1번
음이항분포 \(NB(r, p)\)의 확률밀도함수 \(f(x)\)가 \(\sum_{x=r}^\infty f(x) = 1\)을 만족함을 음이항정리를 이용하여 증명하시오
음이항정리
\(\dfrac{1}{1-q} = \sum_{x=0}^{\infty} \begin{pmatrix} x+1-1 \\ 1-1 \end{pmatrix} q^x (|x|<1)\)
\(\dfrac{1}{(1-q)^2} = \dfrac{d}{dq} (\dfrac{1}{1-q}) = \sum_{x=1}^\infty xq^{x-1} = \sum_{x=0}^\infty (x+1)q^x = \sum_{x=0}^{\infty} \begin{pmatrix} x+2-1 \\ 2-1 \end{pmatrix} q^x (|x|<1)\)
수학적 귀납법을 통해
\((1-q)^{-r} = \sum_{x=0}^\infty \begin{pmatrix} x+r-1 \\ r-1 \end{pmatrix} q^x\) 임을 구할 수 있다. -(A)
\(\sum_{x=r}^\infty f(x)\)
= \(\sum_{x=r}^\infty \begin{pmatrix} x-r \\ r-1 \end{pmatrix} p^r (1-p)^{x-r} (x=r, r+1, \dots)\)
= \(p^r \sum_{x=r}^\infty \begin{pmatrix} x-r \\ r-1 \end{pmatrix} (1-p)^{x-r}\) 이고
\(x-r=t\)로 치환하면
= \(p^r \sum_{t=0}^\infty \begin{pmatrix} t+r-1 \\ r-1 \end{pmatrix} (1-p)^t\) 이다.
(A)식을 이용해
= \(p^r(1-q)^{-r} = p^r p^{-r} = 1\) 을 증명할 수 있다.
이항정리 \((x+y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k}\)
일반화된 이항정리 \((x+y)^a = \sum_{k=0}^\infty \begin{pmatrix} a \\ k \end{pmatrix} a^k y^{n-k}, a\in \mathbb{R}, -1<x<1\)
즉, \((1-w)^{-r}= \sum_{k=0}^\infty \begin{pmatrix} -r \\ k \end{pmatrix} (-w)^k\)
=\(\sum_{k=0}^\infty \dfrac{-r \cdot (-r-1) \dots \cdot (-r-k+1) }{k!}(-1)^k w^k\)
=\(\sum_{k=0}^\infty \dfrac{r \cdot (r+1) \dots \cdot (r+k-1) }{k!}(-1)^{2k} w^k\)
\(\sum_{k=0}^\infty \dfrac{-r \cdot (-r-1) \dots \cdot (-r-k+1) }{k!}(-1)^k w^k\)
\(\sum_{k=0}^\infty\begin{pmatrix} r+k-1 \\ k \end{pmatrix} w^k\)
\(x=r+k\)로 치환하면
=\(\sum_{x=r}^\infty \begin{pmatrix} x-1 \\ r-1 \end{pmatrix} w^{x-r}\)
\(s.t \sum \begin{pmatrix} x-1 \\ r-1 \end{pmatrix} p^r (1-p)^{x-r} = p^r \sum \begin{pmatrix} x-1 \\ r-1 \end{pmatrix} (1-p)^{x-r} = p^r {1-(1-p)}^{-r} = 1\)
2번
음이항분포 \(NB(r, p)\)의 적률생성함수를 구하시오
\(Y_i\)가 기하분포를 따르면 \(\sum_{i=1}^r Y_i = X~NB(r,p)\)이다.
\(M_X(t)=E(e^{tX})=E(e^{tY_1+\dots+tY_r})\)
\(Y_i\)는 각각 독립이므로,
=\(E(e^{tY_1}) \dots E(e^{tY_r})\)
=\((\dfrac{pe^t}{1-qe^t})^r\)
3번
베타분포 \(BETA(a, b)\)의 \(k\)차적률이 \(E(X^k) = \dfrac{Γ(a + k)Γ(a + b)}{Γ(a)Γ(a + b + k)}\)임을 보이시오.
