Adam’s Law with Extra Conditioning¶

Theorem¶

For random variables $X$ , $Y$ and $Z$ with $\mathbb{E}|Y|<\infty$ $\infty$ ,

\begin{aligned} \mathbb{E}[\mathbb{E}[Y\mid X,Z]\mid Z]=\mathbb{E}[Y\mid Z]. \end{aligned}

This is Adam's Law with $Z$ kept as background information. Conditioning on $(X,Z)$ retains all the information in $Z$ , and no independence assumption is needed.

Taking $Y=\indicator_{A}$ for an event $A$ , letting $X$ encode a finite or countable partition $\{B_i\}$ and letting $Z$ indicate an event $C$ gives LOTP with extra conditioning: once we condition on $C$ , we treat $C$ as the new background information and apply the ordinary LOTP inside that conditional world. Assume $\mathbb{P}(C)>0$ and omit terms with $\mathbb{P}(B_i\cap C)=0$ $0$ . For the remaining terms, $\mathbb{P}(A\mid B_i,C)$ means $\mathbb{P}(A\mid B_i\cap C)$ $C)$ . The events $B_i$ still form a partition, but all probabilities are now computed under the assumption that $C$ has occurred:

\begin{aligned} \mathbb{P}(A\mid C) &= \sum_i \mathbb{P}(A\mid B_i,C)\mathbb{P}(B_i\mid C). \end{aligned}

Interpretation¶

Adam’s law still holds when some background information $Z$ is kept fixed throughout: averaging the finer prediction $\mathbb{E}[Y\mid X,Z]$ over $X$ , with $Z$ held fixed, gives the coarser prediction $\mathbb{E}[Y\mid Z]$ $Z]$ . The extra information is averaged out, the background is kept.

Three parallel lines against z: the predictions 1 + z and 3 + z for X = 1 and X = 3, and between them their average 2 + z, the prediction given Z alone.

Let $Y=X+Z$ $Z$ , where $X$ is independent of $Z$ and takes $1$ and $3$ with equal probability. For each $z$ , the two blue lines $1+z$ and $3+z$ are the predictions $\mathbb{E}[Y\mid X,Z]=X+Z$ $Z$ , and the red line halfway between them is their average $\mathbb{E}[\mathbb{E}[Y\mid X,Z]\mid Z]=\mathbb{E}[Y\mid Z]=2+Z$ $Z$ .

On both sides the result is a function of $Z$ alone. Conditioning first on more information and then on less leaves the smaller information: for each value $z$ , the left side is $\displaystyle \sum_x\mathbb{E}[Y\mid X=x,Z=z]\,\mathbb{P}(X=x\mid Z=z)$ for discrete $X$ and $Z$ . No independence is needed, but the weights are the conditional probabilities of $X$ given $Z=z$ $z$ ; in the figure they are $1/2$ and $1/2$ only because $X$ is independent of $Z$ .

With $Z$ constant the theorem is Adam’s law, and with indicators it becomes total probability with extra conditioning, as the Theorem explains.

Averaging over $X$ with its unconditional distribution is the usual slip. If $X=Z$ and $Y=X$ $X$ , then $\mathbb{E}[Y\mid X,Z]=Z$ and both sides equal $Z$ , whereas weighting by the distribution of $X$ alone would give the number $\mathbb{E}X$ .

Key Results¶

For integrable $Y$ , $\mathbb{E}[\mathbb{E}[Y\mid X,Z]\mid Z]=\mathbb{E}[Y\mid Z]$ almost surely.
Taking $Y$ as an event indicator and $X$ as a partition label yields total probability with extra conditioning.

Examples¶

If $Y=X+Z$ $Z$ , $X$ is independent of $Z$ , and $\mathbb{E}X=2$ $2$ , then $\mathbb{E}[\mathbb{E}[Y\mid X,Z]\mid Z]=2+Z$ $Z$ .
If $Z$ is constant, the identity reduces to $\mathbb{E}[\mathbb{E}[Y\mid X]]=\mathbb{E}Y$ $E Y$ .

Prerequisites¶

Adam’s Law: Total Expectation

Total Probability with Extra Conditioning

Sources¶

Nicky van Foreest · Probability Distributions · Week 6

Symbols: infinity, intersection, sum

Last updated: 20 October 2026 03:02