[{"@context":"http:\/\/schema.org\/","@type":"BlogPosting","@id":"https:\/\/wiki.edu.vn\/en\/wiki6\/law-of-total-expectation-wikipedia\/#BlogPosting","mainEntityOfPage":"https:\/\/wiki.edu.vn\/en\/wiki6\/law-of-total-expectation-wikipedia\/","headline":"Law of total expectation – Wikipedia","name":"Law of total expectation – Wikipedia","description":"before-content-x4 Proposition in probability theory after-content-x4 The proposition in probability theory known as the law of total expectation,[1] the law","datePublished":"2014-01-11","dateModified":"2014-01-11","author":{"@type":"Person","@id":"https:\/\/wiki.edu.vn\/en\/wiki6\/author\/lordneo\/#Person","name":"lordneo","url":"https:\/\/wiki.edu.vn\/en\/wiki6\/author\/lordneo\/","image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"Enzyklop\u00e4die","logo":{"@type":"ImageObject","@id":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","url":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/68baa052181f707c662844a465bfeeb135e82bab","url":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/68baa052181f707c662844a465bfeeb135e82bab","height":"","width":""},"url":"https:\/\/wiki.edu.vn\/en\/wiki6\/law-of-total-expectation-wikipedia\/","wordCount":12161,"articleBody":"     (adsbygoogle = window.adsbygoogle || []).push({});before-content-x4Proposition in probability theory       (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4The 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{displaystyle X} is a random variable whose expected value E\u2061(X){displaystyle operatorname {E} (X)} is defined, and        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4Y{displaystyle Y} is any random variable on the same probability space, thenE\u2061(X)=E\u2061(E\u2061(X\u2223Y)),{displaystyle operatorname {E} (X)=operatorname {E} (operatorname {E} (Xmid Y)),}i.e., the expected value of the conditional expected value of X{displaystyle X} given        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4Y{displaystyle Y} is the same as the expected value of X{displaystyle X}.One special case states that if {Ai}i{displaystyle {left{A_{i}right}}_{i}} is a finite or countable partition of the sample space, thenE\u2061(X)=\u2211iE\u2061(X\u2223Ai)P\u2061(Ai).{displaystyle operatorname {E} (X)=sum _{i}{operatorname {E} (Xmid A_{i})operatorname {P} (A_{i})}.}Note: The conditional expected value E(X | Z) is a random variable whose value depend on the value of Z.  Note that the conditional expected value of X given the event Z = z is a function of z.  If we write E(X | Z = z) = g(z) then the random variable E(X | Z) is g(Z). Similar comments apply to the conditional covariance.Table of ContentsExample[edit]Proof in the finite and countable cases[edit]Proof in the general case[edit]Proof of partition formula[edit]See also[edit]References[edit]Example[edit]Suppose that only two factories supply light bulbs to the market. Factory X{displaystyle X}‘s bulbs work for an average of 5000 hours, whereas factory Y{displaystyle Y}‘s bulbs work for an average of 4000 hours. It is known that factory X{displaystyle 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:E\u2061(L)=E\u2061(L\u2223X)P\u2061(X)+E\u2061(L\u2223Y)P\u2061(Y)=5000(0.6)+4000(0.4)=4600{displaystyle {begin{aligned}operatorname {E} (L)&=operatorname {E} (Lmid X)operatorname {P} (X)+operatorname {E} (Lmid Y)operatorname {P} (Y)\\[3pt]&=5000(0.6)+4000(0.4)\\[2pt]&=4600end{aligned}}}whereThus each purchased light bulb has an expected lifetime of 4600 hours.Proof in the finite and countable cases[edit]Let the random variables X{displaystyle X} and Y{displaystyle Y}, defined on the same probability space, assume a finite or countably infinite set of finite values. Assume that E\u2061[X]{displaystyle operatorname {E} [X]} is defined, i.e. min(E\u2061[X+],E\u2061[X\u2212]){displaystyle Omega }, thenE\u2061(X)=\u2211iE\u2061(X\u2223Ai)P\u2061(Ai).