Chapter 12 Conditional densities

1 Chapter12 Conditional OverviewDensity functions determine continuous distributions. If a continuous distri-bution is calculated conditionally on some information, then the density iscalled aconditional density. When the conditioning information involvesanother random variable with a continuous distribution, the Conditional den-sity can be calculated from the joint density for the two random a jointly continuous distribution with joint den-sityf(x,y). From Chapter 11, you know that the marginal distribution ofXis continuous with densityg(y) = f(x,y) Conditional distribution forYgivenX=xhas a ( Conditional ) density,which I will denote byh(y|X=x), or justh(y|x) if the conditioningvariable is unambiguous, for whichP{y Y y+ |X=x} h(y|X=x),for small > onX=xshould be almost the same as conditioning on theevent{x X x+ }for a very small >0.

2 That is, providedg(x)>0,P{y Y y+ |X=x} P{y Y y+ |x X x+ }=P{y Y y+ ,x X x+ }P{x X x+ } f(x,y) g(x).Statistics 241/541 fall 2014c David Pollard, 18 Nov 2014112. Conditional densities2In the limit, as tends to zero , we are left with f(x,y)/g(x). That is,h(y|X=x) =f(x,y)/g(x)for eachxwithg(x)> formally, the Conditional density ish(y|X=x) =joint (X,Y) density at (x,y)marginalXdensity atxThe first Example illustrates two ways to find a Conditional density: firstby calculation of a joint density followed by an appeal to the formula for theconditional density; and then by a sneakier method where all the randomvariables are built directly using polar < >LetXandYbe independent random variables, eachdistributedN(0,1). DefineR= X2+Y2.

3 Show that, for eachr >0, theconditional distribution ofXgivenR=rhas densityh(x|R=r) =1{|x|< r} r2 x2forr > most famous example of a continuous condition distribution comesfrom pairs of random variables that have a bivariate normal each constant ( 1,+1), thestandard bivariate normal withcorrelation is defined as the joint distribution of a pair of random vari-ables constructed from independent random variablesXandY, each dis-tributedN(0,1). DefineZ= X+ 1 2Y. The pairX,Yhas a jointlycontinuous distribution with densityf(x,y) = (2 ) 1exp( (x2+y2)/2).Apply the result from Example< >with(X,Z) = (X,Y)AwhereA=(1 0 1 2)to deduce thatX,Zhave joint densityf (x,z) =1 1 2exp( x2 2 xz+z21 2).Notice the symmetry inxandz.

4 TheXandZmarginals must be the N(0,1). Alsocov(X,Z) = cov(X, X+ 1 2Y)= cov(X,X) + 1 2cov(X,Y) = .Statistics 241/541 fall 2014c David Pollard, 18 Nov 201412. Conditional two random variablesSandTisdefined ascorr(S,T) =cov(S,T) var(S)var(T).If var(S) = var(T) = 1 the correlation reduces to the construction, the Conditional distribution ofZgivenX=xis justthe Conditional distribution of x+ 1 2 YgivenX=x. IndependenceofXandYthen shows thatZ|X=x N( x,1 2).In particular,E(Z|X=x) = x. By symmetry off , we also haveX|Z=z N( z,1 2), a fact that you could check by dividingf (x,z)by the standard normal density < >LetSdenote the height (in inches) of a randomlychosen father, andTdenote the height (in inches) of his son at each ofSandThas aN( , 2) distribution with = 69 and = also that the standardized variables (S )/ and (T )/ havea standard bivariate normal distribution with correlation =.

5 Sam has a height ofS= 74 inches, what would one predict about theultimate heightTof his young son Tom?For the standard bivariate normal, if the variables are uncorrelated (thatis, if = 0) then the joint density factorizes into the product of twoN(0,1) densities , which implies that the variables are independent. This situationis one of the few where a zero covariance ( zero correlation) implies final Example demonstrates yet another connection between Poissonprocesses and order statistics from a uniform distribution. The argumentsmake use of the obvious generalizations of joint densities and conditionaldensities to more than two that random variablesX,Y,Zhave a jointly continuousdistribution with joint densityf(x,y,z) ifP{(X,Y,Z) A}= Af(x,y,z)dxdy dzfor eachA 241/541 fall 2014c David Pollard, 18 Nov 201412.

