111年成大工程科學碩士班-線代與機率詳解

 國立成功大學111學年度碩士班招生考試

系所: 工程科學系
科目: 線性代數與機率

解答: $$\left( \begin{matrix}1 & a & b & c \\1 & a^2 & b^2 & c^2 \\ 1 & a^3 & b^3 & c^3 \\ 1 & a^4 & b^4 & c^4 \end{matrix} \right) \xrightarrow{-R_1+R_2\to R_2, -R_1+R_3\to R_3, -R_1+R_4\to R_4} \left( \begin{matrix} 1 & a & b & c \\ 0 & a^2-a & b^2-b & c^2-c \\ 0 & a^3-a & b^3-b & c^3-c \\ 0 & a^4-a & b^4-b & c^4-c \end{matrix} \right) \\ \xrightarrow{R_3-(a+1)R_2 \to R_3} \left( \begin{matrix} 1 & a & b & c \\0 & a^2-a & b^2-b & c^2-c \\0 & 0 & b^3-ab^2-b^2+ab & c^3-ac^2-c^2+ac \\ 0 & a^4-a & b^4-b & c^4-c \end{matrix} \right) \\ \xrightarrow{R_4-(a^2+a+1) R_2 \to R_4} \left( \begin{matrix} 1 & a & b & c \\0 & a^2-a & b^2-b & c^2-c \\ 0 & 0 & b^3-ab^2-b^2+ab & c^3-ac^2-c^2+ac \\ 0 & 0 & b^4-a^2b^2- ab^2-b^2+a^2b+ab & c^4-a^2 c^2-ac^2-c^2+a^2c +ac \end{matrix} \right) \\\xrightarrow{R_4-(a +b +1) R_3 \to R_4} \left( \begin{matrix} 1 & a & b & c \\ 0 & a^2-a & b^2-b & c^2-c \\0 & 0 & b^3-ab^2 -b^2+ab & c^3-ac^2-c^2 +ac \\ 0 & 0 & 0 & c^4-ac^3 -bc^3-c^3+ac^2+abc^2+bc^2-abc \end{matrix} \right) \\ \Rightarrow \det(D) = (a^2-a) \color{blue}{(b^3-ab^2-b^2+ab)}  \color{green}{ (c^4-ac^3-bc^3-c^3+ac^2 +abc^2+bc^2-abc)} \\= a(a-1) \color{blue}{b(b-1)(b-a)} \color{green}{ c(c-1)(c-a)(c-b)} \\=\bbox[red, 2pt]{abc(a-1)(b-1)(c-1)(a-b) (b-c)(c-a)}$$

解答: $$\det(D)=1\times 2\times 3\times 4\times 5= \bbox[red, 2pt]{120}$$

解答: $$A=\left(\begin{matrix} 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}\right) \Rightarrow A^T=\left(\begin{matrix} 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1\end{matrix} \right)\\ BA^T=0 \Rightarrow B=0或B=\bbox[red, 2pt]{\left(\begin{matrix} 1 & -1 & 0 & -1 & 1 & 0 & 0 & 0 & 2 & 1 \\ 1 & -1 & 0 & -1 & 1 & 0 & 0 & 0 & 2 & 1 \\ 1 & -1 & 0 & -1 & 1 & 0 & 0 & 0 & 2 & 1 \\ 1 & -1 & 0 & -1 & 1 & 0 & 0 & 0 & 2 & 1\end{matrix}\right)}$$

解答: $$Y=\ln X \Rightarrow X=e^Y \Rightarrow E(X^k)=E(e^{kY}) =\int_{-\infty}^\infty e^{ky}\cdot {1\over \sqrt{2\pi}}e^{-y^2/2}\, dy \\= \int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-{1\over 2}(y-k)^2+k^2/2} \,dy =e^{k^2/2} \int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-{1\over 2}(y-k)^2/2} \,dy = e^{k^2/2} \\ \Rightarrow E(X^k)=e^{k^2/2} \Rightarrow \cases{E(X)=e^{1/2}\\ E(X^2)=e^2} \Rightarrow Var(X)=E(X^2)-(E(X))^2 = \bbox[red, 2pt]{e^2-e}$$

解答: $$\textbf{a.}\;X\sim Geo(p) \Rightarrow p(x)=P(X=x)=q^{x-1}p \\ \Rightarrow P(X\gt n)=P(X=n+1) +P(X=n+2)+ \cdots=q^np +q^{n+1}p+ q^{n+2} p+\cdots \\ =q^np( 1+q+ q^2+ \cdots)= q^np{1 \over 1-q} =q^n\\ P(n\lt X\lt k+n)=P(X=n+1) +P(X=n+2) +\cdots +P(X=k+n-1)\\ =q^{n+1}p+ q^{n+2} p+ \cdots +q^{k+n-2}p =q^{n+1}p(1+q+ \cdots +q^{k-3}) =q^{n+2}p {1-q^{k-2} \over 1-q} \\=q^{n+2}(1-q^{k-2})\\ 因此P(X \lt k+n\mid X\gt n) ={ P(n\lt X\lt k+n )\over P(X\gt n)} ={q^{n+2}(1 -q^{k-2}) \over q^n} =q^2(1-q^{k-2})\\ =q^2-q^k = \bbox[red, 2pt]{(1-p)^2-(1-p)^k}\\ \textbf{b.}\; P(X\lt k)=P(X=1) +P(X=2)+ \cdots +P(X=k-1) =p+qp+ q^2p+\cdots q^{k-2}p \\=p(1+q+ \cdots+q^{k-2}) = p\cdot{ 1-q^{k-1}\over 1-q} =1-q^{k-1} =\bbox[red, 2pt]{1-(1-p)^{k-1}} \\\textbf{c.}\; \cases{P(X\gt a) =q^ap+ q^{a+1}p+ \cdots=q^ap(1+q+q^2+ \cdots) =q^a \\ P(X\gt x+a)=q^{x+a}p+ q^{x+a+1}p+ \cdots = q^{x+a}p (1+q+\cdots )=q^{x+a}} \\ \quad \Rightarrow P(X\gt x+a \mid X\gt a)={P(X\gt x+a)\over P(X\gt a)} ={q^{x+a}\over q^a} =q^x =P(X\gt x) \\ \Rightarrow X \text{ is memoryless},\bbox[red, 2pt]{Q.E.D.}$$

