國立臺灣大學一百零九學年度研究所碩士班入學考試試題：微積分(C)

※ Please show the detailed calculation process for all questions.

1. ($20\%$) For a random variable $x$ defined on the real number line, denote its probability density function and cumulative distribution function as $f\left(x\right)$ and $F_x\left(\alpha\right)$, respectively. By definition,

   $\displaystyle\int_{-\infty}^{\infty}f\left(x\right)dx=1$, $\displaystyle F_x\left(\alpha\right)=\int_{-\infty}^{\alpha}f\left(x\right)dx$,

   and the expectation of any function $g\left(x\right)$ can be derived as $\displaystyle E\left[g\left(x\right)\right]=\int_{-\infty}^{\infty}g\left(x\right)f\left(x\right)dx$, e.g., for $g\left(x\right)=x$, $\displaystyle E\left[x\right]=\int_{-\infty}^{\infty}xf\left(x\right)dx$.
   1. ($10\%$) Define $\displaystyle H_x\left(\eta\right)=\int_{-\infty}^{\eta}F_x\left(\alpha\right)d\alpha$. Prove that $H_x\left(\eta\right)=E\left[\left(\eta-x\right)^+\right]$, where $\left(\eta-x\right)^+=\begin{cases}\eta-x&\text{if}~\eta-x\geq0,\\0&\text{if}~\eta-x<0.\end{cases}$ (Hint: you can consider to change the order of integration.)
   2. ($10\%$) Prove that $H_x\left(\eta\right)-\left(\eta-E\left[x\right]\right)=E\left[\left(x-\eta\right)^+\right]$, where $\left(x-\eta\right)^+=\begin{cases}x-\eta&\text{if}~x-\eta\geq0,\\0&\text{if}~x-\eta<0.\end{cases}$
訣竅
運用 Fubini 定理來處理重積分；隨後藉由直接驗算驗證等式。
解法
1. 直接計算可知

   $\displaystyle H_x\left(\eta\right)=\int_{-\infty}^{\eta}\int_{-\infty}^{\alpha}f\left(x\right)dxd\alpha=\int_{-\infty}^{\eta}\int_x^{\eta}f\left(x\right)d\alpha dx=\int_{-\infty}^{\eta}\left(\eta-x\right)f\left(x\right)dx=\int_{-\infty}^{\infty}\left(\eta-x\right)^+f\left(x\right)dx=E\left[\left(\eta-x\right)^+\right]$

2. 透過(a)的結果可知

   $H_x\left(\eta\right)-\left(\eta-E\left[x\right]\right)=E\left[\left(\eta-x\right)^+\right]-\left(\eta-E\left[x\right]\right)=E\left[\left(\eta-x\right)^+-\eta+x\right]=E\left[\left(x-\eta\right)^+\right]$

   其中我們使用了等式　$\left(\eta-x\right)^+-\eta+x=\left(x-\eta\right)^+$，這是因為：當 $\eta\geq x$ 時有

   $\left(\eta-x\right)^+-\eta+x=\eta-x-\eta+x=0=\left(x-\eta\right)^+$

   而當 $\eta<x$ 時有

   $\left(\eta-x\right)^+-\eta+x=x-\eta=\left(x-\eta\right)^+$

   證明完畢。


2. ($30\%$) The formula for a geometric Asian call option with six parameters $\left(S,X,r,q,\sigma,T\right)$ is as follows.

   $c\left(S,X,r,q,\sigma,T\right)=Se^{\left(a-r\right)T}N\left(d_1\right)-Xe^{-rT}N\left(d_2\right)$,

   where

   $\displaystyle a=\frac12\left(r-q-\frac{\sigma^2}6\right)$, $\displaystyle d_1=\frac{\displaystyle\ln\left(S/X\right)+\frac12\left(r-q+\frac{\sigma^2}6\right)T}{\sigma\sqrt{T/3}}$, $d_2=d_1-\sigma\sqrt{T/3}$,

   and $N\left(\cdot\right)$ is the cumulative distribution function of the standard normal distribution defined as

