포아송 분포
단위시간 내의 성공확률
pmf
p X ( x ) = λ x e − λ x ! , x = 0 , 1 , 2 , … p_X(x)=\frac{\lambda^x e^{-\lambda}}{x!}, x=0,1,2, \ldots p X ( x ) = x ! λ x e − λ , x = 0 , 1 , 2 , … mgf
M ( t ) = e λ ( e t − 1 ) M(t) = e^{\lambda\left(e^t-1\right)} M ( t ) = e λ ( e t − 1 ) 평균과 분산
μ = σ 2 = λ \mu = \sigma^2 = \lambda μ = σ 2 = λ 지수 분포
다음 (성공)사건이 발생할때까지의 대기시간
정규분포
표준정규분포
pdf
f Z ( z ) = ϕ ( z ) = 1 2 π exp ( − 1 2 z 2 ) , − ∞ < z < ∞ f_Z(z)=\phi(z)=\frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{1}{2} z^2\right),-\infty<z<\infty f Z ( z ) = ϕ ( z ) = 2 π 1 exp ( − 2 1 z 2 ) , − ∞ < z < ∞ mgf
M ( t ) = exp ( 1 2 t 2 ) M(t) = \exp \left(\frac{1}{2} t^2\right) M ( t ) = exp ( 2 1 t 2 ) M ′ ( t ) = t exp ( 1 2 t 2 ) M ′ ′ ( t ) = exp ( 1 2 t 2 ) + t 2 exp ( 1 2 t 2 ) \begin{aligned} M^{\prime}(t) & =t \exp \left(\frac{1}{2} t^2\right) \\ M^{\prime \prime}(t) & =\exp \left(\frac{1}{2} t^2\right)+t^2 \exp \left(\frac{1}{2} t^2\right) \end{aligned} M ′ ( t ) M ′ ′ ( t ) = t exp ( 2 1 t 2 ) = exp ( 2 1 t 2 ) + t 2 exp ( 2 1 t 2 ) 표준정규분포를 따르는 Z의 평균과 분산은 각각 0, 1.
cdf
Φ ( z ) = ∫ − ∞ z 1 2 π exp ( − 1 2 w 2 ) d w \Phi(z)=\int_{-\infty}^z \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{1}{2} w^2\right) d w Φ ( z ) = ∫ − ∞ z 2 π 1 exp ( − 2 1 w 2 ) d w 특징
Φ ( − z ) = 1 − Φ ( z ) \Phi(-z)=1-\Phi(z) Φ ( − z ) = 1 − Φ ( z ) 여기에서 선형변환 (X = μ + σ Z X=\mu+\sigma Z X = μ + σ Z
정규분포
pdf
1 2 π σ 2 exp { − 1 2 σ 2 ( x − μ ) 2 } , − ∞ < x < ∞ \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left\{-\frac{1}{2 \sigma^2}(x-\mu)^2\right\}, \quad-\infty<x<\infty 2 π σ 2 1 exp { − 2 σ 2 1 ( x − μ ) 2 } , − ∞ < x < ∞ mgf
M ( t ) = exp ( μ t + 1 2 σ 2 t 2 ) M(t) = \exp \left(\mu t+\frac{1}{2} \sigma^2 t^2\right) M ( t ) = exp ( μ t + 2 1 σ 2 t 2 ) cdf
F X ( x ) = P ( X ≤ x ) = P ( Z ≤ x − μ σ ) = Φ ( x − μ σ ) = ∫ − ∞ x f X ( x ) d x . F_X(x) = P(X \leq x)=P\left(Z \leq \frac{x-\mu}{\sigma}\right) ={\Phi}\left(\frac{x-\mu}{\sigma}\right)=\int_{-\infty}^x f_X(x) d x . F X ( x ) = P ( X ≤ x ) = P ( Z ≤ σ x − μ ) = Φ ( σ x − μ ) = ∫ − ∞ x f X ( x ) d x . "독립적인" 정규분포의 합
Y = ∑ i = 1 n a i X i Y=\sum_{i=1}^n a_i X_i Y = i = 1 ∑ n a i X i 평균과 분산
