Arasındaki fark ne ve ?


18

Genel olarak, E ( X | Y )E(X|Y) ve E ( X | Y = y ) arasındaki farkE(X|Y=y) nedir?

Eski işlevi, ikincisi işlevi ? Çok kafa karıştırıcı ..y yxx


Hmmm ... İkincisi x'in bir fonksiyonu değil bir sayı olmalı! Yanlış mıyım?
David

Yanıtlar:


23

Kabaca arasındaki fark E ( X | Y )E(XY) ve E ( X | Y = y )E(XY=y) ikinci ise önceki rastgele değişken (bir anlamda), bir gerçekleşme olmasıdır E ( X | Y )E(XY) . Örneğin, ( X , Y ) ~ N ( 0 , ( 1 p, p, 1 ) )

(X,Y)N(0,(1ρρ1))
daha sonra E ( X | Y )E(XY)R değişkeni E ( X Y ) = ρ Y'dir .
E(XY)=ρY.
Tersine, Y = yY=y gözlendiğinde, bir skaler olan E ( X Y = y ) = ρ y miktarıyla daha fazla ilgileniriz E(XY=y)=ρy.

Belki bu gereksiz bir komplikasyon gibi görünüyor, ancak E ( X Y )E(XY) 'yi kendi başına rastgele bir değişken olarak görmek, kule kanunu E ( X ) = E [ E ( X Y ) ] gibi şeyleri E(X)=E[E(XY)]mantıklı kılan şeydir - kaşlı ayraçların içindeki şey rastgele, bu yüzden beklentisinin ne olduğunu sorabiliriz, oysa E hakkında rastgele bir şey yok ( X Y = y )E(XY=y) . Çoğu durumda E ( X Y =y ) = x f X Y ( x y ) d x 

E(XY=y)=xfXY(xy) dx

ve sonuçta ortaya çıkan ifadede y yerine rastgele değişkeni Y "takarak" E ( X Y ) elde edilir. Daha önceki bir yorumda ima edildiği gibi, bu şeylerin titizlikle nasıl tanımlandığı ve bunları uygun bir şekilde bağladığı konusunda sürünebilecek biraz incelik vardır. Bu, temel teori ile ilgili bazı teknik sorunlar nedeniyle koşullu olasılıkla gerçekleşme eğilimindedir.E(XY)Yy


8

XX ve Y'ninY rastgele değişkenler olduğunu varsayalım .

Let y 0y0 bir olmak sabit reel sayı, ki y 0 = 1y0=1 . Daha sonra, E [ X | , Y = y 0 ] = E [ X | , Y = 1 ]E[XY=y0]=E[XY=1] a, sayı : o koşullu beklenen değer arasında XX verilen YY değerine sahip 11 . Şimdi, başka bir sabit gerçek sayı y 1y1 için not edin , diyelim y 1 = 1.5y1=1.5 , E [ X Y = y 1 ] = E [ X Y = 1.5 ]E[XY=y1]=E[XY=1.5] , Y = 1.5 (gerçek bir sayı)verilen X'inX koşullu beklenen değeri olacaktır . E [ X Y = 1.5 ] ve E [ X Y = 1 ] ' in aynı değere sahipolduğunu varsaymak için hiçbir neden yoktur. Böylece, E [ X Y = yY=1.5E[XY=1.5]E[XY=1]]E[XY=y] as being a real-valued function g(y)g(y) that maps real numbers yy to real numbers E[XY=y]E[XY=y]. Note that the statement in the OP's question that E[XY=y]E[XY=y] is a function of xx is incorrect: E[XY=y]E[XY=y] is a real-valued function of yy.

On the other hand, E[XY]E[XY] is a random variable ZZ which happens to be a function of the random variable YY. Now, whenever we write Z=h(Y)Z=h(Y), what we mean is that whenever the random variable YY happens to have value yy, the random variable ZZ has value h(y)h(y). Whenever YY takes on value yy, the random variable Z=E[XY]Z=E[XY] takes on value E[XY=y]=g(y)E[XY=y]=g(y). Thus, E[XY]E[XY] is just another name for the random variable Z=g(Y)Z=g(Y). Note that E[XY]E[XY] is a function of YY (not yy as in the statement of the OP's question).

As a a simple illustrative example, suppose that XX and YY are discrete random variables with joint distribution P(X=0,Y=0)=0.1,  P(X=0,Y=1)=0.2,P(X=1,Y=0)=0.3,  P(X=1,Y=1)=0.4.

