6 Jointly continuous random variables

1 6 Jointly continuous random variablesAgain, we deviate from the order in the book for this chapter, so the subsec-tions in this chapter do not correspond to those in the Joint density functionsRecall thatXis continuous if there is a functionf(x) (the density) such thatP(X t) =Zt fX(x)dxWe generalize this to two random random variablesXandYare Jointly continuous if thereis a functionfX,Y(x, y)onR2, called the joint probability density function,such thatP(X s, Y t) =Z Zx s,y tfX,Y(x, y)dxdyThe integral is over{(x, y) :x s, y t}.

2 We can also write the integral asP(X s, Y t) =Zs Zt fX,Y(x, y)dy dx=Zt Zs fX,Y(x, y)dx dyIn order for a functionf(x, y) to be a joint density it must satisfyf(x, y) 0Z Z f(x, y)dxdy= 1 Just as with one random variable , the joint density functioncontains allthe information about the underlying probability measure if we only look atthe random variablesXandY. In particular, we can compute the probabilityof any event defined in terms ofXandYjust usingf(x, y).Here are some events defined in terms ofXandY:{X Y},{X2+Y2 1}, and{1 X 4, Y 0}.

3 They can all be writtenin the form{(X, Y) A}for some R2,P((X, Y) A) =Z ZAf(x, y)dxdyThe two-dimensional integral is over the subsetAofR2. Typically, whenwe want to actually compute this integral we have to write it as an iteratedintegral. It is a good idea to draw a picture ofAto help do rigorous proof of this theorem is beyond the scope of this course. Inparticular we should note that there are issues involving -fields and con-straints onA. Nonetheless, it is worth looking at how the proof might startto get some practice manipulating integrals of joint ( , s] ( , t], then the equation is the definition of jointlycontinuous.))

4 Now supposeA= ( , s] (a, b]. The we can write it asA= [( , s] ( , b]]\[( , s] ( , a]] So we can write the event{(X, Y) A}={(X, Y) ( , s] ( , b]}\{(X, Y) ( , s] ( , a]}MORE !!!!!!!!!Definition:LetA R2. We sayXandYare uniformly distributed onAiff(x) = 1c,if (x, y) A0,otherwisewherecis the area :LetX, Ybe uniform on [0,1] [0,2]. FindP(X+Y 1).Example:LetX, Yhave densityf(x, y) =12 exp( 12(x2+y2))ComputeP(X Y) andP(X2+Y2 1).Example:Now supposeX, Yhave densityf(x, y) = e x yifx, y 00,otherwise2 ComputeP(X+Y t).What does the pdf mean?))))))))))

5 In the case of a single discrete RV, the pmfhas a very concrete (x) is the probability thatX=x. IfXis asingle continuous random variable , thenP(x X x+ ) =Zx+ xf(u)du f(x)IfX, Yare Jointly continuous , thanP(x X x+ , y Y y+ ) 2f(x, y) Independence and marginal distributionsSuppose we know the joint densityfX,Y(x, y) ofXandY. How do we findtheir individual densitiesfX(x),fY(y). These are calledmarginal cdf ofXisFX(x) =P(X x) =P( < X x, < Y < )=Zx Z fX,Y(u, y)dy duDifferentiate this with respect toxand we getfX(x) =Z fX,Y(x, y)dyIn words, we get the marginal density ofXby integratingyfrom to in the joint Jointly continuous with joint densityfX,Y(x, y),then the marginal densities are given byfX(x) =Z fX,Y(x, y)dyfY(y) =Z fX,Y(x, y)dx3We will define independence of two contiunous random variables differ-ently than the book.

6 The two definitions are , Ybe Jointly continuous random variables with jointdensityfX,Y(x, y)and marginal densitiesfX(x),fY(y). We say they areindependent iffX,Y(x, y) =fX(x)fY(y)If we know the joint density ofXandY, then we can use the definitionto see if they are independent. But the definition is often used in a differentway. If we know the marginal densities ofXandYandwe know that theyare independent, then we can use the definition to find their joint :IfXandYare independent random variables and each has thestandard normal distribution, what is their joint density?

7 F(x, y) =12 exp( 12(x2+y2))Example:Suppose thatXandYhave a joint density that is uniform onthe disc centered at the origin with radius 1. Are they independent?Example:IfXandYhave a joint density that is uniform on the square[a, b] [c, d], then they are :Suppose thatXandYhave joint densityf(x, y) = e x yifx, y 00,otherwiseAreXandYindependent?Example: Suppose thatXandYare uniform on [0,1]andYhas the Cauchy density.(a) Find their joint density.(b) ComputeP(0 X 1/2,0 Y 1)(c) ComputeP(Y X). Expected valueIfXandYare Jointly continuously random variables , then the mean ofXis still given byE[X] =Z x fX(x)dxIf we write the marginalfX(x) in terms of the joint density, then this becomesE[X] =Z Z x fX,Y(x, y)dxdyNow suppose we have a functiong(x, y) fromR2toR.

8 Then we can definea new random variable byZ=g(X, Y). In a later section we will see how tocompute the density ofZfrom the joint density ofXandY. We could thencompute the mean ofZusing the density ofZ. Just as in the discrete casethere is a , Ybe Jointly continuous random variables with jointdensityf(x, y). Letg(x, y) :R2 R. Define a new random variable byZ=g(X, Y). ThenE[Z] =Z Z g(x, y)f(x, y)dxdyprovidedZ Z |g(x, y)|f(x, y)dxdy < An important special case is the followingCorollary Jointly continuous random variables anda, bare real numbers, thenE[aX+bY] =aE[X] +bE[Y]Example:XandYhave joint densityf(x, y) = x+yif 0 x 1,0 y 10,otherwiseLetZ=X+Y.

9 Find the mean and variance now consider independence and independent and Jointly continuous , thenE[XY] =E[X]E[Y] they are independent,fX,Y(x, y) =fX(x)fY(y). SoE[XY] =Z Zxy fX(x)fY(y)dxdy= Zx fX(x)dx Zy fY(y)dy =E[X]E[Y] Function of two random variablesSupposeXandYare Jointly continuous random variables . Letg(x, y) be afunction fromR2toR. We define a new random variable byZ=g(X, Y).Recall that we have already seen how to compute the expected value ofZ. Inthis section we will see how to compute the density ofZ. The general strategyis the same as when we considered functions of one random variable : we firstcompute the cumulative distribution :LetXandYbe independent random variables , each of which isuniformly distributed on [0,1].

10 LetZ=XY. First note that the range ofZis [0,1].FZ(z) =P(Z z) =Z ZA1dxdyWhereAis the regionA={(x, y) : 0 x 1,0 y 1, xy z}PICTUREFZ(z) =z+Z1z"Zz/x01dy#dx=z+Z1z"Zz/x01dy#dx6=z+ Z1zzxdx=z+zlnx|1z=z zlnzThis is the cdf ofZ. So we differentiate to get the (z) =ddzz zlnz= 1 lnz z1z= lnzfZ(z) = lnz,if 0 z 10,otherwiseExample:LetXandYbe independent random variables , each of which isexponential with parameter . LetZ=X+Y. Find the density get gamma with same andw= is special case of a much more general result. The sum of gamma( , w1)and gamma( , w2) is gamma( , w1+w2).