Y ∼ N ( ∑ i = 1 n a i μ i , ∑ i = 1 n a i 2 σ i 2 ) Y \sim N\left(\sum_{i=1}^n a_i \mu_i, \sum_{i=1}^n a_i^2 \sigma_i^2\right) Y ∼ N ( i = 1 ∑ n a i μ i , i = 1 ∑ n a i 2 σ i 2 ) mgf
M ( t ) = exp { ( ∑ i = 1 n a i μ i ) t + 1 2 ( ∑ i = 1 n a i 2 σ i 2 ) t 2 } M(t) = \exp \left\{\left(\sum_{i=1}^n a_i \mu_i\right) t+\frac{1}{2}\left(\sum_{i=1}^n a_i^2 \sigma_i^2\right) t^2\right\} M ( t ) = exp { ( i = 1 ∑ n a i μ i ) t + 2 1 ( i = 1 ∑ n a i 2 σ i 2 ) t 2 } 다변량 정규분포
Z ∼ N n ( 0 , I n ) \mathbf{Z} \sim N_n\left(\mathbf{0}, \mathbf{I}_n\right) Z ∼ N n ( 0 , I n ) Z = ( Z 1 , Z 2 , ⋯ , Z n ) ′ \mathbf{Z}=\left(Z_1, Z_2, \cdots, Z_n\right)^{\prime} Z = ( Z 1 , Z 2 , ⋯ , Z n ) ′ 각 Zi는 표준정규분포를 따르는 iid한 확률변수이고 Z는 수직 벡터이다.
pdf
f Z ( z ) = ( 1 2 π ) n / 2 exp { − 1 2 z ′ z } f_{\mathbf{Z}}(\mathbf{z}) = \left(\frac{1}{2 \pi}\right)^{n / 2} \exp \left\{-\frac{1}{2} \mathbf{z}^{\prime} \mathbf{z}\right\} f Z ( z ) = ( 2 π 1 ) n / 2 exp { − 2 1 z ′ z } 평균과 분산
E [ Z ] = 0 , Cov [ Z ] = I n E[\mathbf{Z}]=\mathbf{0}, \quad \operatorname{Cov}[\mathbf{Z}]=\mathbf{I}_n E [ Z ] = 0 , C o v [ Z ] = I n 분산이 공분산 행렬이다.
mgf
M Z ( t ) = exp { 1 2 t ′ t } M_{\mathbf{Z}}(\mathbf{t}) =\exp \left\{\frac{1}{2} \mathbf{t}^{\prime} \mathbf{t}\right\} M Z ( t ) = exp { 2 1 t ′ t } X ∼ N n ( μ , Σ ) \mathbf{X} \sim N_n(\boldsymbol{\mu}, \boldsymbol{\Sigma}) X ∼ N n ( μ , Σ ) 더 일반적인 경우를 정의
시그마는 n*n 행렬 - 공분산 행렬임. n은 정규분포를 따르는 변수의 개수, 대각성분에는 각 변수의 분산(σ i 2 \sigma^2_i σ i 2 I n I_n I n
Σ \Sigma Σ Σ \Sigma Σ
Σ = Γ ′ Λ Γ \boldsymbol{\Sigma}=\boldsymbol{\Gamma}^{\prime} \boldsymbol{\Lambda} \Gamma Σ = Γ ′ Λ Γ 확률벡터를 X = Σ 1 / 2 Z + μ \mathbf{X}=\boldsymbol{\Sigma}^{1 / 2} \mathbf{Z}+\boldsymbol{\mu} X = Σ 1 / 2 Z + μ
평균과 분산
E [ X ] = μ , Cov [ X ] = Σ 1 / 2 Σ 1 / 2 = Σ E[\mathbf{X}]=\boldsymbol{\mu}, \quad \operatorname{Cov}[\mathbf{X}]=\boldsymbol{\Sigma}^{1 / 2} \boldsymbol{\Sigma}^{1 / 2}=\boldsymbol{\Sigma} E [ X ] = μ , C o v [ X ] = Σ 1 / 2 Σ 1 / 2 = Σ mgf
M X ( t ) = exp { t ′ μ } exp { ( 1 / 2 ) t ′ Σ t } M_{\mathbf{X}}(\mathbf{t}) = \exp \left\{\mathbf{t}^{\prime} \boldsymbol{\mu}\right\} \exp \left\{(1 / 2) \mathbf{t}^{\prime} \boldsymbol{\Sigma} \mathbf{t}\right\} M X ( t ) = exp { t ′ μ } exp { ( 1 / 2 ) t ′ Σ t } mgf가 위와 같이 계산되면 n차원 확률벡터 X가 다변량 정규분포를 따른다고 한다.
