跳至主要内容

隨機變量的變換（Transformations of Random Variables）

單變量變換​

EX X∼B(n,p)X\sim B(n,p), let Y=n−X=h(X)Y=n-X=h(X)

  ⟹  fY(y)=P(Y=y)=P(n−X=y)=P(X=n−y)=(nn−y)pn−y(1−p)y=(ny)py(1−p)n−y\implies f_Y(y)=P(Y=y)=P(n-X=y)=P(X=n-y)=\binom{n}{n-y}p^{n-y}(1-p)^y=\binom{n}{y}p^y(1-p)^{n-y}

i.e. Y∼B(n,p)Y\sim B(n,p)

EX X∼N(μ,σ2)X\sim N(\mu,\sigma^2)

Y={1if X≥μ−1if X<μY=\begin{cases} 1 & \text{if } X\ge \mu\\ -1 & \text{if } X<\mu \end{cases}   ⟹  fY(y)=P(Y=y)={P(X≥μ)if y=1P(X<μ)if y=−1=12\implies f_Y(y)=P(Y=y)= \begin{cases} P(X\ge \mu) & \text{if } y=1\\ P(X<\mu) & \text{if } y=-1 \end{cases}=\frac{1}{2}

對於連續的隨機變量變換, X∼f(x),Y=h(x)X\sim f(x), Y=h(x)

  ⟹  \implies cdf of YY is FY(y)=P(Y≤y)FY(y)=P(Y\le y) and pdf of YY is fY(y)=ddyFY(y)f_Y(y)=\frac{d}{dy}F_Y(y)

如果 h(x)h(x) 是單調遞增，則 h−1(y)h^{-1}(y) 存在，並且微分為正

Fy(y)≜P(Y≤y)=P(h(X)≤y)=P(X≤h−1(y))=FX(h−1(y))F_y(y)\triangleq P(Y\le y)=P(h(X)\le y)=P(X\le h^{-1}(y))=F_X(h^{-1}(y))   ⟹  fY(y)=ddyFY(y)=ddyFX(h−1(y))=fX(h−1(y))ddyh−1(y)⏟>0\implies f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}F_X(h^{-1}(y))=f_X(h^{-1}(y))\underbrace{\frac{d}{dy}h^{-1}(y)}_{>0}

如果 h(x)h(x) 是單調遞減，則 h−1(y)h^{-1}(y) 存在，並且微分為負

Fy(y)≜P(Y≤y)=P(h(X)≤y)=P(X≥h−1(y))=1−FX(h−1(y))F_y(y)\triangleq P(Y\le y)=P(h(X)\le y)=P(X\ge h^{-1}(y))=1-F_X(h^{-1}(y))   ⟹  fY(y)=ddyFY(y)=ddy(1−FX(h−1(y)))=−fX(h−1(y))ddyh−1(y)⏟<0\implies f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}(1-F_X(h^{-1}(y)))=-f_X(h^{-1}(y))\underbrace{\frac{d}{dy}h^{-1}(y)}_{<0}
Theorem

如果 XX 有連續的 pdf fX(x)f_X(x)， 並且 hh 是 1-1 ，且 h′(x)h'(x) 存在連續導數，則 Y=h(X)Y=h(X) 有 pdf

fY(y)=fX(h−1(y))∣ddyh−1(y)∣,y∈Yf_Y(y)=f_X(h^{-1}(y))\left|\frac{d}{dy}h^{-1}(y)\right|, y\in \mathscr{Y}

其中 Y=h(X),X={ x;fX(x)>0 }\mathscr{Y}=h(\mathscr{X}), \mathscr{X}=\set{x;f_X(x)>0}

EX: X∼N(μ,σ2)X\sim N(\mu,\sigma^2), let Y=h(X)=X−μσ  ⟺  X=σY+μ=h−1(y)  ⟹  ddyh−1(y)=σY=h(X)=\frac{X-\mu}{\sigma}\iff X=\sigma Y+\mu=h^{-1}(y)\implies \frac{d}{dy}h^{-1}(y)=\sigma

fY(y)=fX(σy+μ)∣ddy(σy+μ)∣=12πe−y22∀y∈R\begin{align*} f_Y(y)&=f_X(\sigma y+\mu)\left|\frac{d}{dy}(\sigma y+\mu)\right|\\ &=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\quad \forall y\in\R \end{align*}