6 Conditional densities4As in one and two dimensions, joint densities are typically calculated bylooking at small regions: for a small region around (x0,y0,z0)P{(X,Y,Z) } (volume of ) f(x0,y0,z0).Similarly, the joint density for (X,Y) Conditional onZ=zis defined as thefunctionh(x,y|Z=z) for whichP{(X,Y) B|Z=z}= I{(x,y) B}h(x,y|Z=z)dxdyfor each subsetBofR2. It can be calculated, atzvalues where the marginaldensity forZ,g(z) = R2f(x,y,z)dxdy,is strictly positive, by yet another small-region calculation. If is a smallsubset containing (x0,y0) then, for small >0,P{(X,Y) |Z=z0} P{(X,Y) |z0 Z z0+ }=P{(X,Y) , z0 Z z0+ }P{z0 Z z0+ } ((area of ) )f(x0,y0,z0) g(z0)= (area of )f(x0,y0,z0)g(z0). the identification of the set of points (x,y,z) inR3for which (x,y) andz0 z z0+ as a small region with volumeequal to (area of ).

7 That is, the Conditional (joint) distribution of (X,Y) givenZ=zhas densityh(x,y|Z=z) =f(x,y,z)g(z)providedg(z)> authors (including me) like to abbreviateh(x,y|Z=z)toh(x,y|z). Many others run out of symbols and writef(x,y|z) forthe Conditional (joint) density of (X,Y) givenZ=z. This notation isdefensible if one can somehow tell which values are being conditioned a problem with lots of conditioning it can get confusing to rememberwhichfis the joint density and which is Conditional on something. Toavoid confusion, some authors write things likefX,Y|Z(x,y|z) for theconditional density andfX(x) for theX-marginal density, at the costof more cumbersome 241/541 fall 2014c David Pollard, 18 Nov 201412. Conditional densities5 Example< >LetTidenote the time to theith point in a Poissonprocess with rate on [0, ).]

8 Find the joint distribution of (T1,T2) condi-tional the result in the previous Example, you should be able to de-duce that, Conditional onT3=t3for a givent3>0, the random vari-ables (T1/T3,T2/T3) are uniformly distributed over the triangular region{(u1u2) R2: 0< u1< u2<1}.HW11 will step you through an analogous result for order Examples for Chapter 12< > independent random variables, each distributedN(0,1). DefineR= X2+Y2. For eachr >0, find the density for theconditional distribution ofXgivenR= joint density for (X,Y) equalsf(x,y) = (2 ) 1exp( (x2+y2)/2).To find the Conditional density forXgivenR=r, first I ll find the jointdensity forXandR, then I ll calculate itsXmarginal, and then I ll divideto get the Conditional density.

9 A simpler method is described at the end ofthe need to calculateP{x0 X x0+ , r0 R r0+ }for small,positive and . For|x0|< r0, the event corresponds to the two smallregions in the (X,Y)-plane lying between the linesx=x0andx=x0+ ,and between the circles centered at the origin with radiir0andr0+ .radius r0+ radius r0x0+ x0x0x0+ y0+ = (r0+ )2-x02y0 = r02-x02By symmetry, both regions contribute the same probability. Consider theupper region. For small and , the region is approximately a parallelogram,Statistics 241/541 fall 2014c David Pollard, 18 Nov 201412. Conditional densities6with side length = (r0+ )2 x20 r20 x20and width . We couldexpand the expression for as a power series in by multiple applicationsof Taylor s theorem.

10 It is easier to argue less directly, starting from theequalitiesx20+ (y0+ )2= (r0+ )2andx20+y20= differences to deduce that 2y0 + 2= 2r0 + 2. Ignore the lowerorder terms 2and 2to conclude that (r0 /y0). The upper region hasapproximate arear0 /y0, which impliesP{x0 X x0+ , r0 R r0+ } 2r0 y0f(x0,y0) 2r0 r20 x20exp( r20/2)2 .Thus the random variablesXandRhave joint density (x,r) =rexp( r2/2) r2 x21{|x|< r,0< r}.Once again I have omitted the subscript on the dummy variables, to indicatethat the argument works for everyx,rin the specified >0, the random variableRhas marginal densityg(r) = r r (x,r)dx=rexp( r2/2) r rdx r2 x2putx=rcos =rexp( r2/2) 0 rsin rsin d =rexp( r2/2).The Conditional density forXgivenR=requalsh(x|R=r) = (x,r)g(r)=1 r2 x2for|x|< randr > goodly amount of calculation is easier when expressed in polar coordinates.