5. Random vectors and Joint Probability Distribution s随机向量与联合概率分布5.1 Concept of Joint Probability Distributions(1) Discrete Variables Case 离散型Often, trials are conducted where two random variables are observed simultaneously in order to determine not only their individual behavior but also the degree of relationship between them.( X, Y)For two discrete random variables X and Y, we write the probability that X will take the value x and Y will take the value y as P(X=x, Y=y). Consequently, P(X=x, Y=y) is the probability of the intersection of the events X=x and Y=y.(X=x, Y=y) ------ (X=x)∩(Y=y)The distribution of probability is specified by listing the probabilities associated with all possible pairs of values x and y, either by formula or in a table. We refer to the function p(x, y)=P(X=x, Y=y) and the corresponding possible values (X, Y) as the j oint probability distribution (联合分布)of X and Y.They satisfy(,)0, (,)1xyp x y p x y ≥=∑∑,where the sum is over all possible values of the variable.Example 5.1.1 Calculating probabilities from a discrete joint probability distributionLet X and Y have the joint probability distribution.(a) Find (1)P X Y +>;(b) Find the probability distribution ()()X p x P X x == of the individualrandom variable X . Solution(a) The event 1X Y +>is composed of the pairs of values (l,1), (2,0), and (2,l). Adding their corresponding probabilities(1)(1,1)(2,0)(2,1)0.20.100.3.P X Y p p p +>=++=++=(b) Since the event X =0 is composed of the two pairs of values (0,0) and (0,1), we add their corresponding probabilities to obtain(0)(0,0)(0,1)0.10.20.3P X p p ==+=+=.Continuing, we obtain (1)(1,0)(1,1)0.40.20.6P X p p ==+=+= and(2)(2,0)(2,1)0.100.1P X p p ==+=+=.In summary, (0)0.3X p =, (1)0.6X p = and (2)0.1X p =is the probabilitydistribution of X . Note that the probability distribution ()X p x of appears in the lower margin of this enlarged table. The probability distribution ()Y p y of Y appears in the right-hand margin of the table. Consequently, the individual distributions are called marginal probability distributions .（边缘分布）From the example, we see that for each fixed value of x , the marginalprobability distribution is obtained as()()(,)X yP X x p x p x y ===∑,where the sum is over all possible values of the second variable. Continuing, we obtain()()(,)Y xP Y y p y p x y ===∑.Example 3.5.3Suppose the number X of patent applications （专利申请）submitted by a company during a 1-year period is a random variable having thePoisson distribution with mean λ, (()!n e P X n n λλ-==)and the variousapplications independently have probability (0,1)p ∈ of eventually being approved.Determine the distribution of the number of patent applications during the 1-year period that are eventually approved.先求联合分布密度，再求边缘分布Solution Let Y be the number of patent application being eventually approved during 1-year period. Then the event {}Y k = is the union of mutually exclusive events {,}X n Y k == ()n k ≥.If X n =, then the random variable S has the binomial distribution with parameter n and p :(|)(1)k k n k n P Y k X n C p p -===-. (0)n k ≥≥ Thus(,)()(|)P X n Y k P X n P Y k X n ====== (1)!nk kn k n e C p p n λλ--=⋅⋅-when k>n, P(X=n, Y=k)=0,Hence the distribution of Y is()(,)(,)n n kP Y k P X n Y k P X n Y k ∞∞=========∑∑(1)!nk kn k n n ke C p p n λλ∞--==⋅⋅-∑!(1)!!()!nk n k n k n e p p n k n k λλ∞--==⋅⋅--∑(1)!()!kn kkn k n ke p p k n k λλλ-∞--==⋅⋅--∑()(1)(1)()()!!!mk k p m p p p e e ek m k λλλλλλ∞---=-==∑ ()!k pp e k λλ-= Thus, Y has the Poisson distribution of mean p λ. exercise从1，2，3，4，5五个数中不放回随机的接连地取3个，然后按大小排成123X X X <<，试求13(,)X X 的联合分布，x1,x3 独立吗？Homework Chap 5 1,(2) Continuous Variables Case 连续型随机向量There are many situations in which we describe an outcome by giving the values of several continuous random variables. For instance, we may measure the weight and the hardness of a rock, the pressure and the temperature of a gas. Suppose that X and Y are two continuous random variables. A function (,)f x y is called the joint probability density of these random variables, if the probability that , a X b c Y d ≤≤≤≤ is given by the multiple integral(, )(,)b da cP a X b c Y d f x y dxdy ≤≤≤≤=⎰⎰Thus, a function (,)f x y can serve as a joint probability density if all of the following hold:for all values of x and y , f is integrable on R 2 andTo extend the concept of a cumulative distribution function to the two variables case, we can define F (x , y )(, )(, )F x y P X x Y y =≤≤,and we refer to the corresponding function F as the joint cumulative distribution function of the two random variables.Example 5.1.2If the joint probability density of two random variables is given by236 for 0,0(,)0 elsewherex y e x y f x y --⎧>>=⎨⎩ Find the joint distribution function, and use it to find the probability(2,4)P X Y ≤≤.Solution By definition,23006 for 0, 0(,)(,)0 elsewhere y x yu vxe du e dv x y F x yf u v dudv ---∞-∞⎧>>⎪==⎨⎪⎩⎰⎰⎰⎰Thus,23(1)(1) for >0, >0(,)0 elsewhere x y e e x y F x y --⎧--=⎨⎩.Hence,412(2, 4)(2, 4)(1)(1)0.9817P X Y F e e --≤≤==--=.ExampleIf the joint probability density of two random variables is given by2,1,01(,)0,kxy x y x f x y ⎧≤≤≤≤=⎨⎩其他(a)find the k; (b)find the probability2((,)),{(,)|,01}P X Y D D x y x y x x ∈=≤≤≤≤solutionsince(,)1f x y dxdy ∞∞-∞-∞=⎰⎰24111001(,)()226x x kf x y dxdy dx kxydy k x dx ∞∞-∞-∞==-=⎰⎰⎰⎰⎰ hence k=6.21124001((,))663()4xx DP X Y D xydxdy dx xydy x x x dx ∈===-=⎰⎰⎰⎰⎰joint marginal densities 边缘密度Given the joint probability density of two random variables, the probability density of the X or Y can be obtained by integrating out another variable,The functions f X and f Y respectively are called the marginal density （边缘密度）of X and Y .,ExampleThe joint probability density of two random variables is given by26,1,01(,)0,xy x y x f x y ⎧≤≤≤≤=⎨⎩其他find the marginal density from the joint density when [0,1]x ∈，215()(,)633X xf x f x v dv xydy x x +∞-∞====-⎰⎰[0,1]x ∉，()0X f x =，hence 533,01()0,X x x x f x elsewhere ⎧-≤≤=⎨⎩23,01()0,Y y y f x elsewhere ⎧≤≤=⎨⎩exercises求服从B 上均匀分布的随机向量（X,Y ）的分布密度及分布函数。