  ⟹  Y∼N(0,1)\implies Y\sim N(0,1)

EX: Let X∼Gamma(α,β)X\sim \text{Gamma}(\alpha,\beta) with α>0,β>0\alpha>0,\beta>0 given, and Y=2Xβ=h(X)Y=\frac{2X}{\beta}=h(X)

  ⟹  Y=h(X)=(0,∞)\implies \mathscr{Y}=h(\mathscr{X})=(0,\infty) and X=βY2h−1(Y)X=\frac{\beta Y}{2}h^{-1}(Y)

fY(y)=fX(h−1(y))∣ddyh−1(y)∣,y∈Y=fX(βy2)∣β2∣,y>0=1Γ(α)βα(βy2)α−1exp⁡(−βyβ⋅β2)=1Γ(α)2αyα−1e−y/2,y>0=pdf of Gamma(α,2)\begin{align*} f_Y(y)&=f_X(h^{-1}(y))\left|\frac{d}{dy}h^{-1}(y)\right|, y\in\mathscr{Y}\\ &=f_X\left(\frac{\beta y}{2}\right)\left|\frac{\beta}{2}\right|, y>0\\ &=\frac{1}{\Gamma(\alpha)\beta^\alpha}\left(\frac{\beta y}{2}\right)^{\alpha-1}\exp\left(-\frac{\beta y}{\beta}\cdot\frac{\beta}{2} \right)\\ &=\frac{1}{\Gamma(\alpha)2^\alpha}y^{\alpha-1}e^{-y/2}, y>0\\ &=\text{pdf of Gamma}(\alpha,2) \end{align*}

i.e. X∼Gamma(α,β)  ⟹  β∗Xβ∼Gamma(α,β∗)X\sim \text{Gamma}(\alpha,\beta)\implies \frac{\beta^*X}{\beta}\sim \text{Gamma}(\alpha,\beta^*)

Remark:

  1. X∼Gamma(k2,2)=χk2X\sim\text{Gamma}(\frac{k}{2},2)=\chi^2_k

  2. X∼Gamma(α,β),Y=1XX\sim\text{Gamma}(\alpha,\beta), Y=\frac{1}{X}

    fY(y)=fX(1y)∣−1y2∣=1Γ(α)βα(1y)α+1e−1/βy,y>0\begin{align*} f_Y(y)=f_X(\frac{1}{y})\left|-\frac{1}{y^2}\right|=\frac{1}{\Gamma(\alpha)\beta^\alpha}(\frac{1}{y})^{\alpha+1}e^{-1/\beta y}, y>0 \end{align*}

      ⟹  y\implies y has a inverse Gamma dist

EX: X∼U(a,b)X\sim U(a,b), and Y=X−ab−a=h(X)Y=\frac{X-a}{b-a}=h(X)

  ⟹  Y=(0,1),X=(b−a)Y+a=h−1(y)\implies \mathscr{Y}=(0,1), X=(b-a)Y+a=h^{-1}(y)

  ⟹  fY(y)=fX(h−1(y))∣ddyh−1(y)∣,y∈Y=1b−a⋅(b−1)=1\begin{align*} \implies f_Y(y)&=f_X(h^{-1}(y))\left|\frac{d}{dy}h^{-1}(y)\right|, y\in\mathscr{Y}\\ &=\frac{1}{b-a}\cdot (b-1)=1 \end{align*}

i.e. Y∼U(0,1)Y\sim U(0,1)

EX: U∼U(0,1),X=−2ln⁡U=h(U)∈(0,∞)  ⟹  U=e−X/2=h−1(X)U\sim U(0,1), X=-2\ln U=h(U)\in(0,\infty)\implies U=e^{-X/2}=h^{-1}(X)

