{Y^{-1}(B) \mid B \in \mathcal{B}(\mathbb{R})\}\), the following equation holds.

\[\begin{eqnarray} \forall B \in \mathcal{B}(\mathbb{R}), \ \int_{B} \mathrm{E} \left[ \left. X \right| Y = y \right] \ dP^{Y}(y) & = & \int_{Y^{-1}(B)} X(\omega) \ dP(\omega) \nonumber \\ & = & \int_{Y^{-1}(B)} \mathrm{E} \left[ \left. X \right| Y \right](\omega) \ dP(\omega) \nonumber \end{eqnarray}\]

Hence the difference between conditional expectation of $X$ given $Y$ and conditional expectation of $X$ of given $Y=y$ is just the domain of conditional expecation.

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Proposition

If $X$ and $\mathcal{B}$ is independent, \(\mathrm{E}( \abs{f(X)} ) < \infty\),

\[\mathrm{E} \left[ f(X) \mid \mathcal{B} \right] = \mathrm{E} \left[ f(X) \right] \mathrm{a.s.}\]

proof.

$\Box$

Proposition

(1)

\[\begin{eqnarray} A & := & \{(x, y) \mid f_{X}(x) = 0\} \nonumber \\ P((X, Y) \in A) & = & 0 \nonumber \end{eqnarray}\]

(2)

\[\begin{eqnarray} \mathrm{E} \left[ \left. Y \right| X = a \right] = \int_{\mathbb{R}} y \frac{ f_{(X, Y)}(a, y) }{ f_{X}(a) } \ dy \end{eqnarray}\]

(3)

\[\begin{eqnarray} \mathrm{E} \left[ \left. Y \right| X \right](\omega) = \int_{\mathbb{R}} y \frac{ f_{(X, Y)}(X(\omega), y) }{ f_{X}(X(\omega)) } \ dy \nonumber \end{eqnarray}\]

proof

(1)

\[\begin{eqnarray} P((X, Y) \in A) & = & \int_{A} f_{(X, Y)}(x, y) \ dx dy \nonumber \\ & = & \int_{\{x \mid f_{X}(x) = 0\}} \int_{\mathbb{R}} f_{(X, Y)}(x, y) \ dy dx \nonumber \\ & = & \int_{\{x \mid f_{X}(x) = 0\}} f_{(X, Y)}(x, y) \ dx \nonumber \\ & = & 0 \nonumber \end{eqnarray}\]

(2)

(3)

\[\begin{eqnarray} \mathrm{E} \left[ \mathrm{E} \left[ \left. Y \right| X \right] \right] & = & \mathrm{E} \left[ \int_{\mathbb{R}} y \frac{ f_{(X, Y)}(X, y) }{ f_{X}(X) } \ dy \right] \nonumber \\ & = & \int_{\mathbb{R}} \int_{\mathbb{R}} y \frac{ f_{(X, Y)}(x, y) }{ f_{X}(x) } \ dy f_{X}(x) \ dx \nonumber \\ & = & \int_{\mathbb{R}} \int_{\mathbb{R}} y f_{(X, Y)}(x, y) \ dx \ dy \nonumber \\ & = & \int_{\mathbb{R}} y f_{Y}(x, y) \ dy \nonumber \\ & = & \mathrm{E} \left[ Y \right] \nonumber \end{eqnarray}\]
$\Box$

Remark

If the joint p.d.f. of $(X, Y)$ exists, the p.d.f. of \(\mathrm{E}\left[ Y \mid X \right]\) is

\[p_{Y \mid X}(y) := \frac{ f_{(X, Y)}(X, y) }{ f_{X}(X) } .\]
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Proposition 1

(1)

\[\begin{equation} \mathrm{E} \left[ \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \right] = \mathrm{E} \left[ X \right] \end{equation}\]

(2)

\[\begin{eqnarray} \mathrm{E} \left[ \left. aX + bY \right| \mathcal{G} \right] = a \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] + b \mathrm{E} \left[ \left. Y \right| \mathcal{G} \right] \end{eqnarray}\]

(3)

\[\begin{eqnarray} X \ge 0, \ \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \ge 0 \end{eqnarray}\]

(4) If $0 \le X_{n}$, $X_{n} \nearrow X (n \rightarrow \infty)$, then

\[\begin{eqnarray} \mathrm{E} \left[ \left. X_{n} \right| \mathcal{G} \right] \nearrow \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \text{-a.s.} \end{eqnarray}\]

(5) If $X_{n} \ge 0$,

\[\begin{eqnarray} \mathrm{E} \left[ \left. \liminf_{n \rightarrow \infty} X_{n} \right| \mathcal{G} \right] \le \liminf_{n \rightarrow \infty} \mathrm{E} \left[ \left. X_{n} \right| \mathcal{G} \right] \text{-a.s.} \end{eqnarray}\]

(6) If $c: \mathbb{R} \rightarrow \mathbb{R}$ is a convex function,

\[\begin{equation} c \left( \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \right) \le \mathrm{E} \left[ \left. c(X) \right| \mathcal{G} \right] \text{-a.s.} \end{equation}\]

(7)

\[\begin{eqnarray} \mathrm{E} \left[ \left. \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \right| \mathcal{H} \right] = \mathrm{E} \left[ \left. X \right| \mathcal{H} \right] \text{-a.s.} \end{eqnarray}\]

(8) If $Z$ is $\mathcal{G}$ measurable,

\[\begin{eqnarray} \mathrm{E} \left[ \left. Z X \right| \mathcal{G} \right] = Z \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \text{-a.s.} \end{eqnarray}\]

(9) If $\mathcal{H}$ is independent on $\sigma(X) \vee \mathcal{G}$,

\[\begin{eqnarray} \mathrm{E} \left[ \left. X \right| \mathcal{G} \vee \mathcal{H} \right] = \mathrm{E} \left[ \left. X \right| \mathcal{G} \right] \text{-a.s.} \end{eqnarray}\]

proof

$\Box$

Reference