Law of total expectation

Informal proof

When a joint probability density function is well defined and the expectations are integrable, we write for the general case $${\displaystyle \begin{array}{cc}\mathrm{E}[X]& =\int xPr[X=x]dx\\ \mathrm{E}[X\mid Y=y]& =\int xPr[X=x\mid Y=y]dx\\ \mathrm{E}[\mathrm{E}[X\mid Y]]& =\int \left(\int xPr[X=x\mid Y=y]dx\right)Pr[Y=y]dy\\ & =\iint xPr[X=x,Y=y]dxdy\\ & =\int x\left(\int Pr[X=x,Y=y]dy\right)dx\\ & =\int xPr[X=x]dx\\ & =\mathrm{E}[X].\end{array}}$$ A similar derivation works for discrete distributions using summation instead of integration. For the specific case of a partition, give each cell of the partition a unique label and let the random variable Y be the function of the sample space that assigns a cell's label to each point in that cell.

Proof in the general case

Let ${\displaystyle (\mathrm{\Omega },\mathcal{F},Pr)}$ be a probability space on which two sub σ-algebras ${\displaystyle {\mathcal{G}}_1\subseteq {\mathcal{G}}_2\subseteq \mathcal{F}}$ are defined. For a random variable ${\displaystyle X}$ on such a space, the smoothing law states that if ${\displaystyle \mathrm{E}[X]}$ is defined, i.e. ${\displaystyle min(\mathrm{E}[X_{+}],\mathrm{E}[X_{-}])<\mathrm{\infty }}$, then

$${\displaystyle \mathrm{E}[\mathrm{E}[X\mid {\mathcal{G}}_2]\mid {\mathcal{G}}_1]=\mathrm{E}[X\mid {\mathcal{G}}_1]\text{(a.s.)}.}$$

Proof. Since a conditional expectation is a Radon–Nikodym derivative, verifying the following two properties establishes the smoothing law:

• ${\displaystyle \mathrm{E}[\mathrm{E}[X\mid {\mathcal{G}}_2]\mid {\mathcal{G}}_1]\text{ is }{\mathcal{G}}_1}$-measurable
• ${\displaystyle {\int }_{G_1}\mathrm{E}[\mathrm{E}[X\mid {\mathcal{G}}_2]\mid {\mathcal{G}}_1]d\mathrm{P}={\int }_{G_1}Xd\mathrm{P},}$ for all ${\displaystyle G_1\in {\mathcal{G}}_1.}$

The first of these properties holds by definition of the conditional expectation. To prove the second one,

$${\displaystyle \begin{array}{cc}min\left({\int }_{G_1}{X}_{+}d\mathrm{P},{\int }_{G_1}{X}_{-}d\mathrm{P}\right)& \le min\left({\int }_{\mathrm{\Omega }}{X}_{+}d\mathrm{P},{\int }_{\mathrm{\Omega }}{X}_{-}d\mathrm{P}\right)\\ & =min(\mathrm{E}[X_{+}],\mathrm{E}[X_{-}])<\mathrm{\infty },\end{array}}$$

so the integral ${\int }_{G_1}Xd\mathrm{P}$ is defined (not equal ${\displaystyle \mathrm{\infty }-\mathrm{\infty }}$).

The second property thus holds since ${\displaystyle G_1\in {\mathcal{G}}_1\subseteq {\mathcal{G}}_2}$ implies $${\displaystyle {\int }_{G_1}\mathrm{E}[\mathrm{E}[X\mid {\mathcal{G}}_2]\mid {\mathcal{G}}_1]d\mathrm{P}={\int }_{G_1}\mathrm{E}[X\mid {\mathcal{G}}_2]d\mathrm{P}={\int }_{G_1}Xd\mathrm{P}.}$$

Corollary. In the special case when ${\displaystyle {\mathcal{G}}_1=\{\mathrm{\varnothing },\mathrm{\Omega }\}}$ and ${\displaystyle {\mathcal{G}}_2=\sigma (Y)}$, the smoothing law reduces to $${\displaystyle \mathrm{E}[\mathrm{E}[X\mid Y]]=\mathrm{E}[X].}$$

Alternative proof for ${\displaystyle \mathrm{E}[\mathrm{E}[X\mid Y]]=\mathrm{E}[X].}$

This is a simple consequence of the measure-theoretic definition of conditional expectation. By definition, ${\displaystyle \mathrm{E}[X\mid Y]:=\mathrm{E}[X\mid \sigma (Y)]}$ is a ${\displaystyle \sigma (Y)}$-measurable random variable that satisfies $${\displaystyle {\int }_A\mathrm{E}[X\mid Y]d\mathrm{P}={\int }_{A}Xd\mathrm{P},}$$ for every measurable set ${\displaystyle A\in \sigma (Y)}$. Taking ${\displaystyle A=\mathrm{\Omega }}$ proves the claim.