  ⟹  fX(x)=fU(h−1(x))∣ddxh−1(x)∣=1⋅∣−12e−x/2∣=12e−x/2,x>0\implies f_X(x)=f_U(h^{-1}(x))\left|\frac{d}{dx}h^{-1}(x)\right|=1\cdot\left|\frac{-1}{2}e^{-x/2}\right|=\frac{1}{2}e^{-x/2}, x>0

i.e. X∼χ22  ⟹  −2ln⁡U∼χ22X\sim\chi^2_2\implies -2\ln U\sim \chi^2_2

EX: Zsin⁡N(0,1),Y=Z2=h(Z)Z\sin N(0,1), Y=Z^2=h(Z)，變換並不是 1-1

h={z2z≥0z2z<0  ⟹  h−1={zz≥0−zz<0h=\begin{cases} z^2 & z\ge 0\\ z^2 & z<0 \end{cases}\implies h^{-1}=\begin{cases} \sqrt{z} & z\ge 0\\ -\sqrt{z} & z<0 \end{cases}   ⟹  FY(y)=P(Y≤y)=P(Z2≤y)=P(−y≤Z≤y)=Φ(y)−Φ(−y)\implies F_Y(y)=P(Y\le y)=P(Z^2\le y)=P(-\sqrt{y}\le Z\le \sqrt{y})=\Phi(\sqrt{y})-\Phi(-\sqrt{y}) fY(y)=ddyFY(y)=ddy[Φ(y)−Φ(−y)]=ϕ(y)ddyy−ϕ(−y)ddy(−y)=2ϕ(y)ddyy=2⋅12πe−y/2⋅12y−1/2=1π21/2y12−1e−y/2,y>0∝pdf of Gamma(12,2)\begin{align*} f_Y(y)&=\frac{d}{dy}F_Y(y)\\ &=\frac{d}{dy}[\Phi(\sqrt{y})-\Phi(-\sqrt{y})]\\ &=\phi(\sqrt{y})\frac{d}{dy}\sqrt{y}-\phi(-\sqrt{y})\frac{d}{dy}(-\sqrt{y})\\ &=2\phi(\sqrt{y})\frac{d}{dy}\sqrt{y}=2\cdot\frac{1}{\sqrt{2\pi}}e^{-y/2}\cdot\frac{1}{2}y^{-1/2}\\ &=\frac{1}{\sqrt{\pi}2^{1/2}}y^{\frac{1}{2}-1}e^{-y/2}, y>0\\ &\propto \text{pdf of Gamma}(\frac{1}{2},2) \end{align*}

i.e. Z2∼Gamma(12,2)=χ12Z^2\sim\text{Gamma}(\frac{1}{2},2)=\chi^2_1

Remark: 如果發現兩個 pdf 函數成比例（變數相關部分相同 ∀x\forall x），則兩個 pdf 相同。

因為 pdf 積分為 1，如果變數部分相同，則常數部分一定相同。

Theorem

Let XX have a conti. pdf fX(x)f_X(x) and X={ x:f(x)>0 }\mathscr{X}=\set{x:f(x)>0} be the support of XX, Y=h(X)Y=h(X).

Suppose ∃{ A0,A1,,⋅,Ak }\exist\set{A_0,A_1,,\cdot,A_k} :partition of X\mathscr{X} s.t. P(X∈A0)=0P(X\in A_0)=0 and suppose ∃hi\exist h_i on Ai,i=1,2,⋯ ,kA_i,i=1,2,\cdots,k with

  1. h(x)=hi(x),∀x∈Aih(x)=h_i(x),\forall x\in A_i

  2. hih_i is monotone on AiA_i

  3. hi(Ai)={ y:y=hi(x),x∈Ai },∀ih_i(A_i)=\set{y:y=h_i(x),x\in A_i},\forall i

    Y=h(X)={ y:y=h(x),x∈X }\mathscr{Y}=h(\mathscr{X})=\set{y:y=h(x),x\in\mathscr{X}}

  4. hi−1(y)h_i^{-1}(y) has a conti derivative

then pdf of YY is

fY(y)=∑i=1kfX(hi−1(y))∣ddyhi−1(y)∣,y∈Yf_Y(y)=\sum_{i=1}^k f_X(h_i^{-1}(y))\left|\frac{d}{dy}h_i^{-1}(y)\right|, y\in\mathscr{Y}

將 XX 的 pdf 分段，使得每段有其對應的 1-1 變換函數 hih_i。將每段的 pdf 分別變換，然後合併。

Idea: Y=FX(X)∈[0,1]Y=F_X(X)\in[0,1]

FY(y)=P(Y≤y)=P(FX(X)≤y)=P(X≤FX−1(y))=FX(FX−1(y))=yF_Y(y)=P(Y\le y)=P(F_X(X)\le y)=P(X\le F_X^{-1}(y))=F_X(F_X^{-1}(y))=y   ⟹  fY(y)=ddyFY(y)=1,y∈[0,1]\implies f_Y(y)=\frac{d}{dy}F_Y(y)=1, y\in[0,1]