\(E(X^k)=\int_{-\infty}^{\infty}x^k \dfrac{1}{B(a,b)}x^{a-1}(1-x)^{b-1}I(0<x<1)dx\)
= \(\dfrac{1}{B(a,b)}\int_0^1 x^{a+k-1}(1-x)^{b-1}dx\)
= \(\dfrac{B(a+k,b)}{B(a,b)}= \dfrac{\Gamma(a+k)\Gamma(b)}{\Gamma(a+k+b)} \dfrac{\Gamma(a+B)}{\Gamma(a)\Gamma(b)}=\dfrac{\Gamma(a+k) \Gamma(a+b)}{\Gamma(a) \Gamma(a+b+k)}\)
과제 3의 9번 문제도 이 공식을 이용해 구할 수 있다.
4번
감마분포 \(GAM(α, β)\)의 평균과 분산의 기댓값 정의를 이용하여 구하시오.
\(E(X)=\dfrac{1}{\Gamma(\alpha) \beta^\alpha}\int_0^\infty x x^{\alpha -1} e^{-x/\beta}\)
\(=\dfrac{\Gamma(\alpha+1) \beta^{\alpha+1}}{\Gamma(\alpha) \beta^\alpha} \int_0^\infty \dfrac{1}{\Gamma(\alpha+1)\beta^{\alpha+1}} x^{(a+1)-1}e^{-x/\beta}\)
\(\because \int_0^\infty \dfrac{1}{\Gamma(\alpha+1)\beta^{\alpha+1}} x^{(a+1)-1}e^{-x/\beta} =1\)
\(=\dfrac{\Gamma(\alpha+1) \beta^{\alpha+1}}{\Gamma(\alpha) \beta^\alpha} = \dfrac{\alpha \Gamma(\alpha) \beta^{\alpha +1}}{\Gamma(\alpha) \beta^\alpha} = \alpha\beta\)
\(E(X^2)=\dfrac{1}{\Gamma(\alpha) \beta^\alpha}\int_0^\infty x^2 x^{\alpha -1} e^{-x/\beta}\)
\(=\dfrac{\Gamma(\alpha+2) \beta^{\alpha+2}}{\Gamma(\alpha) \beta^\alpha} \int_0^\infty \dfrac{1}{\Gamma(\alpha+2)\beta^{\alpha+2}} x^{(a+2)-1}e^{-x/\beta}\)
\(=\dfrac{\Gamma(\alpha+2) \beta^{\alpha+2}}{\Gamma(\alpha) \beta^\alpha} = \dfrac{\Gamma(\alpha+1)\alpha \beta^{\alpha +2} \Gamma(\alpha)}{\Gamma(\alpha) \beta^\alpha} = (\alpha+1)\alpha\beta^2\)
\(Var(X)=E(X^2)-[E(X)]^2= (\alpha+1)\alpha\beta^2 - (\alpha\beta)^2 = \alpha\beta^2\)
5번
정규분포 \(N(µ, σ^2)\)의 적률생성함수를 구하시오.
\(M_X(t)=E(e^{tX})=\int_{-\infty}^\infty e^{tx} \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-\dfrac{(x-\mu)^2}{2\sigma^2}}dx\)
(\(\dfrac{x-\mu}{\sigma}=u\) 치환하면, \(dx=\sigma du\), \(x=\mu+\sigma u\))
= \(\int_{-\infty}^\infty e^{t(\mu+\sigma u)} \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-u^2/2} \sigma du\)
= \(\dfrac{e^{t\mu}}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-\dfrac{1}{2}(u-t \mu)^2 + -\dfrac{1}{2}t^2\sigma^2}du\)
=\(\dfrac{e^{t\mu+ \dfrac{1}{2}t^2 \sigma^2}}{\sqrt{2\pi}} \int_{-\infty}^\infty e ^{-\dfrac{(u-t\mu)^2}{2}}du\)
(\(u-t\sigma=v\)로 치환하면, \(du=dv, u=v+t\sigma\))
=\(\dfrac{e^{t\mu+ \dfrac{1}{2}t^2 \sigma^2}}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\dfrac{v}{2}}dv\)
\(\because \int_{-\infty}^\infty e^{-\dfrac{v}{2}}dv = \sqrt{2\pi}\)
=\(e^{t\mu+ \dfrac{1}{2}t^2 \sigma^2}\)
6번
이변량정규분포 \(BVN(µ_X, µ_Y , σ_X, σ_Y , ρ)\)를 따르는 확률벡터 \((X,Y)\)의 확률밀도함수를 다변량정규분포의 확률밀도함수의 2차원형임을 이용하여 구하시오.
