<jupyter_start><jupyter_text>**Review****程远星**$\DeclareMathOperator*{\argmin}{argmin}\DeclareMathOperator*{\argmax}{argmax}\DeclareMathOperator*{\plim}{plim}\newcommand{\using}[1]{\stackrel{\mathrm{1}}{=}}\newcommand{\ffrac}{\displaystyle \frac}\newcommand{\asim}{\overset{\text{a}}{\sim}}\newcommand{\space}{\text{ }}\newcommand{\bspace}{\;\;\;\;}\newcommand{\QQQ}{\boxed{?\:}}\newcommand{\void}{\left.\right.}\newcommand{\Tran}[1]{{1}^{\mathrm{T}}}\newcommand{\d}[1]{\displaystyle{1}}\newcommand{\CB}[1]{\left\{ 1 \right\}}\newcommand{\SB}[1]{\left[ 1 \right]}\newcommand{\P}[1]{\left( 1 \right)}\newcommand{\abs}[1]{\left| 1 \right|}\newcommand{\norm}[1]{\left\| 1 \right\|}\newcommand{\dd}{\mathrm{d}}\newcommand{\Exp}{\mathrm{E}}\newcommand{\RR}{\mathbb{R}}\newcommand{\EE}{\mathbb{E}}\newcommand{\II}{\mathbb{I}}\newcommand{\NN}{\mathbb{N}}\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\QQ}{\mathbb{Q}}\newcommand{\PP}{\mathbb{P}}\newcommand{\AcA}{\mathcal{A}}\newcommand{\FcF}{\mathcal{F}}\newcommand{\AsA}{\mathscr{A}}\newcommand{\FsF}{\mathscr{F}}\newcommand{\Var}[2][\,\!]{\mathrm{Var}_{1}\left[2\right]}\newcommand{\Avar}[2][\,\!]{\mathrm{Avar}_{1}\left[2\right]}\newcommand{\Cov}[2][\,\!]{\mathrm{Cov}_{1}\left(2\right)}\newcommand{\Corr}[2][\,\!]{\mathrm{Corr}_{1}\left(2\right)}\newcommand{\I}[1]{\mathrm{I}\left( 1 \right)}\newcommand{\N}[1]{\mathcal{N} \left( 1 \right)}\newcommand{\ow}{\text{otherwise}}\newcommand{\FSD}{\text{FSD}}\void^\dagger$  Question 1Let $S_t$ denote the stock price at time $t$. Under the physical (objective) measure $P$, the stock price $S_t$  follows the Black-Schole model given by:$$\dd S_t = \mu S_t \dd t + \sigma S_t \dd W_t$$A financial institution plans to offer a security that pays off a dollar amount equal to $S^2_T$ at time $T$.$\P{1}$ Use the risk-neutral valuation method to derive the pricing formula of the security at time $t$ (in terms of the expectation).>Under risk neutral world this process is a martingale, which could help us find the risk neutral measure (this is a time-continuous world). Under $\PP$ measure we have:>>$$\dd \tilde S_t = \P{\mu - r} \tilde S_t \dd t + \sigma \tilde S_t \dd W_t$$>>Then transfer to $\QQ$ measure we find the standard brownian motion $\widetilde W_t$ under $\QQ$, so that the drift is $0$, which leads to $\dd \tilde S_t = \sigma \tilde S_t \dd \widetilde W_t$.>>Balence the two equations we have>>$$ \dd \widetilde W_t = \dd W_t + \ffrac{\mu - r}{\sigma}\dd t $$>>Then using the Girsanov's Theorem, we have>>$$\left.\ffrac{\dd \QQ}{\dd \PP}\right|_{\FcF_t} = \exp\CB{-\ffrac{\mu - r}{\sigma}W_t - \ffrac{1}{2} \P{\ffrac{\mu - r}{\sigma}}^2t }$$>>With $\QQ$, we now can write>>$$V_t = \Exp^\QQ\SB{e^{-r\P{T-t}} S_T^2\mid \FcF_t} = e^{-r\P{T-t}}\Exp^{\QQ}_t\SB{S_T^2}$$$\P{2}$ Under the risk-neutral measure, according to the corresponding stochastic differential equation for $S_t$, solve $S_T$ in terms of $S_t$.>To solve the differential equation under risk-neutral measure, we first write the equation and separate the variables by taking the $\log$ function:>>$$ \dd \tilde S_t = \sigma \tilde S_t \dd \widetilde W_t \Longrightarrow \dd S_t = rS_t\dd t + \sigma S_t \dd \widetilde W_t \Longrightarrow\dd\log S_t \using{ \text{Itô's formula} } \P{r-\ffrac{\sigma^2}{2}}\dd t + \sigma \dd \widetilde W_t$$>>Take the integral from $t$ to $T$, we have>>$$S_T = S_t \cdot\exp\CB{\P{r-\ffrac{\sigma^2}{2}}\P{T-t} +\sigma \P{\widetilde W_T-\widetilde W_t}}$$$\P{3}$ Assume that the formula $\Exp^\PP \SB{\exp\CB{\alpha W_t}} = \exp\CB{\ffrac{\alpha^2}{2}t}$ is known and a similar result holds under the risk-neutral measure. According to the derived pricing formula in $\P{1}$ by making use of $\P 2$, calculate the price of the security at time $t$.