是 U(0,1)U(0,1) 的 pdf

Theorem

U∼U(0,1)U\sim U(0,1) 並且 FF 是 cdf （不要求連續，但單調遞增），定義 F−1(y)=inf⁡{ x;F(x)≥y }F^{-1}(y)=\inf\set{x;F(x)\ge y}

  ⟹  X∼F\implies X\sim F

Lemma

上面定義的 F−1F^{-1} 會有以下性質：

  1. F−1F^{-1} 是單調遞增的（不一定嚴格遞增）
  2. F(F−1(Y))≥y,∀y∈(0,1)F(F^{-1}(Y))\ge y,\forall y\in(0,1)
  3. F−1(F(t))≤t,∀tF^{-1}(F(t))\le t,\forall t
  4. F−1(y)≤t  ⟺  y≤F(t),∀y∈(0,1),∀t∈RF^{-1}(y)\le t\iff y\le F(t),\forall y\in(0,1),\forall t\in\R

如果進一步， FF 是連續的，則 F(F−1(t))=t,∀tF(F^{-1}(t))=t,\forall t

Proof:

P(X≤X)=P(F−1(u)≤x)=P(u≤F(x))=F(x)P(X\le X)=P(F^{-1}(u)\le x)=P(u\le F(x))=F(x)

i.e. X 的 cdf 是 FF

Theorem

The prob integral transformation

XX 有連續的 cdf FXF_X，令 Y=FX(X)  ⟹  Y∼U(0,1)Y=F_X(X)\implies Y\sim U(0,1)

FY(y)≜P(Y≤y)=P(FX(X)≤y)=P(FX−1(FX(X))≤FX−1(y))=P(FX−1(FX(X))⏟≤X≤FX−1(y),X≤FX−1(y))+P(FX−1(FX(X))⏟≤X≤FX−1(y),X>FX−1(y))=P(X≤FX−1(y))+0=FX(FX−1(y))=y∵F is conti.\begin{align*} F_Y(y)&\triangleq P(Y\le y)\\ &=P(F_X(X)\le y)\\ &=P(F_X^{-1}(F_X(X))\le F_X^{-1}(y))\\ &=P(\underbrace{F_X^{-1}(F_X(X))}_{\le X}\le F_X^{-1}(y), X\le F_X^{-1}(y))+P(\underbrace{F_X^{-1}(F_X(X))}_{\le X}\le F_X^{-1}(y), X> F_X^{-1}(y))\\ &=P(X\le F_X^{-1}(y))+0\\ &=F_X(F_X^{-1}(y))=y \quad\because F \text{ is conti.} \end{align*}

Remark: U∼U(0,1)  ⟹  1−U∼U(0,1)U\sim U(0,1)\implies 1-U\sim U(0,1)

−2ln⁡U∼χ22  ⟹  −2ln⁡(1−U)∼χ22-2\ln U\sim \chi^2_2\implies -2\ln(1-U)\sim \chi^2_2

i.e. ∀F\forall F 是連續的 cdf

−2ln⁡F(X)∼χ22−2ln⁡(1−F(X))∼χ22-2\ln F(X)\sim \chi^2_2\qquad -2\ln(1-F(X))\sim \chi^2_2

多變量變換​

Theorem

Let XX be a k-dim r.v. with conti. pdf fX(x)f_X(x) and Y=h(X)=(Y1,⋯ ,Yk)t=(h1(X),⋯ ,hk(X))tY=h(X)=(Y_1,\cdots,Y_k)^t=\left(h_1(X),\cdots,h_k(X)\right)^t

Suppose h:X→1−1Yh:\mathscr{X}\xrightarrow{1-1}\mathscr{Y} with X={ x:fX(x)>0 }\mathscr{X}=\set{x:f_X(x)>0} and Y=h(X)\mathscr{Y}=h(\mathscr{X}) s.t. X=h−1(y)=(g1(y),⋯ ,gk(y))tX=h^{-1}(y)=\left(g_1(y),\cdots,g_k(y)\right)^t exists and ∂∂yigj(y)\frac{\partial}{\partial y_i}g_j(y) exist and conti. ∀i,j\forall i,j