\(\mu =\begin{pmatrix} \mu _{x} \\ \mu _{y} \end{pmatrix}\)
\(\sum =\begin{pmatrix} \sigma _{x}^{2} & \sigma _{xy} \\ \sigma _{xy} & \sigma _{y}^{2} \end{pmatrix}= \begin{pmatrix} \sigma _{x}^{2} & \rho \sigma _{x}\sigma _{y} \\ \rho \sigma _{x}\sigma _{y} & \sigma _{y}^{2} \end{pmatrix}\)
\(f_{X,Y}(x,y)=\dfrac{1}{|2\pi \sum|^{1/2}} exp\left[ -\dfrac{1}{2}\left( \begin{pmatrix} x \\ y \end{pmatrix}-\begin{pmatrix} u_{x} \\ u_{y} \end{pmatrix}\right) ^{t}\Sigma ^{-1}\left( \begin{pmatrix} x \\ y \end{pmatrix}-\begin{pmatrix} \mu _{x} \\ \mu _{y} \end{pmatrix}\right) \right]\)
먼저, \(\dfrac{1}{|2\pi\sum|^{1/2}}=\dfrac{1}{2\pi|\sum|^{1/2}}=\dfrac{1}{2\pi(\sigma_x^2 \sigma_y^2 - \rho^2 \sigma_x^2 \sigma_y^2)^{1/2}}=\dfrac{1}{2\pi \sigma_x \sigma_y \sqrt{1-\rho^2}}\)
\(exp\left[ -\dfrac{1}{2}\left( \begin{pmatrix} x \\ y \end{pmatrix}-\begin{pmatrix} \mu_{x} \\ \mu_{y} \end{pmatrix}\right) ^{t}\Sigma ^{-1}\left( \begin{pmatrix} x \\ y \end{pmatrix}-\begin{pmatrix} \mu _{x} \\ \mu _{y} \end{pmatrix}\right) \right]\)
= \(exp \left[-\dfrac{1}{2(\sigma_x^2 \sigma_y^2 - \rho^2 \sigma_x^2 \sigma_y^2)} \begin{pmatrix} x-\mu_x & y-\mu_y \end{pmatrix} \begin{pmatrix} \sigma _{y}^{2} & - \rho \sigma _{x}\sigma _{y} \\ - \rho \sigma _{x}\sigma _{y} & \sigma _{x}^{2} \end{pmatrix} \begin{pmatrix} x-\mu_x \\ y-\mu_y \end{pmatrix} \right]\)
= \(exp \left[ -\dfrac{1}{2(\sigma_x^2 \sigma_y^2 - \rho^2 \sigma_x^2 \sigma_y^2)} ( (x-\mu_x)^2 \sigma_y^2 + (y-\mu_y)^2 \sigma_x^2 - 2\rho \sigma_x \sigma_y (x-\mu_x)(y-\mu_y)) \right]\)
=\(exp \left[-\dfrac{\sigma_x^2 \sigma_y^2}{2\sigma_x^2 \sigma_y^2(1- \rho^2)} \left( \left(\dfrac{x-\mu_x}{\sigma_x}\right)^2 + \left(\dfrac{y-\mu_y}{\sigma_y}\right)^2 - 2\rho \left(\dfrac{x-\mu_x}{\sigma_x}\right) \left(\dfrac{y-\mu_y}{\sigma_y}\right) \right) \right]\)
=\(exp \left[-\dfrac{1}{2(1- \rho^2)} \left( \left(\dfrac{x-\mu_x}{\sigma_x}\right)^2 + \left(\dfrac{y-\mu_y}{\sigma_y}\right)^2 - 2\rho \left(\dfrac{x-\mu_x}{\sigma_x}\right) \left(\dfrac{y-\mu_y}{\sigma_y}\right) \right) \right]\)
즉,
\(f_{X,Y}(x,y)=\dfrac{1}{2\pi \sigma_x \sigma_y \sqrt{1-\rho^2}}exp \left[-\dfrac{1}{2(1- \rho^2)} \left( \left(\dfrac{x-\mu_x}{\sigma_x}\right)^2 + \left(\dfrac{y-\mu_y}{\sigma_y}\right)^2 - 2\rho \left(\dfrac{x-\mu_x}{\sigma_x}\right) \left(\dfrac{y-\mu_y}{\sigma_y}\right) \right) \right]\)
7번
이변량정규분포 \(BVN(µ_X, µ_Y , σ_X, σ_Y , ρ)\)의 확률밀도함수로부터 \(X\)의 주변확률밀도함수를 구하시오.