{displaystyle operatorname {E} (X)=sum _{i}{operatorname {E} (Xmid A_{i})operatorname {P} (A_{i})}.}Proof.E\u2061(E\u2061(X\u2223Y))=E\u2061[\u2211xx\u22c5P\u2061(X=x\u2223Y)]=\u2211y[\u2211xx\u22c5P\u2061(X=x\u2223Y=y)]\u22c5P\u2061(Y=y)=\u2211y\u2211xx\u22c5P\u2061(X=x,Y=y).{displaystyle {begin{aligned}operatorname {E} left(operatorname {E} (Xmid Y)right)&=operatorname {E} {Bigg [}sum _{x}xcdot operatorname {P} (X=xmid Y){Bigg ]}\\[6pt]&=sum _{y}{Bigg [}sum _{x}xcdot operatorname {P} (X=xmid Y=y){Bigg ]}cdot operatorname {P} (Y=y)\\[6pt]&=sum _{y}sum _{x}xcdot operatorname {P} (X=x,Y=y).end{aligned}}}If the series is finite, then we can switch the summations around, and the previous expression will become\u2211x\u2211yx\u22c5P\u2061(X=x,Y=y)=\u2211xx\u2211yP\u2061(X=x,Y=y)=\u2211xx\u22c5P\u2061(X=x)=E\u2061(X).{displaystyle {begin{aligned}sum _{x}sum _{y}xcdot operatorname {P} (X=x,Y=y)&=sum _{x}xsum _{y}operatorname {P} (X=x,Y=y)\\[6pt]&=sum _{x}xcdot operatorname {P} (X=x)\\[6pt]&=operatorname {E} (X).end{aligned}}}If, on the other hand, the series is infinite, then its convergence cannot be conditional, due to the assumption that min(E\u2061[X+],E\u2061[X\u2212])[X+]{displaystyle operatorname {E} [X_{+}]} and E\u2061[X\u2212]{displaystyle operatorname {E} [X_{-}]} are finite, and diverges to an infinity when either E\u2061[X+]{displaystyle operatorname {E} [X_{+}]} or E\u2061[X\u2212]{displaystyle operatorname {E} [X_{-}]} is infinite.  In both scenarios, the above summations may be exchanged without affecting the sum.Proof in the general case[edit]Let (\u03a9,F,P){displaystyle (Omega ,{mathcal {F}},operatorname {P} )} be a probability space on which two sub \u03c3-algebras G1\u2286G2\u2286F{displaystyle {mathcal {G}}_{1}subseteq {mathcal {G}}_{2}subseteq {mathcal {F}}} are defined. For a random variable X{displaystyle X} on such a space, the smoothing law states that if E\u2061[X]{displaystyle operatorname {E} [X]} is defined, i.e.min(E\u2061[X+],E\u2061[X\u2212])[E\u2061[X\u2223G2]\u2223G1]=E\u2061[X\u2223G1](a.s.).{displaystyle operatorname {E} [operatorname {E} [Xmid {mathcal {G}}_{2}]mid {mathcal {G}}_{1}]=operatorname {E} [Xmid {mathcal {G}}_{1}]quad {text{(a.s.)}}.}Proof. Since a conditional expectation is a Radon\u2013Nikodym derivative, verifying the following two properties establishes the smoothing law:E\u2061[E\u2061[X\u2223G2]\u2223G1]\u00a0is\u00a0G1{displaystyle operatorname {E} [operatorname {E} [Xmid {mathcal {G}}_{2}]mid {mathcal {G}}_{1}]{mbox{ is }}{mathcal {G}}_{1}}-measurable\u222bG1E\u2061[E\u2061[X\u2223G2]\u2223G1]dP=\u222bG1XdP,{displaystyle int _{G_{1}}operatorname {E} [operatorname {E} [Xmid {mathcal {G}}_{2}]mid {mathcal {G}}_{1}]doperatorname {P} =int _{G_{1}}Xdoperatorname {P} ,} for all G1\u2208G1.{displaystyle G_{1}in {mathcal {G}}_{1}.}The first of these properties holds by definition of the conditional expectation. To prove the second one,min(\u222bG1X+dP,\u222bG1X\u2212dP)\u2264min(\u222b\u03a9X+dP,\u222b\u03a9X\u2212dP)=min(E\u2061[X+],E\u2061[X\u2212])G1XdP{displaystyle textstyle int _{G_{1}}X,doperatorname {P} } is defined (not equal \u221e\u2212\u221e{displaystyle infty -infty }).The second property thus holds sinceG1\u2208G1\u2286G2{displaystyle G_{1}in {mathcal {G}}_{1}subseteq {mathcal {G}}_{2}} implies\u222bG1E\u2061[E\u2061[X\u2223G2]\u2223G1]dP=\u222bG1E\u2061[X\u2223G2]dP=\u222bG1XdP.