Penurunan perubahan variabel dari fungsi kepadatan probabilitas?

15

Dalam pengenalan pola buku dan pembelajaran mesin (rumus 1.27), itu memberi

di manax=g(y),px(x)adalah pdf yang sesuai denganpy(y)

py(y)=px(x)|dxdy|=px(g(y))|g(y)|
x=g(y)px(x)py(y) sehubungan dengan perubahan variabel.

Buku-buku mengatakan itu karena pengamatan jatuh dalam kisaran akan, untuk nilai-nilai kecil δ x , ditransformasikan menjadi kisaran ( y , y + δ y ) .(x,x+δx)δx(y,y+δy)

Bagaimana ini diturunkan secara formal?


Pembaruan dari Dilip Sarwate

Hasilnya hanya berlaku jika adalah peningkatan atau penurunan fungsi monoton secara ketat.g


Beberapa perubahan kecil pada jawaban LV Rao maka jika g

P(Yy)=P(g(X)y)={P(Xg1(y)),if g is monotonically increasingP(Xg1(y)),if g is monotonically decreasing
g secara monoton secara monoton meningkatkan f Y ( y ) = f X ( g - 1 ( y ) ) d
FY(y)=FX(g1(y))
jika secara monoton menurunkan FY(y)=1-FX(g-1(y))fY(y)=-fX(g-1(y))d
fY(y)=fX(g1(y))ddyg1(y)
FY(y)=1FX(g1(y))
fY(y)=fX(g1(y))ddyg1(y)
fY(y)=fX(g1(y))|ddyg1(y)|
dontloo
sumber
1
The result holds only if g is a strictly monotone increasing or decreasing function. Draw a graph of g and puzzle it out using the basic idea behind the definition of the derivative (not the formal definition with epsilon and delta). Also, there is an answer by @whuber on this site which spells out the details; that is, this should be closed as a duplicate.
Dilip Sarwate
Your book's explanation is reminiscent of the one I offered at stats.stackexchange.com/a/14490/919. I also posted a general algebraic method at stats.stackexchange.com/a/101298/919 and a geometric explanation at stats.stackexchange.com/a/4223/919.
whuber
1
@DilipSarwate thanks for your explanation, I think I understand the intuition, but I'm more interested in how it can be derived using the existing rules and theorems :)
dontloo

Jawaban:

14

Suppose X is a continuous random variable with pdf f(x). If we define Y=g(X), where g() is a monotone function, then the pdf of Y is obtained as follows:

P(Yy)=P(g(X)y)=P(Xg1(y))orFY(y)=FX(g1(y)),by the definition of CDF
By differentiating the CDFs on both sides w.r.t. y, we get the pdf of Y. The function g() could be either monotonically increasing or monotonically decreasing. If the function g() is monotonically increasing, then the pdf of Y is given by
fY(y)=fX(g1(y))ddyg1(y)
and other hand, if it is monotonically decreasing, then the pdf of Y is given by
fY(y)=fX(g1(y))ddyg1(y)
The above two equations can be combined into a single equation:
fY(y)=fX(g1(y))|ddyg1(y)|
L.V.Rao
sumber
But as the integral over fx must sum to 1 and fy is a scaled version of fx, doesn't that mean fy is not a proper pdf, unless the jacobian in the abs() is 1 or -1?
Chris
@Chris The Jacobian of g1 is not necessarily a constant function, so it can be >1 in some places and <1 in others.
Yatharth Agarwal