Then the pdf of YY is

fY(y)=fY(y1,⋯ ,yk)=fX(h−1(y))∣J∣where J=det⁡[∂g1∂y1⋯∂g1∂yk⋮⋱⋮∂gk∂y1⋯∂gk∂yk]k×kf_Y(y)=f_Y(y1,\cdots,y_k)=f_X(h^{-1}(y))|J| \quad\text{where } J=\det\begin{bmatrix} \frac{\partial g_1}{\partial y_1} & \cdots & \frac{\partial g_1}{\partial y_k}\\ \vdots & \ddots & \vdots\\ \frac{\partial g_k}{\partial y_1} & \cdots & \frac{\partial g_k}{\partial y_k} \end{bmatrix}_{k\times k}

EX: X=(X1,X2,X3,X4)t∼fX(x1,x2,x3,x4)=24e−x1−x2−x3−x4X=(X_1,X_2,X_3,X_4)^t\sim f_X(x_1,x_2,x_3,x_4)=24e^{-x_1-x_2-x_3-x_4} with 0<x1<x2<x3<x4<∞0<x_1<x_2<x_3<x_4<\infty, let

Y1=X1Y2=X2−X1Y3=X3−X2Y4=X4−X3  ⟹  X1=Y1X2=Y1+Y2X3=Y1+Y2+Y3X4=Y1+Y2+Y3+Y4\begin{align*} Y_1&=X_1\\ Y_2&=X_2-X_1\\ Y_3&=X_3-X_2\\ Y_4&=X_4-X_3 \end{align*}\implies \begin{align*} X_1&=Y_1\\ X_2&=Y_1+Y_2\\ X_3&=Y_1+Y_2+Y_3\\ X_4&=Y_1+Y_2+Y_3+Y_4 \end{align*} J=det⁡[1000110011101111]=1≠0J=\det\begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{bmatrix}=1\neq 0 fY(y1,y2,y3,y4)=fX(y1,y1+y2,y1+y2+y3,y1+y2+y3+y4)∣J∣=4e−4y13e−3y22e−2y3e−y4\begin{align*} f_Y(y_1,y_2,y_3,y_4)&=f_X(y_1,y_1+y_2,y_1+y_2+y_3,y_1+y_2+y_3+y_4)|J|\\ &=4e^{-4y_1}3e^{-3y_2}2e^{-2y_3}e^{-y_4} \end{align*} fY1=4e−4y1,y1>0Y1∼Gamma(1,14)fY2=3e−3y2,y2>0Y2∼Gamma(1,13)fY3=2e−2y3,y3>0Y3∼Gamma(1,12)fY4=e−y4,y4>0Y4∼Gamma(1,1)\begin{align*} f_{Y_1}&=4e^{-4y_1}, &y_1>0\quad &Y_1\sim\text{Gamma}(1,\frac{1}{4})\\ f_{Y_2}&=3e^{-3y_2}, &y_2>0\quad &Y_2\sim\text{Gamma}(1,\frac{1}{3})\\ f_{Y_3}&=2e^{-2y_3}, &y_3>0\quad &Y_3\sim\text{Gamma}(1,\frac{1}{2})\\ f_{Y_4}&=e^{-y_4}, &y_4>0\quad &Y_4\sim\text{Gamma}(1,1) \end{align*}

and Y1,Y2,Y3,Y4Y_1,Y_2,Y_3,Y_4 are independent

EX: Given X∼Gamma(α1,β),Y∼Gamma(α2,β)X\sim\text{Gamma}(\alpha_1,\beta), Y\sim\text{Gamma}(\alpha_2,\beta) and X⊥Y  ⟺  fX,Y(x,y)=fX(x)fY(y)X\perp Y\iff f_{X,Y}(x,y)=f_X(x)f_Y(y)

