Law of total expectation

Template:Short description The proposition in probability theory known as the law of total expectation,^[1] the law of iterated expectations^[2] (LIE), Adam's law,^[3] the tower rule,^[4] and the smoothing property of conditional expectation,^[5] among other names, states that if $X$ is a random variable whose expected value $E [X]$ is defined, and $Y$ is any random variable on the same probability space, then

$E [X] = E [E [X ∣ Y]],$

i.e., the expected value of the conditional expected value of $X$ given $Y$ is the same as the expected value of $X$ .

The conditional expected value $E [X ∣ Y]$ , with $Y$ a random variable, is not a simple number; it is a random variable whose value depends on the value of $Y$ . That is, the conditional expected value of $X$ given the event $Y = y$ is a number and it is a function of $y$ . If we write $g (y)$ for the value of $E [X ∣ Y = y]$ then the random variable $E [X ∣ Y]$ is $g (Y)$ .

One special case states that if ${A_{i}}$ is a finite or countable partition of the sample space, then

$E [X] = \sum_{i} E [X ∣ A_{i}] Pr (A_{i}) .$

Example

Suppose that only two factories supply light bulbs to the market. Factory Template:Mvar's bulbs work for an average of 5000 hours, whereas factory Template:Mvar's bulbs work for an average of 4000 hours. It is known that factory Template:Mvar supplies 60% of the total bulbs available. What is the expected length of time, Template:Mvar, that a purchased bulb will work for?

Applying the law of total expectation, we have:

$\begin{matrix} E [L] & = E [L ∣ X] Pr (X) + E [L ∣ Y] Pr (Y) \\ = 5000 (0.6) + 4000 (0.4) \\ = 4600 \end{matrix}$

where

$E [L]$ is the expected life of the bulb;
$Pr (X) = \frac{6}{10}$ is the probability that the purchased bulb was manufactured by factory $X$ ;
$Pr (Y) = \frac{4}{10}$ is the probability that the purchased bulb was manufactured by factory $Y$ ;
$E [L ∣ X] = 5000$ is the expected lifetime of a bulb manufactured by $X$ ;
$E [L ∣ Y] = 4000$ is the expected lifetime of a bulb manufactured by $Y$ .

Thus each purchased light bulb has an expected lifetime of 4600 hours.

Informal proof

When a joint probability density function is well defined and the expectations are integrable, we write for the general case $\begin{matrix} E [X] & = \int x Pr [X = x] d x \\ E [X ∣ Y = y] & = \int x Pr [X = x ∣ Y = y] d x \\ E [E [X ∣ Y]] & = \int (\int x Pr [X = x ∣ Y = y] d x) Pr [Y = y] d y \\ = \iint x Pr [X = x, Y = y] d x d y \\ = \int x (\int Pr [X = x, Y = y] d y) d x \\ = \int x Pr [X = x] d x \\ = E [X] . \end{matrix}$ 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 $(Ω, ℱ, Pr)$ be a probability space on which two sub σ-algebras $𝒢_{1} \subseteq 𝒢_{2} \subseteq ℱ$ are defined. For a random variable $X$ on such a space, the smoothing law states that if $E [X]$ is defined, i.e. $min (E [X_{+}], E [X_{-}]) < \infty$ , then

$E [E [X ∣ 𝒢_{2}] ∣ 𝒢_{1}] = E [X ∣ 𝒢_{1}] (a.s.) .$

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

$E [E [X ∣ 𝒢_{2}] ∣ 𝒢_{1}] is 𝒢_{1}$ -measurable
$\int_{G_{1}} E [E [X ∣ 𝒢_{2}] ∣ 𝒢_{1}] d P = \int_{G_{1}} X d P,$ for all $G_{1} \in 𝒢_{1} .$

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

$\begin{matrix} min (\int_{G_{1}} X_{+} d P, \int_{G_{1}} X_{-} d P) & \leq min (\int_{Ω} X_{+} d P, \int_{Ω} X_{-} d P) \\ = min (E [X_{+}], E [X_{-}]) < \infty, \end{matrix}$

so the integral $\int_{G_{1}} X d P$ is defined (not equal $\infty - \infty$ ).

The second property thus holds since $G_{1} \in 𝒢_{1} \subseteq 𝒢_{2}$ implies $\int_{G_{1}} E [E [X ∣ 𝒢_{2}] ∣ 𝒢_{1}] d P = \int_{G_{1}} E [X ∣ 𝒢_{2}] d P = \int_{G_{1}} X d P .$

Corollary. In the special case when $𝒢_{1} = {\emptyset, Ω}$ and $𝒢_{2} = σ (Y)$ , the smoothing law reduces to $E [E [X ∣ Y]] = E [X] .$

Alternative proof for $E [E [X ∣ Y]] = E [X] .$

This is a simple consequence of the measure-theoretic definition of conditional expectation. By definition, $E [X ∣ Y] := E [X ∣ σ (Y)]$ is a $σ (Y)$ -measurable random variable that satisfies $\int_{A} E [X ∣ Y] d P = \int_{A} X d P,$ for every measurable set $A \in σ (Y)$ . Taking $A = Ω$ proves the claim.

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Law of total expectation

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Example

Informal proof

Proof in the general case

See also

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Further reading

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Law of total expectation

Example

Informal proof

Proof in the general case

See also

References

Further reading

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