Penurunan perubahan variabel dari fungsi kepadatan probabilitas?

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

di mana,adalah pdf yang sesuai dengan

p_{y} (y) = p_{x} (x) | \frac{d x}{d y} | = p_{x} (g (y)) | g^{'} (y) |

$p_y(y)=p_x(x) \left | \frac{d x}{d y} \right |=p_x(g(y)) | g'(y) |$

x = g (y)

$x=g(y)$

p_{x} (x)

$p_x(x)$

p_{y} (y)

$p_y(y)$ sehubungan dengan perubahan variabel.

Buku-buku mengatakan itu karena pengamatan jatuh dalam kisaran akan, untuk nilai-nilai kecil , ditransformasikan menjadi kisaran . $(x, x + \delta x)$ $\delta x$ $(y, y + \delta y)$

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

P (Y \leq y) = P (g (X) \leq y) = {\begin{cases} P (X \leq g^{- 1} (y)), & if g is monotonically increasing \\ P (X \geq g^{- 1} (y)), & if g is monotonically decreasing \end{cases}

$\begin{equation} P(Y\le y) = P(g(X)\le y)= \begin{cases} P(X\le g^{-1}(y)), & \text{if}\ g \text{ is monotonically increasing} \\ P(X\ge g^{-1}(y)), & \text{if}\ g \text{ is monotonically decreasing} \end{cases} \end{equation}$

g

$g$ secara monoton secara monoton meningkatkan

F_{Y} (y) = F_{X} (g^{- 1} (y))

$F_{Y}(y)=F_{X}(g^{-1}(y))$

jika secara monoton menurunkan

f_{Y} (y) = f_{X} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y)

$f_{Y}(y)= f_{X}(g^{-1}(y))\cdot \frac{d}{dy}g^{-1}(y)$

F_{Y} (y) = 1 - F_{X} (g^{- 1} (y))

$F_{Y}(y)=1-F_{X}(g^{-1}(y))$

f_{Y} (y) = - f_{X} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y)

$f_{Y}(y)=- f_{X}(g^{-1}(y))\cdot \frac{d}{dy}g^{-1}(y)$

∴ f_{Y} (y) = f_{X} (g^{- 1} (y)) \cdot | \frac{d}{d y} g^{- 1} (y) |

$\therefore f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left | \frac{d}{dy}g^{-1}(y) \right |$

— dontloo
sumber

The result holds only if

g

$g$ is a strictly monotone increasing or decreasing function. Draw a graph of

g

$g$ 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

@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

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:

\begin{array}{rcl} P (Y \leq y) & = & P (g (X) \leq y) \\ = & P (X \leq g^{- 1} (y)) \\ o r F_{Y} (y) & = & F_{X} (g^{- 1} (y)), by the definition of CDF \end{array}

$\begin{eqnarray*} P(Y\le y) &=& P(g(X)\le y)\\ &=& P(X\le g^{-1}(y))\\ or\;\;F_{Y}(y)&=& F_{X}(g^{-1}(y)),\quad \mbox{by the definition of CDF}\\ \end{eqnarray*}$ By differentiating the CDFs on both sides w.r.t.

y

$y$ , we get the pdf of

Y

$Y$ . The function g() could be either monotonically increasing or monotonically decreasing. If the function g() is monotonically increasing, then the pdf of

Y

$Y$ is given by

f_{Y} (y) = f_{X} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y)

$\begin{equation*} f_{Y}(y)= f_{X}(g^{-1}(y))\cdot \frac{d}{dy}g^{-1}(y) \end{equation*}$ and other hand, if it is monotonically decreasing, then the pdf of

Y

$Y$ is given by

f_{Y} (y) = - f_{X} (g^{- 1} (y)) \cdot \frac{d}{d y} g^{- 1} (y)

$\begin{equation*} f_{Y}(y)= - f_{X}(g^{-1}(y))\cdot \frac{d}{dy}g^{-1}(y) \end{equation*}$ The above two equations can be combined into a single equation:

∴ f_{Y} (y) = f_{X} (g^{- 1} (y)) \cdot | \frac{d}{d y} g^{- 1} (y) |

$\begin{equation*} \therefore f_{Y}(y) = f_{X}(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)| \end{equation*}$

— 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

g^{- 1}

$g^{-1}$ is not necessarily a constant function, so it can be >1 in some places and <1 in others.

— Yatharth Agarwal