다변량 정규확률벡터의 선형변환
X ∼ N n ( μ , Σ ) X \sim N_n(\boldsymbol{\mu}, \boldsymbol{\Sigma}) X ∼ N n ( μ , Σ ) Y = A X + b \mathbf{Y}=\mathbf{A} \mathbf{X}+\mathbf{b} Y = A X + b N m ( A μ + b , A Σ A ′ ) N_m\left(\mathbf{A} \boldsymbol{\mu}+\mathbf{b}, \mathbf{A} \boldsymbol{\Sigma} \mathbf{A}^{\prime}\right) N m ( A μ + b , A Σ A ′ )
평균과 분산
Y ∼ N m ( A μ + b , A Σ A ′ ) Y \sim N_m\left(\mathbf{A} \boldsymbol{\mu}+\mathbf{b}, \mathbf{A} \boldsymbol{\Sigma} \mathbf{A}^{\prime}\right) Y ∼ N m ( A μ + b , A Σ A ′ ) mgf
M Y ( t ) = exp { t ′ ( A μ + b ) + 1 2 t ′ ( A Σ A ′ ) t } M_{\mathbf{Y}}(\mathbf{t}) = \exp \left\{\mathbf{t}^{\prime}(\mathbf{A} \boldsymbol{\mu}+\mathbf{b})+\frac{1}{2} \mathbf{t}^{\prime}\left(\mathbf{A} \boldsymbol{\Sigma} \mathbf{A}^{\prime}\right) \mathbf{t}\right\} M Y ( t ) = exp { t ′ ( A μ + b ) + 2 1 t ′ ( A Σ A ′ ) t } 다변량 확률변수에서의 주변분포
X가 N n ( μ , Σ ) N_n(\boldsymbol{\mu}, \boldsymbol{\Sigma}) N n ( μ , Σ )
X = [ X 1 X 2 ] ∼ N n ( [ μ 1 μ 2 ] , [ Σ 11 Σ 12 Σ 21 Σ 22 ] ) . \mathbf{X}=\left[\begin{array}{l} \mathbf{X}_1 \\ \mathbf{X}_2 \end{array}\right] \sim N_n\left(\left[\begin{array}{l} \boldsymbol{\mu}_1 \\ \boldsymbol{\mu}_2 \end{array}\right],\left[\begin{array}{ll} \boldsymbol{\Sigma}_{11} & \boldsymbol{\Sigma}_{12} \\ \boldsymbol{\Sigma}_{21} & \boldsymbol{\Sigma}_{22} \end{array}\right]\right) . X = [ X 1 X 2 ] ∼ N n ( [ μ 1 μ 2 ] , [ Σ 1 1 Σ 2 1 Σ 1 2 Σ 2 2 ] ) . 이때 X1은 N m ( μ 1 , Σ 11 ) N_m\left(\boldsymbol{\mu}_1, \boldsymbol{\Sigma}_{11}\right) N m ( μ 1 , Σ 1 1 )
이때, X1과 X2는 Σ 12 = O \boldsymbol{\Sigma}_{12}=\mathbf{O} Σ 1 2 = O 이다.
