- ($30\%$) Suppose a random value $y$ follows the $g$ distribution, i.e., $y$ can be expressed as a function of a normally distributed variable as follows.
$\displaystyle y=\delta+\lambda\frac{\exp\left(gz\right)-1}g$,
where $\delta$ and $g$ are real numbers, $\lambda$ is a positive real number, and $z$ follows the normal distribution with mean to be $\mu$ and standard deviation to be $\sigma$. Given the constraint $\left[g\left(y-\delta\right)/\lambda+1\right]>0$, solve the following problems.- ($4\%$) Express $z$ in terms of $y$, i.e., find the function $h\left(y\right)=z$.
- ($4\%$) With the function $h\left(y\right)$, derive the density function of $y$, $f\left(y\right)$, according to the definition $\displaystyle f\left(y\right)=\frac1{\sigma\sqrt{2\pi}}\left|h'\left(y\right)\right|e^{-\frac1{2\sigma^2}\left(h\left(y\right)-\mu\right)^2}$, where $\mu$ and $\sigma$ are the mean and the standard deviation of $z$.
- ($8\%$) Derive the expectation of $y$, $E(y)$.
- ($8\%$) Derive the variance of $y$, $\text{var}(y)$.
- ($6\%$) Show that when $g$ is zero, $y$ follows a normal distribution. [Hint: Consider the result of $\displaystyle\lim_{g\to0}\frac{\exp\left(gz\right)-1}g$.]
- 移項便有
$\displaystyle \frac{g\left(y-\delta\right)}{\lambda}+1=\exp\left(gz\right)$
由於 $\left[g\left(y-\delta\right)\lambda+1\right]>0$，故可同取自然對數便有 $\displaystyle gz=\ln\left(1+\frac{g\left(y-\delta\right)}{\lambda}\right)$，故得$\displaystyle z=\frac1g\ln\left(1+\frac{g\left(y-\delta\right)}{\lambda}\right):=h\left(y\right)$
- 先求 $h$ 的導函數有
$\displaystyle h'\left(y\right)=\frac1g\cdot\frac{g}{\lambda+g\left(y-\delta\right)}=\frac1{\lambda+g\left(y-\delta\right)}$
由於 $g\left(y-\delta\right)/\lambda+1>0$，故 $\lambda+g\left(y-\delta\right)>0$，故 $\left|h'\left(y\right)\right|=h'\left(y\right)$。從而 $y$ 的機率密度函數為$\displaystyle f\left(y\right)=\frac1{\sigma\sqrt{2\pi}}\cdot\frac1{\lambda+g\left(y-\delta\right)}\exp\left[-\frac1{2\sigma^2}\left(\frac{\ln\left(\lambda+g\left(y-\delta\right)\right)-\ln\lambda}g-\mu\right)^2\right]$
其中 $y\in\left(y^*,\infty\right)$，$y^*=-\lambda/g+\delta$。 - 那麼由期望值的定義可知
$\displaystyle E\left(y\right)=\int_{y^*}^{\infty}yf\left(y\right)dy$
令 $\displaystyle t=\frac{h\left(y\right)-\mu}{\sigma\sqrt2}$，那- 當 $y\to y^*$ 時有 $t\to-\infty$；
- 當 $y\to\infty$ 時有 $t\to\infty$；
- 求導有 $\displaystyle dt=\frac{h'\left(y\right)}{\sigma\sqrt2}dy$。再者整理有 $y=h^{-1}\left(\mu+t\sigma\sqrt2\right)$。
$\displaystyle\begin{aligned}E\left(y\right)&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}h^{-1}\left(\mu+t\sigma\sqrt2\right)e^{-t^2}dt=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left[\delta-\frac{\lambda}g+\frac{\lambda}g\exp\left(g\mu+tg\sigma\sqrt2\right)\right]e^{-t^2}dt\\&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left(\delta-\frac{\lambda}g\right)e^{-t^2}dt+\frac{\lambda e^{g\mu}}{g\sqrt\pi}\int_{-\infty}^{\infty}\exp\left(-t^2+tg\sigma\sqrt2\right)dt\\&=\delta-\frac{\lambda}g+\frac{\lambda e^{g\mu+g^2\sigma^2/2}}{g}\end{aligned}$
- 按照變異數的定義有 $\text{var}\left(y\right)=E\left(y^2\right)-\left(E\left(y\right)\right)^2$，故我們僅需計算 $E\left(y^2\right)$，即計算