{displaystyle int _{G_{1}}operatorname {E} [operatorname {E} [Xmid {mathcal {G}}_{2}]mid {mathcal {G}}_{1}]doperatorname {P} =int _{G_{1}}operatorname {E} [Xmid {mathcal {G}}_{2}]doperatorname {P} =int _{G_{1}}Xdoperatorname {P} .}Corollary. In the special case when G1={\u2205,\u03a9}{displaystyle {mathcal {G}}_{1}={emptyset ,Omega }} and G2=\u03c3(Y){displaystyle {mathcal {G}}_{2}=sigma (Y)}, the smoothing law reduces toE\u2061[E\u2061[X\u2223Y]]=E\u2061[X].{displaystyle operatorname {E} [operatorname {E} [Xmid Y]]=operatorname {E} [X].}Alternative proof for E\u2061[E\u2061[X\u2223Y]]=E\u2061[X].{displaystyle operatorname {E} [operatorname {E} [Xmid Y]]=operatorname {E} [X].}This is a simple consequence of the measure-theoretic definition of conditional expectation.  By definition, E\u2061[X\u2223Y]:=E\u2061[X\u2223\u03c3(Y)]{displaystyle operatorname {E} [Xmid Y]:=operatorname {E} [Xmid sigma (Y)]} is a \u03c3(Y){displaystyle sigma (Y)}-measurable random variable that satisfies\u222bAE\u2061[X\u2223Y]dP=\u222bAXdP,{displaystyle int _{A}operatorname {E} [Xmid Y]doperatorname {P} =int _{A}Xdoperatorname {P} ,}for every measurable set A\u2208\u03c3(Y){displaystyle Ain sigma (Y)}. Taking A=\u03a9{displaystyle A=Omega } proves the claim.Proof of partition formula[edit]\u2211iE\u2061(X\u2223Ai)P\u2061(Ai)=\u2211i\u222b\u03a9X(\u03c9)P\u2061(d\u03c9\u2223Ai)\u22c5P\u2061(Ai)=\u2211i\u222b\u03a9X(\u03c9)P\u2061(d\u03c9\u2229Ai)=\u2211i\u222b\u03a9X(\u03c9)IAi(\u03c9)P\u2061(d\u03c9)=\u2211iE\u2061(XIAi),{displaystyle {begin{aligned}sum limits _{i}operatorname {E} (Xmid A_{i})operatorname {P} (A_{i})&=sum limits _{i}int limits _{Omega }X(omega )operatorname {P} (domega mid A_{i})cdot operatorname {P} (A_{i})\\&=sum limits _{i}int limits _{Omega }X(omega )operatorname {P} (domega cap A_{i})\\&=sum limits _{i}int limits _{Omega }X(omega )I_{A_{i}}(omega )operatorname {P} (domega )\\&=sum limits _{i}operatorname {E} (XI_{A_{i}}),end{aligned}}}where IAi{displaystyle I_{A_{i}}} is the indicator function of the set Ai{displaystyle A_{i}}.If the partition {Ai}i=0n{displaystyle {{A_{i}}}_{i=0}^{n}} is finite, then, by linearity, the previous expression becomesE\u2061(\u2211i=0nXIAi)=E\u2061(X),{displaystyle operatorname {E} left(sum limits _{i=0}^{n}XI_{A_{i}}right)=operatorname {E} (X),}and we are done.If, however, the partition {Ai}i=0\u221e{displaystyle {{A_{i}}}_{i=0}^{infty }} is infinite, then we use the dominated convergence theorem to show thatE\u2061(\u2211i=0nXIAi)\u2192E\u2061(X).{displaystyle operatorname {E} left(sum limits _{i=0}^{n}XI_{A_{i}}right)to operatorname {E} (X).}Indeed, for every n\u22650{displaystyle ngeq 0},|\u2211i=0nXIAi|\u2264|X|I\u22c3i=0n\u2061Ai\u2264|X|.{displaystyle left|sum _{i=0}^{n}XI_{A_{i}}right|leq |X|I_{mathop {bigcup } limits _{i=0}^{n}A_{i}}leq |X|.}Since every element of the set \u03a9{displaystyle Omega } falls into a specific partition Ai{displaystyle A_{i}}, it is straightforward to verify that the sequence {\u2211i=0nXIAi}n=0\u221e{displaystyle {left{sum _{i=0}^{n}XI_{A_{i}}right}}_{n=0}^{infty }} converges pointwise to X{displaystyle X}. By initial assumption, E\u2061|X|     (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4"},{"@context":"http:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"item":{"@id":"https:\/\/wiki.edu.vn\/en\/wiki6\/#breadcrumbitem","name":"Enzyklop\u00e4die"}},{"@type":"ListItem","position":2,"item":{"@id":"https:\/\/wiki.edu.vn\/en\/wiki6\/law-of-total-expectation-wikipedia\/#breadcrumbitem","name":"Law of total expectation – Wikipedia"}}]}]