國立臺灣大學一百零八學年度研究所碩士班入學考試試題：微積分(C)

※ Please write down the detailed calculation process for all questions.

1. ($20\%$) The Black-Scholes formula for a call option with six input parameters $\left(S,X,r,q,\sigma.T\right)$ is as follows.

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

   where

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

   and $N\left(\cdot\right)$ si 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. ($10\%$) Derive and express $\displaystyle\frac{\partial c}{\partial X}$ as the form of $AN\left(B\right)$. What are $A$ and $B$?
   2. ($10\%$) Derive and express $\displaystyle\frac{\partial c}{\partial T}$ as the form of $Cn\left(D\right)+EN\left(D\right)+FN\left(G\right)$. What are $C$, $D$, $E$, $F$, and $G$?
訣竅
運用多變數函數的連鎖律計算即可。
解法
1. 對 $X$ 求導便有

   $\displaystyle\frac{\partial c}{\partial X}=Se^{-qT}n\left(d_1\right)\cdot-\frac1{X\sigma\sqrt{T}}-e^{-rt}N\left(d_2\right)-Xe^{-rT}n\left(d_2\right)\cdot-\frac1{X\sigma\sqrt{T}}$

   其中觀察到

   $\displaystyle d_1+d_2=\frac{2\left[\ln\left(S/X\right)+\left(r-q\right)T\right]}{\sigma\sqrt{T}}=\frac{2\left[\ln\left(S/X\right)+\left(r-q\right)T\right]}{d_1-d_2}$

   故 $\displaystyle\frac{d_1^2-d_2^2}2=\ln\left(S/X\right)+\left(r-q\right)T$，從而有 $\displaystyle e^{\frac{d_1^2-d_2^2}2}=\frac{S}Xe^{\left(r-q\right)T}$，即得 $Se^{-qT}n\left(d_1\right)=Xe^{-rT}n\left(d_2\right)$。由此可得 $\displaystyle\frac{\partial c}{\partial X}=-e^{-rT}N\left(d_2\right)$，因此 $A=-e^{-rT}$、$B=d_2$。
2. 對 $T$ 求導便有

   $\displaystyle\begin{aligned}\frac{\partial c}{\partial T}=&-Sqe^{-qT}N\left(d_1\right)+Se^{-qT}n\left(d_1\right)\left[-\frac{\ln\left(S/X\right)}{2\sigma T\sqrt{T}}+\frac{r-q+\sigma^2/2}{2\sigma\sqrt{T}}\right]\\&+rXe^{-rT}N\left(d_2\right)-Xe^{-rT}n\left(d_2\right)\cdot\left[-\frac{\ln\left(S/X\right)}{2\sigma T\sqrt{T}}+\frac{r-q-\sigma^2/2}{2\sigma\sqrt{T}}\right]\end{aligned}$

   利用 $Se^{-qT}n\left(d_1\right)=Xe^{-rT}n\left(d_2\right)$，如此有

   $\displaystyle\frac{\partial c}{\partial T}=\frac{S\sigma e^{-qT}}{2\sqrt{T}}n\left(d_1\right)-Sqe^{-qT}N\left(d_1\right)+rXe^{-rT}N\left(d_2\right)$

   取 $\displaystyle C=\frac{S\sigma e^{-qT}}{2\sqrt{T}}$、$D=d_1$、$E=-Sqe^{-qT}$、$F=rXe^{-rT}$、$G=d_2$。


2. ($10\%$) Consider that $X$ follows a zero-mean normal distribution, i.e., $X\sim ND\left(0,\sigma^2\right)$. Evaluate the expectation $E\left[Xe^{cX}\right]$, where $c$ is a constant real number. (Hint: The probability density function for $ND\left(\mu,\sigma^2\right)=\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}}$.)
訣竅
按定義演算之。
解法

由定義可知

$\displaystyle E\left[Xe^{cX}\right]=\int_{-\infty}^{\infty}xe^{cx}\cdot\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx=\frac{e^{c^2/2}}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}x\exp\left(-\frac{\left(x-c\sigma^2\right)^2}{2\sigma^2}\right)dx$

令 $\displaystyle t=\frac{x-c\sigma^2}{\sqrt{2\sigma}}$，那麼 $dx=\sqrt{2\sigma}dt$，如此所求可改寫並計算如下

$\displaystyle E\left[Xe^{cX}\right]=\frac{e^{c^2\sigma^2/2}}{\sqrt{\pi}}\int_{-\infty}^{\infty}\left(c+t\sqrt{2\sigma}\right)e^{-t^2}dt=ce^{c^2\sigma^2/2}+\frac{\sqrt{2\sigma}e^{c^2/2}}{\sqrt{\pi}}\int_{-\infty}^{\infty}te^{-t^2}dt=ce^{c^2\sigma^2/2}$