$\displaystyle E\left(y^2\right)=\int_{y^*}^{\infty}y^2f\left(y\right)dy$
沿用前一小題的代換可知$\displaystyle\begin{aligned}E\left(y^2\right)&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left[\delta-\frac{\lambda}g+\frac{\lambda}g\exp\left(g\mu+tg\sigma\sqrt2\right)\right]^2e^{-t^2}dt\\&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left(\delta-\frac{\lambda}g\right)^2e^{-t^2}dt+\frac{2\lambda e^{g\mu}}{g\sqrt\pi}\left(\delta-\frac{\lambda}g\right)\int_{-\infty}^{\infty}\exp\left(-t^2+tg\sigma\sqrt2\right)dt+\frac{\lambda^2e^{2g\mu}}{g^2\sqrt\pi}\int_{-\infty}^{\infty}\exp\left(-t^2+2tg\sigma\sqrt2\right)dt\\&=\left(\delta-\frac{\lambda}g\right)^2+\frac{2\lambda e^{g\mu+g^2\sigma^2/2}}g\left(\delta-\frac{\lambda}g\right)+\frac{\lambda^2e^{2g\mu+2g^2\sigma^2}}{g^2}\end{aligned}$
因此所求為$\displaystyle\begin{aligned}\text{var}\left(y\right)&=\left[\left(\delta-\frac{\lambda}g\right)^2+\frac{2\lambda e^{g\mu+g^2\sigma^2/2}}g\left(\delta-\frac{\lambda}g\right)+\frac{\lambda^2e^{2g\mu+2g^2\sigma^2}}{g^2}\right]-\left(\delta-\frac{\lambda}g+\frac{\lambda e^{g\mu+g^2\sigma^2/2}}g\right)^2\\&=\frac{\lambda^2}{g^2}e^{2g\mu+g^2\sigma^2}\left(e^{g^2\sigma^2}-1\right)\end{aligned}$
- 由於當 $g$ 趨於零時可以發現 $\displaystyle\lim_{g\to0}\frac{\exp\left(gz\right)-1}g=\lim_{g\to0}\frac{z\exp\left(gz\right)}1=z$。故當 $g=0$ 時有 $y=\delta+\lambda z$，此表明經由線性轉換能使 $y$ 成為平均為 $\delta+\lambda\mu$，而標準差則為 $\lambda\sigma$。由 (c) 與 (d) 能知 $\displaystyle\lim_{g\to0}E\left(y\right)=\delta+\lambda\mu$ 且 $\displaystyle\lim_{g\to0}\text{var}\left(y\right)=\lambda^2\sigma^2$。
- ($8\%$) Find the maximum and minimum values of $f\left(x,y,z\right)=x+3y-z$ subject to $4x^2+2y^2+z^2=4$.
- ($12\%$) A $10,000$-cubic-foot-room has $500$ smoke particles per cubic foot. A ventilation system is turned on that each minute brings in $500$ cubic feet of smoke-free air, while an equal volume of air leaves the room. Also, during each minute, smoker in the room add a total of $10,000$ particles of smoke to the room. Assume that the air in the room mixes thoroughly.
- ($4\%$) Find a differential equation and initial condition that govern the total number $y\left(t\right)$ of smoke particles in the room after $t$ minutes.
- ($4\%$) Solve this differential equation.
- ($4\%$) Find how soon the smoke level will fall to $100$ smoke particles per cubic foot. (If the answers are with decimal numbers, please round to the nearest hundredth.)
- 由於 $y\left(t\right)$ 表在時刻 $t$ 分鐘後的煙霧粒子的個數，那麼 $y'\left(t\right)$ 便表示在 $t$ 分鐘後增加煙霧粒子的瞬間增加量，按照題意可知
$\displaystyle y'\left(t\right)=10000-\frac{y\left(t\right)}{10000}\cdot500$
再者一開始房間內有 $500\cdot10^4$ 個煙霧粒子，因此 $y\left(0\right)=5\cdot10^6$。 - 經由移項可知
$\displaystyle\frac{dy}{200000-y}=\frac{dt}{20}$
同取積分有$\displaystyle-\ln\left|200000-y\right|=\frac{t}{20}+C$
透過初始條件可知 $C=-\ln4800000$，從而所求為$y\left(t\right)=200000+4800000e^{-t/20}$
- 要達到濃度為每立方英尺有 $100$ 顆煙霧粒子，則該房間要有總數 $10^6$ 顆煙霧粒子。因此解方程式 $y\left(t\right)=10^6$，即 $2+48e^{-t/20}=10$，故有 $t=20\ln6$。
- Evaluate:
- ($5\%$) $\displaystyle\phi\left(\alpha\right)=\int_0^{\infty}\alpha e^{-\alpha x}dx$ for $\alpha>0$.