P(X=0,Y=0)P(X=1,Y=0)=0.1,  P(X=0,Y=1)=0.2,=0.3,  P(X=1,Y=1)=0.4.
Note that XX and YY are (dependent) Bernoulli random variables with parameters 0.70.7 and 0.60.6 respectively, and so E[X]=0.7E[X]=0.7 and E[Y]=0.6E[Y]=0.6. Now, note that conditioned on Y=0Y=0, XX is a Bernoulli random variable with parameter 0.750.75 while conditioned on Y=1Y=1, XX is a Bernoulli random variable with parameter 2323. If you cannot see why this is so immediately, just work out the details: for example P(X=1Y=0)=P(X=1,Y=0)P(Y=0)=0.30.4=34,P(X=0Y=0)=P(X=0,Y=0)P(Y=0)=0.10.4=14,
P(X=1Y=0)=P(X=1,Y=0)P(Y=0)=0.30.4=34,P(X=0Y=0)=P(X=0,Y=0)P(Y=0)=0.10.4=14,
and similarly for P(X=1Y=1)P(X=1Y=1) and P(X=0Y=1)P(X=0Y=1). Hence, we have that E[XY=0]=34,E[XY=1]=23.
E[XY=0]=34,E[XY=1]=23.
Thus, E[XY=y]=g(y)E[XY=y]=g(y) where g(y)g(y) is a real-valued function enjoying the properties: g(0)=34,g(1)=23.
g(0)=34,g(1)=23.

On the other hand, E[XY]=g(Y)E[XY]=g(Y) is a random variable that takes on values 3434 and 2323 with probabilities 0.4=P(Y=0)0.4=P(Y=0) and 0.6=P(Y=1)0.6=P(Y=1) respectively. Note that E[XY]E[XY] is a discrete random variable but is not a Bernoulli random variable.

As a final touch, note that E[Z]=E[E[XY]]=E[g(Y)]=0.4×34+0.6×23=0.7=E[X].

E[Z]=E[E[XY]]=E[g(Y)]=0.4×34+0.6×23=0.7=E[X].
That is, the expected value of this function of YY, which we computed using only the marginal distribution of YY, happens to have the same numerical value as E[X]E[X] !! This is an illustration of a more general result that many people believe is a LIE: E[E[XY]]=E[X].
E[E[XY]]=E[X].

Sorry, that's just a small joke. LIE is an acronym for Law of Iterated Expectation which is a perfectly valid result that everyone believes is the truth.


3

E(X|Y)E(X|Y) is the expectation of a random variable: the expectation of XX conditional on YY. E(X|Y=y)E(X|Y=y), on the other hand, is a particular value: the expected value of XX when Y=yY=y.

Think of it this way: let XX represent the caloric intake and YY represent height. E(X|Y)E(X|Y) is then the caloric intake, conditional on height - and in this case, E(X|Y=y)E(X|Y=y) represents our best guess at the caloric intake (XX) when a person has a certain height Y=yY=y, say, 180 centimeters.


4
I believe your first sentence should replace "distribution" with "expectation" (twice).
Glen_b -Reinstate Monica

4
E(XY)E(XY) isn't the distribution of XX given YY; this would be more commonly denotes by the conditional density fXY(xy)fXY(xy) or conditional distribution function. E(XY)E(XY) is the conditional expectation of XX given Y, which is a Y-measurable random variable. E(XY=y) might be thought of as the realization of the random variable E(XY) when Y=y is observed (but there is the possibility for measure-theoretic subtlety to creep in).
guy

1
@guy Your explanation is the first accurate answer yet provided (out of three offered so far). Would you consider posting it as an answer?
whuber

@whuber I would but I'm not sure how to strike the balance between accuracy and making the answer suitably useful to OP and I'm paranoid about getting tripped up on technicalities :)
guy

@Guy I think you have already done a good job with the technicalities. Since you are sensitive about communicating well with the OP (which is great!), consider offering a simple example to illustrate--maybe just a joint distribution with binary marginals.
whuber

1

E(X|Y) is expected value of values of X given values of Y E(X|Y=y) is expected value of X given the value of Y is y

Generally P(X|Y) is probability of values X given values Y, but you can get more precise and say P(X=x|Y=y), i.e. probability of value x from all X's given the y'th value of Y's. The difference is that in the first case it is about "values of" and in the second you consider a certain value.

You could find the diagram below helpful.

Bayes theorem diagram form Wikipedia


This answer discusses probability, while the question asks about expectation. What is the connection?
whuber
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