   $\displaystyle N\left(d\right)=\int_{-\infty}^dn\left(x\right)dx=\int_{-\infty}^d\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}dx$,

   where $n\left(\cdot\right)$ is the probability density function of the standard normal distribution.
   1. ($10\%$) Prove that $Se^{\left(a-r\right)T}n\left(d_1\right)=Xe^{-rT}n\left(d_2\right)$.
   2. ($6\%$) Derive and express $\displaystyle\frac{\partial c}{\partial S}$ as the form of $AN\left(B\right)$. What are $A$ and $B$?
   3. ($8\%$) Derive and express $\displaystyle\frac{\partial d_1}{\partial\sigma}$ and $\displaystyle\frac{\partial d_2}{\partial\sigma}$ as $C+Dd_2$ and $E+Fd_2$, respectively. What are $C$, $D$, $E$, and $F$?
   4. ($6\%$) Derive and express $\displaystyle\frac{\partial c}{\partial\sigma}$ as the form of $Gn\left(H\right)$. What are $G$ and $H$?
訣竅
運用多變數函數的連鎖律求解即可。
解法
1. 首先可以發現 $\displaystyle d_1+d_2=\frac{2\ln\left(S/X\right)+\left(r-q\right)T}{\sigma\sqrt{T/3}}-\frac{\sigma\sqrt{3T}}6$，同乘以 $\left(d_1-d_2\right)/2=\sigma\sqrt{T/3}/2$ 便有

   $\displaystyle\frac{d_1^2-d_2^2}2=\ln\frac{S}X+\frac{\left(r-q\right)T}2-\frac{\sigma^2T}{12}=\ln\frac{S}X+aT$

   同取自然指數便有

   $\displaystyle e^{\frac{d_1^2-d_2^2}2}=\frac{S}Xe^{aT}$

   整理移項便有

   $\displaystyle\begin{aligned}Se^{\left(a-r\right)T}n\left(d_1\right)&=Se^{\left(a-r\right)T}\cdot\frac1{\sqrt{2\pi}}e^{-\frac{d_1^2}2}\\&=Se^{\left(a-r\right)T}\cdot\frac1{\sqrt{2\pi}}\cdot\left(e^{-\frac{d_2^2}2}\frac{X}Se^{-aT}\right)=Xe^{-rT}\cdot\frac1{\sqrt{2\pi}}e^{-\frac{d_2^2}2}=Xe^{-rT}n\left(d_2\right)\end{aligned}$

2. 運用連鎖律對 $S$ 求導並搭配(a)的結果有

   $\displaystyle\frac{\partial c}{\partial S}=e^{\left(a-r\right)T}N\left(d_1\right)+Se^{\left(a-r\right)T}n\left(d_1\right)\cdot\frac1{S\sigma\sqrt{T/3}}-Xe^{-rT}n\left(d_2\right)\cdot\frac1{S\sigma\sqrt{T/3}}=e^{\left(a-r\right)T}N\left(d_1\right)$

   故 $A=e^{\left(a-r\right)T}$、$B=d_1$。
3. 直接計算偏導有

   $\displaystyle\begin{aligned}\frac{\partial d_1}{\partial\sigma}&=-\frac{\displaystyle\ln\frac{S}X+\frac12\left(r-q+\frac{\sigma^2}6\right)T}{\sigma^2\sqrt{T/3}}+\frac{\sigma T/6}{\sigma\sqrt{T/3}}=-\frac{\displaystyle\ln\frac{S}X+\frac12\left(r-q-\frac{\sigma^2}6\right)T}{\sigma^2\sqrt{T/3}}\\&=-\frac{\displaystyle\ln\frac{S}X+\frac12\left(r-q-\frac{\sigma^2}2\right)T}{\sigma^2\sqrt{T/3}}-\frac{\sigma\sqrt{3T}}6=-\frac{\sqrt{3T}}6-\frac{d_2}{\sigma}\end{aligned}$

   因此 $\displaystyle C=-\frac{\sqrt{3T}}6$、$\displaystyle D=-\frac1{\sigma}$。類似地，有

   $\displaystyle\frac{\partial d_2}{\partial\sigma}=\frac{\partial d_1}{\partial\sigma}-\sqrt{\frac{T}3}=-\frac{\sqrt{3T}}2-\frac{d_2}{\sigma}$

   故 $\displaystyle E=-\frac{\sqrt{3T}}2$、$\displaystyle F=-\frac1{\sigma}$。
4. 運用(a)(c)的結果進行偏導計算有