\(f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy\)
\(= \int_{-\infty}^\infty \dfrac{1}{2\pi \sigma_x \sigma_y \sqrt{1-\rho^2}}exp \left[-\dfrac{1}{2(1- \rho^2)} \left( \left(\dfrac{x-\mu_x}{\sigma_x}\right)^2 + \left(\dfrac{y-\mu_y}{\sigma_y}\right)^2 - 2\rho \left(\dfrac{x-\mu_x}{\sigma_x}\right) \left(\dfrac{y-\mu_y}{\sigma_y}\right) \right) \right]dy\)
\(v=\dfrac{y-\mu_y}{\sigma_y}\)치환하면, \(dv=\dfrac{1}{\sigma_y}dy\)
\(=\int_{-\infty}^\infty \dfrac{1}{2 \pi \sigma_x \sqrt{1-\rho^2}} exp \left[ -\dfrac{1}{2} \left(\dfrac{x-\mu_x}{\sigma_x}\right)^2 - \dfrac{1}{2(1-\rho^2)} \left( v-\rho \left( \dfrac{x-\mu_x}{\sigma_x}\right) \right)^2 \right]dy\)
\(u=\left( v-\rho \left( \dfrac{x-\mu_x}{\sigma_x}\right) \right)\) 치환하면, \(du=\dfrac{dv}{\sqrt{1-\rho^2}}\)
\(=\int_{-\infty}^\infty \dfrac{1}{\sqrt{2\pi}\sigma_x} exp \left[ -\dfrac{1}{2} \left( \dfrac{x-\mu_x}{\sigma_x}\right)^2 - \dfrac{u^2}{2}\right]du\)
\(=\dfrac{1}{\sqrt{2\pi}\sigma_x} exp \left[ -\dfrac{1}{2} \left( \dfrac{x-\mu_x}{\sigma_x}\right)^2 \right] \int_{-\infty}^\infty e^{-u^2/2}du\)
\(\because \int_{-\infty}^\infty e^{-u^2/2}du=1\)
\(=\dfrac{1}{\sqrt{2\pi}\sigma_x} exp \left[ -\dfrac{1}{2} \left( \dfrac{x-\mu_x}{\sigma_x}\right)^2 \right]\)
8번
\((X_1, X_2, X_3) ∼ MULT(n, p_1, p_2, p_3)\)일 때 \(Cov(X_1, X_2)\)를 구하시오
\(Var(X_1+X_2)=n(p_1+p_2)(1-(p_1+p_2))\)
\(Var(X_1+X_2)=Var(X_1)+Var(X_2)+2Cov(X_1,X_2)=np_1(1-p_1)+np_2(1-p_2)+2Cov(X_1,X_2)\)
1과 2를 연립하면,
\(Cov(X_1,X_2)=-np_1p_2\)
\(f_{X_1,X_2,X_3}(x_1,x_2,x_3)\begin{pmatrix} n \\ x_1 \ x_2\ x_3 \end{pmatrix} p_1^{x_1} p_2^{x_2} p_3^{x_3}\) , 단 \(\begin{pmatrix} 0\leq x_{1}\leq 1 & \\ 0\leq x_{2}\leq 1 & \\ 0\leq x_{3}\leq 1 & \\ x_{1}+x_{2}+x_{3}= & 1 \end{pmatrix}\)
\(Cov(X_1, X_2) = E(X_1X_2) - E(X_1)E(X_2)\)
\(X_1\) ~ \(B(n,p_1)\)이므로 \(E(X_1)=np_1, E(X_2)=np_2\)
\(E(X_1X_2)=E(E(X_1X_2|X_2))\) 이고 \(X_2\)가 주어졌을 때의 조건부서식이므로 \(X_2\)는 밖으로 나갈 수 있다.