- ($10\%$) Evaluate $\displaystyle\int_0^1e^{x^2}dx$, approximately by using Taylor's theorem of the mean and estimate the maximum error.
- 直接計算瑕積分可知
$\displaystyle\phi\left(\alpha\right)=-e^{-\alpha x}\Big|_0^{\infty}=1$
- 回憶自然指數的泰勒展開式為
$\displaystyle e^x=\sum_{k=0}^n\frac{x^k}{k!}+\frac{e^{\xi}}{\left(n+1\right)!}x^{n+1}$
此處 $\xi$ 介於 $0$ 與 $x$ 之間。現用 $x^2$ 取代 $x$ 有$\displaystyle e^{x^2}=\sum_{k=0}^n\frac{x^{2k}}{k!}+\frac{e^{\eta^2}}{\left(n+1\right)!}x^{2n+2}$
而此時的 $\eta$ 介於 $0$ 與 $x$ 之間。由此計算定積分可知其近似值為$\displaystyle\int_0^1e^{x^2}dx\approx\int_0^1\sum_{k=0}^n\frac{x^{2k}}{k!}dx=\sum_{k=0}^n\frac1{k!\left(2k+1\right)}$
而其最大的誤差估計為$\displaystyle\left|\int_0^1e^{x^2}dx-\sum_{k=0}^n\frac1{k!\left(2k+1\right)}\right|\leq\int_0^1\frac{e^{\eta^2}x^{2n+2}}{\left(n+1\right)!}dx=\frac{e^{\eta^2}}{\left(n+1\right)!\left(2n+3\right)}\leq\frac{e}{\left(n+1\right)!\left(2n+3\right)}$
這是因為 $\eta\in\left[0,x\right]\subseteq\left[0,1\right]$，因此 $e^{\eta^2}\leq e$。 - Let $\varphi$ be a continuous real function on $\left[0,\alpha\right]$. Assume that $\varphi\left(x\right)>0$ if $0<x\leq\alpha$ and that $\varphi\left(x\right)\sim Ax^r$ as $x\to0$ ($A>0$, $r\geq0$). For all $t>0$ set
$\displaystyle F\left(t\right)=\int_0^{\alpha}\frac{dx}{t+\varphi\left(x\right)}$.
- ($10\%$) Show that if $r>1$, then as $t\to0$
$\displaystyle F\left(t\right)\sim\frac{\pi}{rA^{1/r}\sin\left(\pi/r\right)}\cdot\frac1{t^{1-1/r}}$.
- ($10\%$) Show that if $r=1$, then as $t\to0$
$\displaystyle F\left(t\right)\sim\frac1A\log\frac1t$.
- ($10\%$) Show that if $r>1$, then as $t\to0$
- 給定正數 $\varepsilon<1$，那麼存在 $\delta>0$使得 $0<x<\delta$ 蘊含 $Ax^r\left(1-\varepsilon\right)<\varphi\left(x\right)<Ax^r\left(1+\varepsilon\right)$。據此我們可寫 $F$ 如下
$\displaystyle F\left(t\right)=\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}+\int_{\delta}^{\alpha}\frac{dx}{t+\varphi\left(x\right)}$
對於第二項可得如下的估算$\displaystyle0<\int_{\delta}^{\alpha}\frac{dx}{t+\varphi\left(x\right)}\leq\int_{\delta}^{\alpha}\frac{dx}{\displaystyle\min_{\left[\delta,\alpha\right]}\varphi}=\frac{\alpha-\delta}{\displaystyle\min_{\left[\delta,\alpha\right]}\varphi}<\infty$
故有 $t_1^*$ 使得 $0<t<t_1^*$ 蘊含$\displaystyle\int_{\delta}^{\alpha}\frac1{t+\varphi\left(x\right)}<\frac{\varepsilon}{2rA^{1/r}\left(1+\varepsilon\right)^{1/r}t^{1-1/r}}$
至於第一項則容易知道$\displaystyle\frac1t\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}=\int_0^{\delta}\frac{dx}{t+Ax^r\left(1+\varepsilon\right)}<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\int_0^{\delta}\frac{dx}{t+Ax^r\left(1-\varepsilon\right)}=\frac1t\int_0^{\delta}\frac{dx}{1+Ax^r\left(1-\varepsilon\right)/t}$
首先我們處理積分 $\displaystyle\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}$，令 $y=1+Ax^r\left(1+\varepsilon\right)/t$，那麼有- 當 $x=0$ 時有 $y=1$；
- 當 $x=\delta$ 時有 $y=1+A\delta^r\left(1+\varepsilon\right)/t$；
- 整理有 $\displaystyle x=\left(\frac{t\left(y-1\right)}{A\left(1+\varepsilon\right)}\right)^{1/r}$，求導有 $\displaystyle dx=\frac1r\left(\frac{t}{A\left(1+\varepsilon\right)}\right)^{1/r}\left(y-1\right)^{\left(1-r\right)/r}dy$。