   $\displaystyle\begin{aligned}\frac{\partial c}{\partial\sigma}&=Se^{\left(a-r\right)T}N\left(d_1\right)\cdot T\frac{\partial a}{\partial\sigma}+Se^{\left(a-r\right)T}n\left(d_1\right)\cdot\frac{\partial d_1}{\partial\sigma}-Xe^{-rT}n\left(d_2\right)\cdot\frac{\partial d_2}{\partial\sigma}\\&=-\frac{S\sigma Te^{\left(a-r\right)T}}6N\left(d_1\right)+\frac{Se^{\left(a-r\right)T}\sqrt{3T}}3n\left(d_1\right)\end{aligned}$

   似乎不太可能將之簡化為與題目給定之形式，故無解。


3. ($10\%$) Given $x\cos y+y\cos x=1$, find $\displaystyle\frac{dy}{dx}$.
訣竅
運用隱函數微分求解。
解法
將給定的方程對 $x$ 微分可得
$\displaystyle\cos y-x\sin y\cdot\frac{dy}{dx}+\frac{dy}{dx}\cdot\cos x-y\sin x=0$
如此有 $\displaystyle\frac{dy}{dx}=\frac{y\sin x-\cos y}{\cos x-x\sin y}$。


4. ($10\%$) Find the area of the region bounded by the two curves $y=x^3-6x^2+8x$ and $y=x^2-4x$.
訣竅
找出兩曲線的交點後運用積分表示面積並計算即可。
解法

先解方程式 $x^3-6x^2+8x=x^2-4x$，可得 $x^3-7x^2+12x=x\left(x-3\right)\left(x-4\right)$，如此所求的面積為

$\displaystyle\begin{aligned}A&=\int_0^3\left[\left(x^3-6x^2+8x\right)-\left(x^2-4x\right)\right]dx+\int_3^4\left[\left(x^2-4x\right)-\left(x^3-6x^2+8x\right)\right]dx\\&=\int_0^3\left(x^3-6x^2+12x\right)dx+\int_3^4\left(-x^3+7x^2-12x\right)dx\\&=\left.\left(\frac{x^4}4-\frac{7x^3}3+6x^2\right)\right|_0^3+\left.\left(-\frac{x^4}4+\frac{7x^3}3-6x^2\right)\right|_3^4=\frac{71}6\end{aligned}$



5. ($10\%$) Find $\displaystyle\int\sin3x\cos2xdx$.
訣竅
運用三角恆等式進行適當的改寫後即可積分。
解法

由和差角公式可知

$\begin{aligned}\sin5x&=\sin\left(3x+2x\right)=\sin3x\cos2x+\cos3x\sin2x\\\sin x&=\sin\left(3x-2x\right)=\sin3x\cos2x-\cos3x\sin2x\end{aligned}$

兩式相加除以二可知

$\displaystyle\int\sin3x\cos2xdx=\frac12\int\left(\sin5x+\sin x\right)dx=-\frac{5\cos x+\cos5x}{10}+C$



6. ($10\%$) Determine whether the infinite series

   $\displaystyle\sum_{n=1}^{+\infty}\frac1{\left(n^2+2\right)^{1/3}}$

   is convergent or divergent.
訣竅
運用極限比較審歛法。
解法
設 $\displaystyle a_n=\frac1{\left(n^2+2\right)^{1/3}}$ 與 $\displaystyle b_n=\frac1{n^{2/3}}$，那麼容易知道 $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=1$，故級數 $\displaystyle\sum_{n=1}^{\infty}a_n$ 與 $\displaystyle\sum_{n=1}^{\infty}b_n$ 有相同的歛散性。而後者為 $p$ 級數在 $\displaystyle p=\frac23<1$ 的情形，故發散，因此原給定的級數也發散。


7. ($10\%$) Determine the interval of convergence of the power series $\displaystyle\sum_{n=1}^{+\infty}n\left(x-2\right)^n$.
訣竅
運用比值審歛法確定收斂半徑，隨後檢驗端點的歛散性。
解法
由比值審歛法的概念有
$\displaystyle R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim_{n\to\infty}\frac{n}{n+1}=1$
因此收歛半徑為 $1$。而當 $x=2+1=3$ 或 $x=2-1=1$ 時，級數可分別寫為 $\displaystyle\sum_{n=1}^{\infty}n$ 或 $\displaystyle\sum_{n=1}^{\infty}\left(-1\right)^nn$，而其一般項皆不趨於零，故收斂區間為 $\left(1,3\right)$。