4.2 Conditional Distributions and Independence

1 Conditional Distributions and IndependenceDefinition (X, Y) be a discrete bivariate random vector with joint pmff(x, y) andmarginal pmfsfX(x) andfY(y). For anyxsuch thatP(X=x) =fX(x)>0, the conditionalpmf ofYgiven thatX=xis the function ofydenoted byf(y|x) and defined byf(y|x) =P(Y=y|X=x) =f(x, y)fX(x).For anyysuch thatP(Y=y) =fY(y)>0, the Conditional pmf ofXgiven thatY=yisthe function ofxdenoted byf(x|y) and defined byf(x|y) =P(X=x|Y=y) =f(x, y)fY(y).It is easy to verify thatf(y|x) andf(x|y) are indeed Distributions .

2 First,f(y|x) 0 foreveryysincef(x, y) 0 andfX(x)>0. Second, yf(y|x) = yf(x, y)fX(x)=fX(x)fX(x)= (Calculating Conditional probabilities)Define the joint pmf of (X, Y) byf(0,10) =f(0,20) =218, f(1,10) =f(1,30) =318, f(1,20) =418, f(2,30) = Conditional probabilityfY|X(10|0) =f(0,10)fX(0)=f(0,10)f(0,10) +f(0,20)= (X, Y) be a continuous bivariate random vector with joint pdff(x, y) and marginal pdfsfX(x) andfY(y). For anyxsuch thatfX(x)>0, the Conditional pdf ofYgiven thatX=xis the function ofydenoted byf(y|x) and defined byf(y|x) =f(x, y)fX(x).

3 1 For anyysuch thatfY(y)>0, the Conditional pdf ofXgiven thatY=yis the function ofxdenoted byf(x|y) and defined byf(x|y) =f(x, y)fy(y).Ifg(Y) is a function ofY, then the Conditional expected value ofg(Y) given thatX=xisdenoted byE(g(Y)|x) and is given byE(g(Y)|x) = yg(y)f(y|x) andE(g(Y)|x) = g(y)f(y|x)dyin the discrete and continuous cases, (Calculating Conditional pdfs)Let the continuous random vector (X, Y) have joint pdff(x, y) =e y,0< x < y < .The marginal ofXisfX(x) = f(x, y)dy= xe ydy=e6 , marginally,Xhas an exponential distribution.

4 The Conditional distribution ofYisf(y|x) =f(x, y)fX(x)= e ye x=e (y x),ify > x,0e x= 0,ify xThe mean of the Conditional distribution isE(Y|X=x) = xye (y x)dy= 1 + variance of the Conditional distribution isVar (Y|x) =E(Y2|x) (E(Y|x))2= xy2e (y x)dy ( xye (y x))2= 12In all the previous examples, the Conditional distribution ofYgivenX=xwas differentfor different values ofx. In some situations, the knowledge thatX=xdoes not give usany more information aboutYthan we already had. This important relationship betweenXandYis called (X, Y) be a bivariate random vector with joint pdf or pmff(x, y) andmarginal pdfs or pmfsfX(x) andfY(y).

5 ThenXandYare called independent randomvariables if, forEVERYx Randy mR,f(x, y) =fX(x)fY(y).IfXandYare independent, the Conditional pdf ofYgivenX=xisf(y|x) =f(x, y)fX(x)=fX(x)fY(y)fX(x)=fY(y)regardless of the value (X, Y) be a bivariate random vector with joint pdf or pmff(x, y). ThenXandYare independent random variables if and only if there exist functionsg(x) andh(y)such that, for everyx Randy R,f(x, y) =g(x)h(y).Proof:The only if part is proved by definingg(x) =fX(x) andh(y) =fY(y). To provedthe if part for continuous random variables, suppose thatf(x, y) =g(x)h(y).

6 Define g(x)dx=cand h(y)dy=d,where the constantscanddsatisfycd= ( g(x)dx)( h(y)dy)= g(x)h(y)dxdy= f(x, y)dxdy= 13 Furthermore, the marginal pdfs are given byfX(x) = g(x)h(y)dy=g(x)dandfY(y) = g(x)h(y)dx=h(y) , we havef(x, y) =g(x)h(y) =g(x)h(y)cd=fX(x)fY(y),showing thatXandYare independent. Replacing integrals with sums proves the lemmafor discrete random vectors. Example (Checking Independence )Consider the joint pdff(x, y) =1384x2y2e y (x/2),x >0 andy >0. If we defineg(x) = x2e x/2x >00x 0andh(y) = y4e y/384y >00y 0thenf(x, y) =g(x)h(y) for allx Rand ally R.

7 By Lemma , we conclude thatXandYare independent random independent random variables.(a) For anyA RandB R,P(X A, Y B) =P(X A)P(Y B); that is, theevents{X A}and{Y B}are independent events.(b) Letg(x) be a function only ofxandh(y) be a function only ofy. ThenE(g(X)h(Y)) = (Eg(X))(Eh(Y)).4 Proof:For continuous random variables, part (b) is proved by noting thatE(g(X)h(Y)) = g(x)h(y)f(x, y)dxdy= g(x)h(y)fX(x)fY(y)dxdy= ( g(x)fX(x)dx)( h(y)fY(y)dy)= (Eg(X))(Eh(Y)).The result for discrete random variables is proved bt replacing integrals by (a) can be proved similarly.

8 Letg(x) be the indicator function of the setA. leth(y)be the indicator function of the setB. Note thatg(x)h(y) is the indicator function of thesetC R2defined byC={(x, y) :x A, y B}. Also note that for an indicator functionsuch asg(x),Eg(X) =P(X A). Thus,P(X A, Y B) =P((X, Y) C) =E(g(X)h(Y))= (Eg(X))(Eh(Y)) =P(X A)P(Y B). Example (Expectations of independent variables)LetXandYbe independent exponential(1) random variables. SoP(X 4, Y 3) =P(X 4)P(Y 3) =e 4(1 e 3)/Lettingg(x) =x2andh(y) =y, we haveE(X2Y) =E(X2)E(Y) = (2)(1) = independent random variables with moment generatingfunctionsMX(t) andMY(t).

9 Then the moment generating function of the random variableZ=X+Yis given byMZ(t) =MX(t)MY(t).5 Proof:MZ(t) =Eet(X+Y)= (EetX)(EetY) =MX(t)MY(t). Theorem N( , 2) andY N( , 2) be independent normal random vari-ables. Then the random variableZ=X+Yhas aN( + , 2+ 2) :Using Theorem , we haveMZ(t) =MX(t)MY(t) = exp{( + )t+ ( 2+ 2)t2/2}.Hence,Z N( + , 2+ 2). 6