여기서 이변량정규분포의 조건부분포를 유도할 수 있다.
조건부분포
평균과 분산
X 1 ∣ X 2 ∼ N m ( μ 1 + Σ 12 Σ 22 − 1 ( X 2 − μ 2 ) , Σ 11 − Σ 12 Σ 22 − 1 Σ 21 ) \mathbf{X}_1 \mid \mathbf{X}_2 \sim N_m\left(\boldsymbol{\mu}_1+\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1}\left(\mathbf{X}_2-\boldsymbol{\mu}_2\right), \boldsymbol{\Sigma}_{11}-\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21}\right) X 1 ∣ X 2 ∼ N m ( μ 1 + Σ 1 2 Σ 2 2 − 1 ( X 2 − μ 2 ) , Σ 1 1 − Σ 1 2 Σ 2 2 − 1 Σ 2 1 ) mgf
M X 1 ∣ x 2 ( t 1 ) = exp ( t 1 ′ ( μ 1 + Σ 12 Σ 22 − 1 ( X 2 − μ 2 ) ) + 1 2 t 1 ′ ( Σ 11 − Σ 12 Σ 22 − 1 Σ 21 ) t 1 ) , M_{\mathbf{X}_1 \mid \mathbf{x}_2}\left(\mathbf{t}_1\right)=\exp \left(\mathbf{t}_1^{\prime}\left(\boldsymbol{\mu}_1+\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1}\left(\mathbf{X}_2-\boldsymbol{\mu}_2\right)\right)+\frac{1}{2} \mathbf{t}_1^{\prime}\left(\boldsymbol{\Sigma}_{11}-\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21}\right) \mathbf{t}_1\right), M X 1 ∣ x 2 ( t 1 ) = exp ( t 1 ′ ( μ 1 + Σ 1 2 Σ 2 2 − 1 ( X 2 − μ 2 ) ) + 2 1 t 1 ′ ( Σ 1 1 − Σ 1 2 Σ 2 2 − 1 Σ 2 1 ) t 1 ) , μ 1 + Σ 12 Σ 22 − 1 ( X 2 − μ 2 ) \boldsymbol{\mu}_1+\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1}\left(\mathbf{X}_2-\boldsymbol{\mu}_2\right) μ 1 + Σ 1 2 Σ 2 2 − 1 ( X 2 − μ 2 ) Σ 11 − Σ 12 Σ 22 − 1 Σ 21 \boldsymbol{\Sigma}_{11}-\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21} Σ 1 1 − Σ 1 2 Σ 2 2 − 1 Σ 2 1
감마 분포
지수분포의 일반화된 형태
감마 함수
감마함수란, α > 0 \alpha>0 α > 0 α \alpha α
Γ ( α ) = ∫ 0 ∞ y α − 1 e − y d y \Gamma(\alpha)=\int_0^{\infty} y^{\alpha-1} e^{-y} d y Γ ( α ) = ∫ 0 ∞ y α − 1 e − y d y
If α = 1 \alpha=1 α = 1 Γ ( 1 ) = ∫ 0 ∞ e − y d y = 1 \Gamma(1)=\int_0^{\infty} e^{-y} d y=1 Γ ( 1 ) = ∫ 0 ∞ e − y d y = 1
If α > 1 \alpha>1 α > 1 Γ ( α ) = ∫ 0 ∞ ( α − 1 ) y α − 2 e − y d y = ( α − 1 ) Γ ( α − 1 ) = ( α − 1 ) ! \Gamma(\alpha)=\int_0^{\infty}(\alpha-1) y^{\alpha-2} e^{-y} d y=(\alpha-1) \Gamma(\alpha-1) = (\alpha-1)! Γ ( α ) = ∫ 0 ∞ ( α − 1 ) y α − 2 e − y d y = ( α − 1 ) Γ ( α − 1 ) = ( α − 1 ) !