=\(E(X_2 E(X_1|X_2))\)를 구하자.
먼저, \(E(X_1|X_2=x_2)\)는 \(f_{X_1|X_2}(x_1|x_2)=\dfrac{f_{X_1,X_2}(x_1,x_2)}{f_{X_2}(x_2)}\)이다.
분자먼저 구하자!
\(f_{X_1,X_2}(x_1,x_2)=P(X_1=x_1,X_2=x_2)\) 이고
이 실험에서는,
\(=P(X_1=x_1,X_2=x_2,X_3=n-x_1-x_2)\)
\(=\dfrac{n!}{x_1! x_2! (n-x_1 -x_2)!}p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}\)
분모는, 이항분포를 따르므로
\(f_{X_2}(x_2) = \dfrac{n!}{x_2!(n-x_2)!}p_2^{x_2}(1-p_2)^{n-x_2}\)
\(f_{X_1|X_2}(x_1|x_2)=\dfrac{f_{X_1,X_2}(x_1,x_2)}{f_{X_2}(x_2)}\)
\(=\dfrac{\dfrac{n!}{x_1! x_2! (n-x_1 -x_2)!}p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}}{\dfrac{n!}{x_2!(n-x_2)!}p_2^{x_2}(1-p_2)^{n-x_2}}\)
\(=\dfrac{(n-x_2)!}{x_1!(n-x_2-x_1)!} \left( \dfrac{p_{1}}{1-p_{2}}\right) ^{x_{1}} \left( \dfrac{p_{3}}{1-p_{2}}\right) ^{n-x_{2}-x_{1}}\)
\(=\begin{pmatrix} n-x_{2} \\ x_{1} \end{pmatrix} \left( \dfrac{p_{1}}{1-p_{2}}\right) ^{x_{1}} \left( 1-\dfrac{p_{1}}{1-p_{2}}\right) ^{n-x_{2}-x_{1}}\)
즉, \(B(n-x_2, \dfrac{p_1}{1-p_2})\)를 따른다.
즉, \(E(X_1X_2)=\dfrac{(n-x_2)p_1}{1-p_2}\)
다시 이중기댓값 정의로 돌아가서..
구하려는 값 : \(E(X_2 E(X_1|X_2))\)
\(E(X_2 E(X_1|X_2))=E(X_2 \cdot \dfrac{(n-x_2)p_1}{1-p_2})\)
\(=\dfrac{p_1}{1-p_2}[E(n_2X_2)-E(X_2^2)]\)
\(\because E(X_2^2) = Var(X_2) + [E(X_2)]^2 = np_2(1-p_2) + n^2 p_2^2\)
\(=\dfrac{p_1}{1-p_2}(n^2p_2 - np_2(1-p_2) - n^2 p_2^2)\)
\(=\dfrac{np_1p_2}{1-p_2}(n-1+p_2-np_2)\)
\(=\dfrac{np_1p_2(1-p_2)(n-1)}{1-p_2}\)
\(=n(n-1)p_1p_2\)
돌고돌아 구하고자 하는 공분산을 구해보자.
\(Cov(X_1, X_2) = E(X_1X_2) - E(X_1)E(X_2)\)
\(=n(n-1)p_1p_2 - n^2p_1p_2\)
\(=-np_1p_2\)