Law of total expectation

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 theorem,^[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}) P (A_{i}) .

Example

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

\begin{aligned} E (L) & = E (L ∣ X) P (X) + E (L ∣ Y) P (Y) \\ = 5000 (0.6) + 4000 (0.4) \\ = 4600 \end{aligned}

where

$E (L)$ is the expected life of the bulb;
$P (X) = \frac{6}{10}$ is the probability that the purchased bulb was manufactured by factory $X$ ;
$P (Y) = \frac{4}{10}$ 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 $\begin{aligned} E (X) & = \int x \Pr [X = x] d x \\ E (X ∣ Y = y) & = \int x \Pr [X = x ∣ Y = y] d x \\ E (E (X ∣ Y)) & = \int (\int x \Pr [X = x ∣ Y = y] d x) \Pr [Y = y] d y \\ = \int \int x \Pr [X = x, Y = y] d x d y \\ = \int x (\int \Pr [X = x, Y = y] d y) d x \\ = \int x \Pr [X = x] d x \\ = E (X) . \end{aligned}$ 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 $(Ω, ℱ, P)$ 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{aligned} \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{aligned}

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.

References

↑ Weiss, Neil A. (2005). A Course in Probability. Boston: Addison–Wesley. pp. 380–383. ISBN 0-321-18954-X.
↑ "Law of Iterated Expectation | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2018-03-28.
↑ "Adam's and Eve's Laws". Adam and Eve's laws (Shiny app). 2024-09-15. Retrieved 2022-09-15.
↑ Rhee, Chang-han (Sep 20, 2011). "Probability and Statistics" (PDF).
↑ Wolpert, Robert (November 18, 2010). "Conditional Expectation" (PDF).

Billingsley, Patrick (1995). Probability and measure. New York: John Wiley & Sons. ISBN 0-471-00710-2. (Theorem 34.4)
Christopher Sims, "Notes on Random Variables, Expectations, Probability Densities, and Martingales", especially equations (16) through (18)

[1] Weiss, Neil A. (2005). A Course in Probability. Boston: Addison–Wesley. pp. 380–383. ISBN 0-321-18954-X.

[2] "Law of Iterated Expectation | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2018-03-28.

[3] "Adam's and Eve's Laws". Adam and Eve's laws (Shiny app). 2024-09-15. Retrieved 2022-09-15.

[4] Rhee, Chang-han (Sep 20, 2011). "Probability and Statistics" (PDF).

[5] Wolpert, Robert (November 18, 2010). "Conditional Expectation" (PDF).

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Law of total expectation

Contents

Example

Informal proof

Proof in the general case

See also

References

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