3. ($20\%$) Consider $\displaystyle J=\sum_{k=1}^NY$ to be a compound Poisson distribution, where $N$ follows a Poisson distribution with the expected number of event occurrences to be $\lambda$, and for each event occurrence, one receives a payoff $Y$, which follows an independent and identical normal distribution, i.e., $Y\sim ND\left(\mu,\sigma^2\right)$. Furthermore, assume $N$ and $Y$ are independent.
   1. ($10\%$) Evaluate $\displaystyle E\left[\prod_{k=1}^NY\right]$.
   2. ($10\%$) Evaluate $\displaystyle E\left[N\prod_{k=1}^NY\right]$.
   (Hint: For $N$, the probability of observing $k$ events is given by $\displaystyle P\left(k~\text{events ocurring}\right)=\frac{e^{-\lambda}\lambda^k}{k!}$.)
訣竅
利用隨機變數的獨立性從而計算之，其中應用條件期望值表達之。
解法

1. 直接列式求解

   $\displaystyle E\left[\prod_{k=1}^{N}Y\right]=\sum_{N=0}^{\infty}P\left(N\right)\mu^N=\sum_{N=0}^{\infty}\frac{e^{-\lambda}\lambda^N}{N!}\mu^N=e^{\lambda\mu-\lambda}\sum_{N=0}^{\infty}\frac{e^{-\lambda\mu}\left(\lambda N\right)^N}{N!}=e^{\lambda\left(\mu-1\right)}$

2. 直接列式求解

   $\displaystyle E\left[N\prod_{k=1}^{N}Y\right]=\sum_{N=0}^{\infty}P\left(N\right)N\mu^N=\sum_{N=1}^{\infty}\frac{e^{-\lambda}\lambda^N}{\left(N-1\right)!}\mu^N=\lambda\mu e^{\lambda\mu-\lambda}\sum_{N=1}^{\infty}\frac{e^{-\lambda\mu}\left(\lambda\mu\right)^{N-1}}{\left(N-1\right)!}=\lambda\mu e^{\lambda\left(\mu-1\right)}$



4. ($10\%$) Evaluate the integral $\displaystyle\int\frac{\sin x}{1+\sin x}dx$.
訣竅
運用半角代換求解即可。
解法

令 $\displaystyle t=\tan\frac{x}2$，那麼有 $x=2\tan^{-1}t$ 且有 $\displaystyle\sin x=\frac{2t}{1+t^2}$ 以及 $\displaystyle dx=\frac{2dt}{1+t^2}$。據此所求的不定積分可改寫並整理如下

$\displaystyle\int\frac{\sin x}{1+\sin x}dx=\int\frac{\displaystyle\frac{2t}{1+t^2}}{\displaystyle1+\frac{2t}{1+t^2}}\cdot\frac{2dt}{1+t^2}=\int\frac{4t}{\left(1+t\right)^2\left(1+t^2\right)}dt$

運用部分分式法改寫並計算如下

$\displaystyle\int\frac{4t}{\left(1+t\right)^2\left(1+t^2\right)}dt=\int\left(\frac2{1+t^2}-\frac2{\left(1+t\right)^2}\right)dt=2\tan^{-1}t+\frac2{1+t}+C=x+\frac2{1+\tan\left(x/2\right)}+C$



5. ($10\%$) Perform the integration $\displaystyle\int\frac{dx}{\sqrt{x}-2}$.
訣竅
運用變數代換法求解即可。
解法

令 $t=\sqrt{x}-2$，那麼 $x=\left(t+2\right)^2$，故 $dx=\left(2t+4\right)dt$。從而給定的不定積分可改寫並計算如下

$\displaystyle\int\frac{dx}{\sqrt{x}-2}=\int\frac{2t+4}{t}dt=\int\left(2+\frac4t\right)dt=2t+4\ln\left|t\right|+C==2\sqrt{x}-4+4\ln\left|\sqrt{x}-2\right|+C$



6. ($10\%$) Find $dy/dx$ for the equation $e^{xy}-xy=0$.
訣竅
運用隱函數微分求導。
解法
對 $x$ 求導便有
$\displaystyle e^{xy}\left(y+x\frac{dy}{dx}\right)-\left(y+x\frac{dy}{dx}\right)=0$
整理便有 $\displaystyle\frac{dy}{dx}=-\frac{y}x$。


7. ($10\%$) Find the derivative of $y$ with respect to $x$ in the function $\displaystyle y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$.
訣竅
使用基本的導函數與除法的微分公式即可。
解法

直接使用公式求導即有

$\displaystyle\frac{dy}{dx}=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}=4\left(e^x+e^{-x}\right)^{-2}$



8. ($10\%$) Solve the differential equation $\left(x^2+1\right)dx+\cos y\,dy=0$.
訣竅
運用分離變量法的概念解微分方程。
解法

直接積分便有

$\frac{x^3}3+x+\sin y=C$