If α = 1 / 2 \alpha=1/2 α = 1 / 2 Γ ( 1 / 2 ) = π \Gamma(1 / 2)=\sqrt{\pi} Γ ( 1 / 2 ) = π
감마 분포
연속확률분포 중 하나로, 두 개의 파라미터, α \alpha α β \beta β
y = x / β y=x/\beta y = x / β
Γ ( α ) = ∫ 0 ∞ ( x β ) α − 1 e − x / β ( 1 β ) d x = ∫ 0 ∞ 1 β α x α − 1 e − x / β d x \begin{aligned} \Gamma(\alpha) & =\int_0^{\infty}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-x / \beta}\left(\frac{1}{\beta}\right) d x \\ & =\int_0^{\infty} \frac{1}{\beta^\alpha} x^{\alpha-1} e^{-x / \beta} d x \end{aligned} Γ ( α ) = ∫ 0 ∞ ( β x ) α − 1 e − x / β ( β 1 ) d x = ∫ 0 ∞ β α 1 x α − 1 e − x / β d x 이때 양변을 Γ ( α ) \Gamma(\alpha) Γ ( α )
1 = ∫ 0 ∞ 1 Γ ( α ) β α x α − 1 e − x / β d x = ∫ 0 ∞ f X ( x ) d x \begin{aligned} 1 & =\int_0^{\infty} \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha-1} e^{-x / \beta} d x \\ & =\int_0^{\infty} f_X(x) d x \end{aligned} 1 = ∫ 0 ∞ Γ ( α ) β α 1 x α − 1 e − x / β d x = ∫ 0 ∞ f X ( x ) d x 이때 X를 감마분포(α , β \alpha, \beta α , β f X ( x ) d x = 1 Γ ( α ) β α x α − 1 e − x / β f_X(x) d x = \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha-1} e^{-x / \beta} f X ( x ) d x = Γ ( α ) β α 1 x α − 1 e − x / β α \alpha α β \beta β
mgf
M ( t ) = ( 1 − β t ) − α M(t) = (1-\beta t)^{-\alpha} M ( t ) = ( 1 − β t ) − α M ′ ( t ) = ( − α ) ( 1 − β t ) − α − 1 ( − β ) ‾ M ′ ′ ( t ) = ( − α ) ( − α − 1 ) ( 1 − β t ) − α − 2 ( − β ) 2 \begin{gathered} M^{\prime}(t)=\underline{(-\alpha)(1-\beta t)^{-\alpha-1}(-\beta)} \\ M^{\prime \prime}(t)=(-\alpha)(-\alpha-1)(1-\beta t)^{-\alpha-2}(-\beta)^2 \end{gathered} M ′ ( t ) = ( − α ) ( 1 − β t ) − α − 1 ( − β ) M ′ ′ ( t ) = ( − α ) ( − α − 1 ) ( 1 − β t ) − α − 2 ( − β ) 2 평균과 분산
μ = α β , σ 2 = α β 2 \mu = \alpha\beta, \sigma^2=\alpha\beta^2 μ = α β , σ 2 = α β 2 특징
감마분포를 따르는 확률변수의 합은 감마분포를 따른다.