$\displaystyle\begin{aligned}\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}&=\frac{t^{1/r}}{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\int_1^{1+A\delta^r\left(1+\varepsilon\right)/t}\frac1{y\left(y-1\right)^{1-1/r}}dy\\&=\frac{t^{1/r}}{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left[\int_1^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy-\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy\right]\end{aligned}$
由於 $\displaystyle\lim_{y\to\infty}\frac{\displaystyle\frac1{y\left(y-1\right)^{1-1/r}}}{\displaystyle\frac1y}=0$，故瑕積分 $\displaystyle\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy$ 有限，從而 $\displaystyle\lim_{t\to0^+}\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy=0$，即當 $t<t_2^*$ 時有$\displaystyle\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy<\frac\varepsilon2$
另一方面，透過變換 $u=y-1$ 可知$\displaystyle\int_1^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy=\int_0^{\infty}\frac{u^{1/r-1}}{1+u}du$
運用複變函數論中的留數定理與路徑積分（可參見「複變數－原理及題解」，中央圖書出版社，吳新生編著，第 403 頁）可知$\displaystyle\int_0^{\infty}\frac{u^{1/r-1}}{1+u}du=\frac{\pi}{\sin\left(\pi/r\right)}$
綜合上述的推導可得$\displaystyle\frac1{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}-\varepsilon\right)\frac1{t^{1-1/r}}<F\left(t\right)$
類似地，我們也有$\displaystyle F\left(t\right)<\frac1{rA^{1/r}\left(1-\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}+\varepsilon\right)\frac1{t^{1-1/r}}$
將兩者結合可知$\displaystyle\frac1{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}-\varepsilon\right)<t^{1-1/r}F\left(t\right)<\frac1{rA^{1/r}\left(1-\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}+\varepsilon\right)$
故取由 $t\to0^+$ 與 $\varepsilon$ 的任意性可知$\displaystyle\lim_{t\to0^+}t^{1-1/r}F\left(t\right)=\frac\pi{rA^{1/r}\sin\left(\pi/r\right)}$
這就給出了$\displaystyle F\left(t\right)\sim\frac\pi{rA^{1/r}\sin\left(\pi/r\right)}\frac1{t^{1-1/r}}$
- 承前述步驟可知
$\displaystyle\int_0^{\delta}\frac{dx}{t+Ax\left(1+\varepsilon\right)}<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\int_0^{\delta}\frac{dx}{t+Ax\left(1-\varepsilon\right)}$
直接計算可給出$\displaystyle\frac1{A\left(1+\varepsilon\right)}\log\left(1+\frac{A\delta\left(1+\varepsilon\right)}t\right)<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\frac1{A\left(1-\varepsilon\right)}\log\left(1+\frac{A\delta\left(1-\varepsilon\right)}t\right)$
同乘以 $\displaystyle\frac{A}{\log\left(1/t\right)}$ 有$\displaystyle\frac1{1+\varepsilon}\frac{\log\left(1+A\delta\left(1+\varepsilon\right)/t\right)}{\log\left(1/t\right)}<\frac{A}{\log\left(1/t\right)}\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\frac1{1-\varepsilon}\frac{\log\left(1+A\delta\left(1-\varepsilon\right)/t\right)}{\log\left(1/t\right)}$
取 $t$ 趨於零且由 $\varepsilon$ 的任意性可知 $\displaystyle\lim_{t\to0^+}\frac{A}{\log\left(1/t\right)}\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}=1$，從而有$\displaystyle\lim_{t\to0^+}\frac{A}{\log\left(1/t\right)}F\left(t\right)=1$
這就給出了$\displaystyle F\left(t\right)\sim\frac1A\log\frac1t$
- Suppose that $a_n\to c$ as $n\to\infty$ and that $\left\{\alpha_i\right\}_{i=1}^{\infty}$ is a sequence of positive terms for which $\displaystyle\sum_{i=1}^na_i\to\infty$ as $n\to\infty$.
