[理工] 找順序統計量的分配
Let Y1<Y2<...<Y10 be the order statistics of a random sample
of size 10 from a distribution of the continuous type having
distribution function F(x). Determine the distribution of V
= F(Y10)-F(Y5)+F(Y2)-F(Y1)
iid
這一題因為F(X1),F(X2),...,F(X10)----->Uni(0,1)
所以F(Y1),F(Y2),...,F(Y10)很好找
所以我就從F(Y10),F(Y5),F(Y2),F(Y1)他們的joint pdf下手去做
另W1=F(Y10)-F(Y5) F(Y5)=W3-W1
|-1 0 1 0|
W2=F(Y2)-F(Y1) F(Y1)=W4-W2 | 0 -1 0 1|
|J|=|| 0 0 1 0||=1
W3=F(Y10) F(Y10)=W3 | 0 0 0 1|
W4=F(Y2) F(Y2)=W4
再來就作變數變換卡在範圍不知道怎麼定...
還是有別的作法可以做出來呢?
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上面的這個證法我有看過,只是證到2以上的就不知道該怎麼證了...
然後後來我換別的轉換來證
F(Y1)=y1 F(Y2)=y2 F(Y5)=y3 F(Y10)=y4
10! 2 4
f(y1,y2,y3,y4)=----*(y3-y2)*(y4-y3) 0<y1<y2<y3<y4<1
2!4!
W1=y4-y3+y2-y1 y1=W4-W3+W2-W1
|-1 1 -1 1|
W2=y2 y2=W2 | 0 1 0 |
|J|=|| 0 0 1 0||=1
W3=y3 y3=W3 | 0 0 0 1|
W4=y4 y4=W4
10! 2 4
f(w1,w2,w3,w4)=----*(w3-w2)*(w4-w3) 0<w4-w3+w2-w1<w2<w3<w4<1
2!4!
1 10! 2 4
f(w1,w2,w3) =∫ ----*(w3-w2)*(w4-w3)dw4
w3 2!4!
10! 2 5
=----*(w3-w2)*(1-w3) 0<-w3+w2-w1<w2<w3<1
2!5
1 10! 2 5
f(w1,w2) =∫ ----*(w3-w2)*(1-w3)dw3
w2 2!5!
10! 8
=----*(1-w2) 0<w2-w1<w2<1
8!
1 10! 8
f(w1) =∫ ----*(1-w2)dw3
w1 8!
10! 9
=----*(1-w1) 0<w1<1 證出來變成Beta(1,10)QQ 請問是哪裡錯了呢?
9!
※ 編輯: tokyo291 來自: 61.227.244.17 (02/19 23:40)
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