Let T=X+Y,W=XX+Y  ⟹  X=TW,Y=T−TWT=X+Y, W=\frac{X}{X+Y}\implies X=TW, Y=T-TW

J=det⁡[WT1−W−T]=−TW−(1−W)T=−TJ=\det\begin{bmatrix} W & T\\ 1-W & -T \end{bmatrix}=-TW-(1-W)T=-T fT,W(t,w)=fX,Y(tw,t−tw)∣J∣=fX(tw)fY(t−tw)t =1Γ(α1)βα1(tw)α1−1e−tw/β⋅1Γ(α2)βα2(t−tw)α2−1e−(t−tw)/β⋅t=1Γ(α1+α2)βα1+α2tα1+α2−1e−t/β⋅Γ(α1+α2)Γ(α1)Γ(α2)wα1−1(1−w)α2−1\begin{align*} f_{T,W}(t,w)&=f_{X,Y}(tw,t-tw)|J|\\ &=f_X(tw)f_Y(t-tw)t\\\ &=\frac{1}{\Gamma(\alpha_1)\beta^{\alpha_1}}(tw)^{\alpha_1-1}e^{-tw/\beta}\cdot\frac{1}{\Gamma(\alpha_2)\beta^{\alpha_2}}(t-tw)^{\alpha_2-1}e^{-(t-tw)/\beta}\cdot t\\ &=\frac{1}{\Gamma(\alpha_1+\alpha_2)\beta^{\alpha_1+\alpha_2}}t^{\alpha_1+\alpha_2-1}e^{-t/\beta}\cdot\frac{\Gamma(\alpha_1+\alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)}w^{\alpha_1-1}(1-w)^{\alpha_2-1} \end{align*}

  ⟹  T⊥W\implies T\perp W and T∼Gamma(α1+α2,β),W∼Beta(α1,α2)T\sim\text{Gamma}(\alpha_1+\alpha_2,\beta), W\sim\text{Beta}(\alpha_1,\alpha_2)

i.e.

⊥<X∼Gamma(α1,β)Y∼Gamma(α2,β)  ⟹  ⊥<X+Y∼Gamma(α1+α2,β)XX+Y∼Beta(α1,α2)\perp\bigg<\begin{align*} X&\sim\text{Gamma}(\alpha_1,\beta)\\ Y&\sim\text{Gamma}(\alpha_2,\beta)\\ \end{align*}\implies\perp\bigg< \begin{align*} X+Y&\sim\text{Gamma}(\alpha_1+\alpha_2,\beta)\\ \frac{X}{X+Y}&\sim\text{Beta}(\alpha_1,\alpha_2) \end{align*}

Xi∼Gamma(αi,β),u=1,⋯ ,kX_i\sim \text{Gamma}(\alpha_i,\beta),u=1,\cdots,k indep.

  ⟹  ∑i=1kXi∼Gamma(∑i=1kαi,β),l<k∑i=1lXi∑i=1kXi∼Beta(∑i=1lαi,∑i=1kαi)\implies\sum_{i=1}^k X_i\sim\text{Gamma}(\sum_{i=1}^k\alpha_i,\beta),\qquad l<k \quad\frac{\sum_{i=1}^l X_i}{\sum_{i=1}^k X_i}\sim\text{Beta}(\sum_{i=1}^l\alpha_i,\sum_{i=1}^k\alpha_i)

Remark X1,⋯ ,Xn∼iidX_1,\cdots,X_n\overset{\text{iid}}{\sim} conti cdf F  ⟹  F(Xi)∼iidU(0,1)F\implies F(X_i)\overset{\text{iid}}{\sim} U(0,1)

  ⟹  −2ln⁡∏i=1nF(Xi)=−2∑i=1nln⁡F(Xi)∼χ2n2\implies -2\ln\prod_{i=1}^n F(X_i)=-2\sum_{i=1}^n\ln F(X_i)\sim\chi^2_{2n}

Remark:

  1. Z1,⋯ ,Zn∼iidN(0,1)  ⟹  ∑i=1nZi2∼χn2Z_1,\cdots,Z_n\overset{\text{iid}}{\sim}N(0,1)\implies\sum_{i=1}^nZ^2_i\sim\chi^2_n

      ⟹  Xi∼N(μi,σi2)\implies X_i\sim N(\mu_i,\sigma^2_i) indep.   ⟹  ∑i=1n(Xi−μiσi)2∼χn2\implies\sum_{i=1}^n\left(\frac{X_i-\mu_i}{\sigma_i}\right)^2\sim\chi^2_n