- ($10\%$) Show that
$\displaystyle\frac{\displaystyle\sum_{i=1}^n\alpha_ia_i}{\displaystyle\sum_{i=1}^n\alpha_i}\to c$ as $n\to\infty$.
In particular, if $\alpha_i=1$ for all $i$, then$\displaystyle\frac1n\sum_{i=1}^na_i\to c$ as $n\to\infty$.
- ($5\%$) Show that the converse of the special case in (a) does not always hold by giving a counterexample of a sequence $\left\{a_n\right\}_{n=1}^{\infty}$ that does not converges, yet $\displaystyle\frac1n\sum_{i=1}^na_i$ converges as $n\to\infty$.
- ($10\%$) Show that
- 設 $\displaystyle b_n=\sum_{i=1}^n\alpha_i$、$\displaystyle c_n=\sum_{i=1}^n\alpha_ia_i$，按題設可知 $\displaystyle\lim_{n\to\infty}b_n=\infty$。再者，有
$\displaystyle\lim_{n\to\infty}\frac{c_{n+1}-c_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\alpha_{n+1}a_{n+1}}{\alpha_{n+1}}=\lim_{n\to\infty}a_{n+1}=c$
那麼根據Stolz－Cesàro定理可知 $\displaystyle\lim_{n\to\infty}\frac{c_n}{b_n}=c$，此即所欲證。而當 $\alpha_i=1$，那麼容易知道 $b_n=n$，$\displaystyle c_n=\sum_{i=1}^na_i$。 - 考慮 $a_n=\left(-1\right)^n$，容易直接計算
$\displaystyle\sum_{i=1}^na_i=\begin{cases}-1,&\text{if}~n~\text{is odd},\\0,&\text{if}~n~\text{is even}.\end{cases}$
由此可知 $\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^na_i=0$，但數列 $\left\{a_i\right\}_{i=1}^{\infty}$ 發散。
訣竅
前兩小題根據題目給定的資訊整理即可，而後三小題則分別按照期望值、變異數與常態分配的定義處理。解法
訣竅
運用初等不等式求極值；亦可使用拉格朗日乘子法求解。解法一
由柯西不等式可知$\displaystyle23=\left[\left(2x\right)^2+\left(\sqrt2y\right)^2+z^2\right]\left[\left(\frac12\right)^2+\left(\frac3{\sqrt2}\right)^2+\left(-1\right)^2\right]\geq\left(x+3y-z\right)^2$
因此函數 $f$ 在給定的限制條件下的最大最小值分別為 $\sqrt{23}$ 與 $-\sqrt{23}$，此時的等號成立條件為 $\displaystyle 4x=\frac{2y}3=-z$，即 $\displaystyle\left(\frac1{\sqrt{23}},\frac6{\sqrt{23}},-\frac4{\sqrt{23}}\right)$ 與 $\displaystyle\left(-\frac1{\sqrt{23}},-\frac6{\sqrt{23}},\frac4{\sqrt{23}}\right)$。解法二
設定拉格朗日乘子函數如下$\displaystyle F\left(x,y,z,\lambda\right)=x+3y-z+\lambda\left(4x^2+2y^2+z^2-4\right)$
據此解下列的聯立方程組$\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=1+8\lambda x=0\\&F_y\left(x,y,z,\lambda\right)=3+4\lambda y=0\\&F_z\left(x,y,z,\lambda\right)=-1+2\lambda z=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=4x^2+2y^2+z^2-4=0\end{aligned}\right.$
由於明顯 $\lambda\neq0$，故 $\displaystyle x=-\frac1{8\lambda}$、$\displaystyle y=-\frac3{4\lambda}$、$\displaystyle z=\frac1{2\lambda}$，由此代入第四式為$\displaystyle\frac4{64\lambda^2}+\frac{2\cdot9}{16\lambda^2}+\frac1{4\lambda^2}=4$
即得 $\lambda^2=\frac{23}{64}$，故有 $\displaystyle\lambda=\pm\frac{\sqrt{23}}8$。如此得兩座標$\displaystyle\left(x,y,z\right)=\frac{\left(1,6,-4\right)}{\sqrt{23}}$, $\displaystyle\left(x,y,z\right)=\frac{\left(-1,-6,4\right)}{\sqrt{23}}$
而最大最小值分別為 $\sqrt{23}$ 與 $-\sqrt{23}$。
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