P 问题1-在线等~~

是 否 +2 论坛币

k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群

赵安豆老师微信：zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

立即领取

感谢您参与论坛问题回答

经管之家送您两个论坛币！

+2 论坛币

1. An insurance company sells its auto insurance product to the family of one, two
and three drivers. In the next year, the company will rebate those families who
le at most one claim last year and the number of drivers is strictly great than
the number of claims. The company will cover at most three claims for one
family. The joint mass probability of the number of drivers X and the number
of claims Y is proportional to 8 + X-2Y. Find the rebate probability.

2. For a given policyholder, the probability that he les claims is given by
P(N = 0) = 0.7; P(N = 1) = 0.2; P(N = 2) = 0.1:
When he les one claim, the amount of the claim follows a uniform distribution
over [0; 60]; if there are two claims, the amount of two claims follows a joint
uniform distribution over [0; 60]  [0; 60]. Calculate the probability that the
total amount of the claims the policyholder files is 48 or less.�-- 2Y . Find the rebate probability.

扫码加我 拉你入群

请注明：姓名-公司-职位

以便审核进群资格，未注明则拒绝

分享0 收藏0 回帖

关键词：在线等 distribution proportional Probability Proportion 在线

自己顶下！

这个第一问，断句是不是有点问题呀。。还有部分地方缩写了。。搞不懂意思
第二问。。两个的时候，用图像法就行了

X可以取1,2,3 Y可以取0,1,2,3  题目很难理解吗？
这种类型的题目出现过啊  一家买了2车保险 一个deductible 1 另一个2 计算总claims少于多少的概率
这里要求Y <=1 且 X>Y  穷举法列出所有可能就行了

第二道也是基本题啊  分N=0, N=1, N=2三种情况啊
P(N=0, Y<=48)=P(N=0)=0.7
P(N=1, Y<=48)=P(N=1) P(Y<=48|N=1) = 0.2 * 48/60 =0.12
P(N=2, Y<=48)=P(N=2) P(Y<=48|N=1) = 0.1 * (0.5* 48 *48)/(60*60)
最好求和就好  
对于均匀分布 一维对应长度 二维对应面积

Ah, X,Y 反了，改了。直接求相对值还是要求和，一样的。
1. P(rebate) = P(Y<=1, X>Y)
    joint pmf f(x,y) = k (8+x-2y) , total probability =1, that is, sum_(x=0)^(3) sum_(y=1)^(3) f(x,y) = 1,
    plug in,  f(x,y) = k (8+x-2y), get:
    k sum_(x=1)^(3) sum_(y=0)^(3) (8+x-2y) = 1,
    notice that:  sum_(x=1)^(3) sum_(y=0)^(3) 8 = 3*4*8=96,  
                          sum_(x=1)^(3) sum_(y=0)^(3) x = 4*sum_(x=1)^(3) x = 4*(1+3)*3/2=24
                          sum_(x=1)^(3) sum_(y=0)^(3) y = 3*sum_(y=0)^(3) y = 3*(0+3)*4/2=18
    therefore, k *(96+24-2*18)=84k =1, thus  k = 1/84  and  the pmf is   f(x,y) = (8+x-2y)/84
    Finally  P(rebate) = P(Y<=1, X>Y) = sum_(y=0,1) sum_(x>y) (8+x-2y)/84
              = (1/84) [sum_(y=0,1) sum_(x>y)8 + sum_(y=0,1) sum_(x>y)x - 2sum_(y=0,1) sum_(x>y)y ]
              = (1/84) { [ sum_(y=0) sum_(x=1,2,3) 8 +sum_(y=1) sum_(x=2,3) 8 ]
                            + [ sum_(y=0) sum_(x=1,2,3) x +sum_(y=1) sum_(x=2,3) x ]
                            - 2 [ sum_(y=0) sum_(x=1,2,3) y +sum_(y=1) sum_(x=2,3) y ] }
              =(1/84) [ 24 + 16 + 6 + 5 - 2*(0+2) ] =47/84


2. let C be the total claim amount,  X be the claim amount when N=1, Y,Z be the claim amounts,  individually, when N=2, then from the question, we know:   X|N=1 ~ Unif(0, 60) ,
   Y,Z|N=2 ~ joint bivariate Uniform on [0,60]x[0,60]
   (and obviously, any other cases have 0 claim amount), so
    P(C<=48) = P(C<=48|N=0)P(N=0) + P(X<=48|N=1)P(N=1) + P(Y+Z<=48|N=2)P(N=2)
= 1*0.7 + (48/60) * 0.2 + [ int_(0<=y<=48) dy int_(0<=z<=48-y) (1/3600) dz } * 0.1

    notice that the double integral is essentially a triangle's area, which is confined by y-axis, z-axis, and the straight line y+z=48, therefore, the double integral is (1/3600)*(1/2)*48*48 = 0.32
    finally,  P(C<=48)=0.7+ 0.8*0.2 + 0.32*0.1 = 0.892

isuck 发表于 2011-7-22 03:00
1. P(rebate) = P(XX)
    joint pmf f(x,y) = k (8+x-2y) , total probability =1, that is, sum_(x=0)^(3) sum_(y=1)^(3) f(x,y) = 1,
    plug in,  f(x,y) = k (8+x-2y), get:
    k sum_(x=0)^(3) sum_(y=1)^(3) (8+x-2y) = 1,
    notice that:  sum_(x=0)^(3) sum_(y=1)^(3) 8 = 4*3*8=96,  
                          sum_(x=0)^(3) sum_(y=1)^(3) x = 3*sum_(x=0)^(3) x = 3*(0+3)*4/2=18
                          sum_(x=0)^(3) sum_(y=1)^(3) y = 4*sum_(y=1)^(3) y = 4*(1+3)*3/2=24
    therefore, k *(96+18-2*24)=66k =1, thus  k = 1/66  and  the pmf is   f(x,y) = (8+x-2y)/66
    Finally  P(rebate) = P(XX) = sum_(x=0,1) sum_(y>x) (8+x-2y)/66
              = (1/66) [sum_(x=0,1) sum_(y>x)8 + sum_(x=0,1) sum_(y>x)x - 2sum_(x=0,1) sum_(y>x)y ]
              = (1/66) { [ sum_(x=0) sum_(y=1,2,3) 8 +sum_(x=1) sum_(y=2,3) 8 ]
                            + [ sum_(x=0) sum_(y=1,2,3) x +sum_(x=1) sum_(y=2,3) x ]
                            - 2 [ sum_(x=0) sum_(y=1,2,3) y +sum_(x=1) sum_(y=2,3) y ] }
              =(1/66) [ 24+ 16 + 3*0+ 2*1 - 2(1*6+1*5) ] =20/66 = 10/33


2. let C be the total claim amount,  X be the claim amount when N=1, Y,Z be the claim amounts,  individually, when N=2, then from the question, we know:   X|N=1 ~ Unif(0, 60) ,
   Y,Z|N=2 ~ joint bivariate Uniform on [0,60]x[0,60]
   (and obviously, any other cases have 0 claim amount), so
    P(C

第一题要求是Y<=1 不是X  而且对于这种不需要求k  最后分子分母都有可以约掉  直接计算相对值就可以
X=1,2,3  Y=0,1,2,3  代入数值就可以
#(1,0)= 9  #(1,1)=7 #(1,2)=5 #(1,3)=3 ......
(Y<=1, X>Y)=(1,0) + (2,0) + (2,1) + (3,0) + (3,1)

thx!