X ∼ Γ ( α , β ) ⇒ c X ∼ Γ ( α , c β ) Γ ( 1 , β ) = d Exp ( 1 β ) Γ ( r 2 , 2 ) = d χ ( r ) 2 \begin{aligned} & X \sim \Gamma(\alpha, \beta) \Rightarrow c X \sim \Gamma(\alpha, c \beta) \\ & \Gamma(1, \beta) \stackrel{d}{=} \operatorname{Exp}\left(\frac{1}{\beta}\right) \\ & \Gamma\left(\frac{r}{2}, 2\right) \stackrel{d}{=} \chi_{(r)}^2 \end{aligned} X ∼ Γ ( α , β ) ⇒ c X ∼ Γ ( α , c β ) Γ ( 1 , β ) = d E x p ( β 1 ) Γ ( 2 r , 2 ) = d χ ( r ) 2 k번째 사건이 일어나기까지의 대기시간은 Γ ( k , 1 / λ ) \Gamma(k, 1 / \lambda) Γ ( k , 1 / λ )
베타 분포
t분포
표준정규분포를 따르는 확률변수 Z 와 카이제곱분포를 따르는 확률변수 V (독립)
T = Z V / r T=\frac{Z}{\sqrt{V / r}} T = V / r Z pdf
자유도: r
f T ( t ) = ∫ 0 ∞ 1 2 π Γ ( r / 2 ) 2 r / 2 u r / 2 − 1 e − u 2 ( 1 + t 2 r ) u r d u = 1 2 π r Γ ( r / 2 ) 2 r / 2 ∫ 0 ∞ u ( r + 1 ) / 2 − 1 e − u 2 ( 1 + t 2 r ) d u = 1 2 π r Γ ( r / 2 ) 2 r / 2 Γ ( r + 1 2 ) { 2 ( 1 + t 2 r ) − 1 } ( r + 1 ) / 2 = Γ ( ( r + 1 ) / 2 ) π r Γ ( r / 2 ) ( 1 + t 2 r ) − ( r + 1 ) / 2 , − ∞ < t < ∞ , \begin{aligned} f_T(t) & =\int_0^{\infty} \frac{1}{\sqrt{2 \pi} \Gamma(r / 2) 2^{r / 2}} u^{r / 2-1} e^{-\frac{u}{2}\left(1+\frac{t^2}{r}\right)} \sqrt{\frac{u}{r}} d u \\ & =\frac{1}{\sqrt{2 \pi r} \Gamma(r / 2) 2^{r / 2}} \int_0^{\infty} u^{(r+1) / 2-1} e^{-\frac{u}{2}\left(1+\frac{t^2}{r}\right)} d u \\ & =\frac{1}{\sqrt{2 \pi r} \Gamma(r / 2) 2^{r / 2}} \Gamma\left(\frac{r+1}{2}\right)\left\{2\left(1+\frac{t^2}{r}\right)^{-1}\right\}^{(r+1) / 2} \\ & =\frac{\Gamma((r+1) / 2)}{\sqrt{\pi r} \Gamma(r / 2)}\left(1+\frac{t^2}{r}\right)^{-(r+1) / 2}, \quad-\infty<t<\infty, \end{aligned} f T ( t ) = ∫ 0 ∞ 2 π Γ ( r / 2 ) 2 r / 2 1 u r / 2 − 1 e − 2 u ( 1 + r t 2 ) r u d u = 2 π r Γ ( r / 2 ) 2 r / 2 1 ∫ 0 ∞ u ( r + 1 ) / 2 − 1 e − 2 u ( 1 + r t 2 ) d u = 2 π r Γ ( r / 2 ) 2 r / 2 1 Γ ( 2 r + 1 ) { 2 ( 1 + r t 2 ) − 1 } ( r + 1 ) / 2 = π r Γ ( r / 2 ) Γ ( ( r + 1 ) / 2 ) ( 1 + r t 2 ) − ( r + 1 ) / 2 , − ∞ < t < ∞ , mgf
E ( T k ) = r k / 2 E ( Z k ) Γ ( ( r − k ) / 2 ) 2 − k / 2 Γ ( r / 2 ) E\left(T^k\right) = r^{k / 2} E\left(Z^k\right) \frac{\Gamma((r-k) / 2) 2^{-k / 2}}{\Gamma(r / 2)} E ( T k ) = r k / 2 E ( Z k ) Γ ( r / 2 ) Γ ( ( r − k ) / 2 ) 2 − k / 2 평균과 분산
E ( T ) = 0 and Var ( T ) = E ( T 2 ) = r r − 2 E(T)=0 \text { and } \operatorname{Var}(T)=E\left(T^2\right)=\frac{r}{r-2} E ( T ) = 0 and V a r ( T ) = E ( T 2 ) = r − 2 r 