[機統]Order statistics:Y1,Y2-Y1獨立=>X~Gamma?

[yueayase (scrya)]

(From Craig Hogg Mathematical statistics 6th, exercise 5.2.16)
Question: Order statistics Y1 < Y2 of a random sample of size 2 from a

          continous type distribution with pdf f(x) such that f(x) > 0

          for x≧0, f(x)=0, elsewhere. Show that if Z1=Y1, Z2=Y2-Y1 are

          independent, f(x) has a Gamma pdf with α=1, β>0(That is, a
          exponential distribution with mean β)


過程:

Step 1: Find the joint pdf of Z1 and Z2

f(y1,y2) = 2f(y1)f(y2) , 0≦y1<y2<∞

Y1 = Z1, Y2 = Z1 + Z2 => ∂Y1/ ∂Z1 = ∂Y2/∂Z2 = ∂Y2/∂Z2=1, ∂Y1/∂Z2=0

=> |J| = |1 0|  = 1
         |1 1|

=> g(z1,z2) = |J|f(z1,z1+z2) = 2f(z1)f(z1+z2), 0<z1<∞, 0 < z2 < ∞


Step 2:

因為Z1, Z2獨立, 所以 g(z1,z2) = g (z1) g (z2),
                                 1      2

           ∞                            ∞
g (z1) = ∫   2f(z1)f(z1+z2)dz2 = 2f(z1)∫ f(u)du = 2f(z1)[1-F(z1)]
 1         0                             z1

           ∞
g (z2) = ∫  2f(z1)f(z1+z2)dz1 = ?
 2         0



g  (z2) 好像沒有漂亮的形式
 2

所以不知道題目的Hint: h(0)h(x+y)=h(x)h(y)有什麼用?


--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.251.164.119
推 rorbert7    :當joint pdf  可拆成f(x,y)=g(x)g(y) and 為積空間
→ rorbert7    :則 X獨立Y  g(x) and g(y)不一定要是PDF
→ yueayase    :但是拆出的2函數不一定能用f(x+y)的f表示?
雖然利用g1(z1)=2f(z1)[1-F(z1)]代回 g(z1,z2) = 2f(z1)f(z1+z2)

可得


f(z1+z2) = [1-F(z1)]g (z2)
                     2


乍看之下的確拆成2個只有單變數的函數.

但如果他的提示是對的,


應該要能夠進一步讓 [1-F(z1)] = c1 f(z1) 且 g (z2) = c2 f(z2)
                                            2

但想不出任何方法可以推出這樣的結果

※ 編輯: yueayase        來自: 111.251.165.155      (09/10 02:16)