解答: $$\textbf{a.}\; X_1 \sim Exp(\lambda_1) \Rightarrow P(X=x)=\lambda_1e^{-\lambda_1 x} ,x\gt 0 \\ \quad \Rightarrow M_{X_1}(t)= E(e^{Xt}) =\int_0^\infty e^{xt}\cdot \lambda_1e^{-\lambda_1 x} \,dx =\int_0^\infty  \lambda_1e^{(t-\lambda_1) x} \,dx =\left. \left[ {\lambda_1\over t-\lambda_1}e^{(t-\lambda_1) x} \right] \right|_0^\infty \\\quad  ={\lambda_1\over \lambda_1-t} \Rightarrow \bbox[red, 2pt]{M_{X_1}(t)= {\lambda_1\over \lambda_1-t}, \lambda_1 \gt t} \\ \textbf{b.}\; P(X\lt x) =1-P(X\ge x) =1-P(\min\{X_1,X_2, \dots, X_n \}\ge x) \\\quad=1-P(X_1\ge x, X_2\ge x,\dots ,X_n\ge x) =1-P(X_1\ge x) P(X_2\ge x)\cdots P(X_n\ge x) \\ \quad =1- \left( \int_x^\infty \lambda_1e^{-\lambda_1 t}dt \right)\left( \int_x^\infty \lambda_2e^{-\lambda_2 t}dt \right) \cdots \left( \int_x^\infty \lambda_ne^{-\lambda_n t}dt \right) \\\quad = 1-e^{-\lambda_1 x} \cdot e^{-\lambda_2 x} \cdots e^{-\lambda_n x}  = \bbox[red, 2pt]{1-e^{-(\lambda_1+ \lambda_2+\cdots +\lambda_n)x}} \\\textbf{c.}\; P(X\lt x)   = 1-e^{-(\lambda_1+ \lambda_2+\cdots +\lambda_n)x} \Rightarrow P(X=x)= {d\over dx} \left(1-e^{-(\lambda_1+ \lambda_2+\cdots +\lambda_n)x} \right) \\\quad = (\lambda_1+\lambda_2 +\cdots +\lambda_n) e^{-(\lambda_1+ \lambda_2+\cdots +\lambda_n)x}\\\quad \Rightarrow M_X(t)= E(e^{tX}) = \int_0^\infty e^{tx} \cdot (\lambda_1+\lambda_2 +\cdots +\lambda_n) e^{-(\lambda_1+ \lambda_2+\cdots +\lambda_n)x} \,dx \\\quad = (\lambda_1+\lambda_2 +\cdots +\lambda_n) \int_0^\infty e^{(t-(\lambda_1+ \lambda_2+\cdots +\lambda_n))x} \,dx \\\quad ={\lambda_1+\lambda_2 +\cdots +\lambda_n \over t-(\lambda_1+\lambda_2 +\cdots +\lambda_n)} \left. e^{(t-(\lambda_1+ \lambda_2+\cdots +\lambda_n))x}\right|_{0}^\infty ={\lambda_1+\lambda_2 +\cdots +\lambda_n \over  (\lambda_1+\lambda_2 +\cdots +\lambda_n)-t} \\\quad =\bbox[red, 2pt]{1+{ t\over  (\lambda_1+\lambda_2 +\cdots +\lambda_n)-t}, \lambda_1+\lambda_2+\cdots +\lambda_n \gt t}$$

解答: $$X\sin B(n=20,p=1/20) \Rightarrow EX= np=\bbox[red, 2pt]1$$

解答: $$X\sim B(n,p) \Rightarrow f(n,p,x)= C^n_x p^xq^{n-x}\\ \Rightarrow M_X(t)= E[e^{tx}] =\sum_{x=0}^n e^{tx}C^n_xp^x q^{n-x} = \sum_{x=0}^n C^n_x (e^tp)^x q^{n-x} =(e^tp+q)^n\\ \Rightarrow {d\over dt} M_X(t) = npe^t(e^tp +q)^{n-1}\\ \Rightarrow {d^2\over dt^2} M_X(t) =npe^t(e^tp +q)^{n-1}  +n(n-1)p^2e^{2t} (e^tp+q)^{n-2} \\ \Rightarrow {d^3\over dt^3} M_X(t) = npe^t(e^tp+ q)^{n-1} +3n(n-1)p^2e^{2t} (e^tp+q)^{n-2} +n(n-1)(n-2) p^3e^{3t}( pe^t+q)^{n-3} \\ \Rightarrow E[X^3] = \left. {d^3\over dt^3} M_X(t) \right|_{t=0} = \bbox[red, 2pt]{np +3n(n-1)p^2 +n(n-1)(n-2)p^3}$$

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