  2. Z∼N(0,1)⊥X∼χm2Z\sim N(0,1)\perp X\sim\chi^2_m

    Let T=ZX/m,W=Xm  ⟹  ∫fT,W(t,w)dw=TT=\frac{Z}{\sqrt{X/m}},W=\sqrt{\frac{X}{m}}\implies \int f_{T,W}(t,w)dw=T 's pdf

    i.e N(0,1)/χm2/m∼tm  ⟹  tm2=(N(0,1)2)χm2/m=χ12/1χm2/m=F1,mN(0,1)/\sqrt{\chi^2_m/m}\sim t_m\implies t_m^2=\frac{(N(0,1)^2)}{\chi^2_m/m}=\frac{\chi^2_1/1}{\chi^2_m/m}=F_{1,m}

EX: X,Y∼iidU(α,β)X,Y\overset{\text{iid}}{\sim} U(\alpha,\beta)，求 X+YX+Y 的分佈

  ⟹  T=X+Y\implies T=X+Y, let W=XW=X

X=WY=T−W  ⟹  J=det⁡[011−1]=−1\begin{align*} X&=W\\ Y&=T-W \end{align*}\implies J=\det\begin{bmatrix} 0 & 1\\ 1 & -1 \end{bmatrix}=-1   ⟹  fT,W(t,w)=fX(w)fY(t−w)=1(β−α)2\implies f_{T,W}(t,w)=f_X(w)f_Y(t-w)=\frac{1}{(\beta-\alpha)^2} α<w<βα<t−w<β  ⟺  α<w<βt−β<w<t−α  ⟺  max⁡{ α,t−β }<w<min⁡{ β,t−α }\begin{align*} \alpha<w<\beta\\ \alpha<t-w<\beta \end{align*}\iff\begin{align*} \alpha<w<\beta\\ t-\beta<w<t-\alpha \end{align*}\iff\begin{align*} \max\set{\alpha,t-\beta}<w<\min\set{\beta,t-\alpha} \end{align*}   ⟹  fT(t)≜∫−∞∞fT,W(t,w)dw=∫max⁡{ α,t−β }min⁡{ β,t−α }1(β−α)2dw=(min⁡{ β,t−α }−max⁡{ α,t−β })(β−α)2={1(β−α)2(t−2α)2α≤t<α+β1(β−α)2(2β−t)α+β≤t<2β\begin{align*} \implies f_T(t)&\triangleq\int_{-\infty}^\infty f_{T,W}(t,w)dw\\ &=\int_{\max\set{\alpha,t-\beta}}^{\min\set{\beta,t-\alpha}}\frac{1}{(\beta-\alpha)^2}dw\\ &=\frac{\left(\min\set{\beta,t-\alpha}-\max\set{\alpha,t-\beta}\right)}{(\beta-\alpha)^2}\\ &=\begin{cases} \frac{1}{(\beta-\alpha)^2}(t-2\alpha) & 2\alpha\le t< \alpha+\beta\\ \frac{1}{(\beta-\alpha)^2}(2\beta-t) & \alpha+\beta\le t< 2\beta \end{cases} \end{align*}

EX: X∼P(λ)⊥Y∼P(θ)X\sim P(\lambda)\perp Y\sim P(\theta)，求 X+YX+Y 的分佈

Let T=X+YT=X+Y

fT(t)=P(X+Y=t)=∑xP(X=x,Y=t−x)=∑x=0te−θθt−x(t−x)!⋅e−λλxx!=e−(λ+θ)⋅1t!∑x=0tt!(t−x)!x!λxθt−x⏟Binomial theorem=e−(λ+θ)(λ+θ)tt!\begin{align*} f_T(t)&=P(X+Y=t)\\ &=\sum_x P(X=x,Y=t-x)\\ &=\sum_{x=0}^t \frac{e^{-\theta}\theta^{t-x}}{(t-x)!}\cdot\frac{e^{-\lambda}\lambda^x}{x!}\\ &=e^{-(\lambda+\theta)}\cdot\frac{1}{t!}\underbrace{\sum_{x=0}^t\frac{t!}{(t-x)!x!}\lambda^x\theta^{t-x}}_{\text{Binomial theorem}}\\ &=\frac{e^{-(\lambda+\theta)}(\lambda+\theta)^t}{t!} \end{align*}

i.e. Xi∼P(λi)X_i\sim P(\lambda_i) indep.   ⟹  ∑Xi∼P(∑λi)\implies \sum X_i\sim P(\sum\lambda_i)