推 yaevna:謝謝你^^ 07/11 15:19
推 alienQbi:B.看不懂~~~~ 07/13 23:48
推 loveliver:用反函數的概念試看看 07/14 23:27
※ 引述《yaevna (yaevna)》之銘言：
: Lex X have a logistic distrubution with p.d.f.
: f(X)=[e^(-x)]/[1+e^(-x)]^2，-∞< x <∞
: A. Find the c.d.f. of X, say F(x)
: B. Let Y=F(X). Find the p.d.f. of Y.
: 麻煩大家了~謝謝
x e^(-u)
A. F(x) = ∫ ----------------- du ( let 1 + e^(-u) = y )
-∞ [1 + e^(-u) ]^2
1+e^(-x) -1
= ∫ -------------- dy
-∞ y^2
1
= ------------
1 + e^(-x)
-1 -1
B. F (y) = P(Y<=y) = P(F(X)<=y) = P(X <= F(y) ) = F (F(y)) = y
Y X
=> Y ~ U ( 0 , 1 )
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