Law of Total Expectation over a Partition¶

Theorem¶

If $\{A_i\}$ is a finite or countable partition of the sample space and $Y$ is integrable, then

\begin{aligned} \mathbb{E}[Y]=\sum_i\mathbb{E}[Y\mid A_i]\,\mathbb{P}(A_{i}), \end{aligned}

where terms with $\mathbb{P}(A_i)=0$ are omitted. The law of total probability (LOTP) is the special case $Y=\indicator_B$ of this partition form of the law of total expectation; the random-variable form $\mathbb{E}[\mathbb{E}[Y\mid X]]=\mathbb{E}[Y]$ is Adam’s Law.

Interpretation¶

The overall average is a weighted average of the averages within disjoint cases: split the sample space into cases, find the mean of $Y$ in each case, and weight each case mean by the probability of the case.

Two bars with widths equal to the probabilities 0.3 and 0.7 of the cases A and its complement, and heights equal to the case means 10 and 2; a dashed line at 4.4 marks the overall mean.

In the diagram, the two cases have probabilities $0.3$ and $0.7$ , drawn as bar widths, and conditional means $10$ and $2$ , drawn as bar heights. Each term $\mathbb{E}[Y\mid A_i]\mathbb{P}(A_i)$ is the area of a bar, $3$ and $1.4$ . The dashed level $4.4$ has the same total area.

The cases must be disjoint so that no outcome is counted twice, and must cover the sample space so that none is missed; that is what a partition guarantees. A case with $\mathbb{P}(A_i)=0$ has no conditional mean, but it also contributes nothing, so it is left out.

The theorem is useful when the cases make $Y$ simple. For the number $N$ of tosses up to and including the first head, with head probability $p$ , condition on the first toss; after a tail the count starts afresh, so $\mathbb{E}N=p\cdot1+(1-p)(1+\mathbb{E}N)$ $E N)$ , giving $\mathbb{E}N=1/p$ $1/ p$ , the mean of the first success distribution. With $Y=\indicator_B$ the theorem becomes the law of total probability, and with the cases $\{X=x\}$ for a discrete $X$ it becomes Adam’s law.

The weights are the case probabilities, not equal weights: the unweighted average $(10+2)/2=6$ is wrong.

Key Results¶

For a partition $\{A_i\}$ , $\displaystyle \mathbb{E}Y=\sum_{i\colon\mathbb{P}(A_i)>0}\mathbb{E}[Y\mid A_i]\mathbb{P}(A_i)$ $A_{i}] P (A_{i})$ .
Putting $Y=\indicator_B$ gives the law of total probability.

Examples¶

If two equally likely customer groups have mean spending $20$ and $50$ , overall mean spending is $35$ .
If $\mathbb{P}(A)=0.3$ $0.3$ , $\mathbb{E}[Y\mid A]=10$ $10$ , and $\mathbb{E}[Y\mid A^c]=2$ $2$ , then $\mathbb{E}Y=0.3(10)+0.7(2)=4.4$ $4.4$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 4

Symbols: sum

Last updated: 20 October 2026 03:02