> $$\begin{align}V_t &= e^{-r\P{T-t}}\Exp^{\QQ}_t\SB{S_T^2} \\&= e^{-r\P{T-t}}\Exp^{\QQ}_t\SB{\P{S_t \cdot\exp\CB{\P{r-\ffrac{\sigma^2}{2}}\P{T-t} +\sigma \P{\widetilde W_T-\widetilde W_t}}}^2}\\&= e^{-r\P{T-t}}\Exp^{\QQ}_t\SB{S_t^2} \cdot \exp\CB{\P{2r - \sigma^2}\P{T-t}} \cdot \Exp_t^\QQ \SB{\exp\CB{2\sigma \P{\widetilde W_T-\widetilde W_t}}}\\&= \exp\CB{\P{-r + 2r - \sigma^2 - \ffrac{4\sigma^2}{2}}\P{T-t}}\cdot S_t^2\\&= \exp\CB{\P{r+\sigma^2}\P{T-t}}\cdot S_t^2\end{align}$$$\P{4}$ Confirm that your price satisfies the differential equation.$$\ffrac{\partial f}{\partial t} + rS_t\ffrac{\partial f}{\partial S_t} + \ffrac{1}{2}\sigma^2S_t^2\ffrac{\partial^2 f}{\partial S_t^2} = rf$$>It's easy to see the result by taking the derivatives term by term, or using the Feynman-Kac formula. First we let $f\P{t} = V_t$>>The second equation is satisfied since the payoff is $\Phi\P{X} = S_T^2 = f\P{T}$. Then the first equation, since $\mu_t = rS_t$, $\sigma_t = \sigma S_t$, it can't be more obvious.  Question 2see HW_04_Question_5 for complete solution. However here we present another method to solve for $h_t$. Still we have the BS equation $\ffrac{\partial c}{\partial t} + \ffrac{1}{2}\sigma^2S_t^2\ffrac{\partial^2 c}{\partial S_t^2} + rS_t\ffrac{\partial c}{\partial S_t} - rc = 0$, then substituting $V = h\P{t,T}S^n$, we have the differential equation$$\ffrac{\partial h\P{t,T}}{\partial t} + \ffrac{h\P{t,T}}{2} \sigma^2 \cdot n\P{n-1} + rh\P{t,T} \cdot n-rh\P{t,T} = 0$$Or, like this: $\ffrac{h_t}{h} = -r\P{n-1} - \ffrac{1}{2}\sigma^2 n\P{n-1}$. The solution to this is$$\ln h = \P{-r\P{n-1} - \ffrac{1}{2}\sigma^2 n\P{n-1}} t + k$$since when $t = T$, $\ln h = 0$ it follows that $k = \P{r\P{n-1} + \ffrac{1}{2}\sigma^2 n\P{n-1}}T$ thus$$\ln h = \P{r\P{n-1} + \ffrac{1}{2}\sigma^2 n\P{n-1}}\P{T-t}$$  Question 4Consider the three-step binomial tree and set $T = 3$, $S_0 = 80$, $u = 1.5$, $d=0.5$, $p_u = 0.6$, $p_d = 0.4$, and for computational simplicity, $r=0$. We consider a European call option on the underlying stock. The date of expiration of the option is $T=3$, and the strike price is $K=80$. Suppose that at time $t = 1$ the stock price has gone up to $120$, and that the market price of the option turns out to be $50.0$. Show explicitly how you can make an arbitrage profit.>Ignore $p_u$ and $p_d$ and things will be easy.  Question 5Consider a two-period Cox-Ross-Rubinstein model with $S_0 = 100$, $u = 1.01$, $d = 1/u$, and $r = 0.005$. Find a replicating portfolio in this model for the put option $g\P{s} = \P{100-s}^+$>There's no authentic answer for this. So the formulas are presented here>>$$q = \ffrac{e^{rT}-1/u}{u-1/u} = \ffrac{ue^{rT}-1}{u^2-1}$$  Question 6Prove directly from the definition of Itô integrals, that$$\int_0^t s\;\dd W_s = tW_t - \int_0^t W_s\;\dd s$$>Split the interval $\SB{0,t}$ into any $n$ subintervals $[t_j,t_{j+1})$, then we obtain>>$$\begin{align}\sum_{j=0}^{n-1} t_j\P{W_{t_{\void_{j+1}}} - W_{t_{\void_j}}}&= \sum_{j=0}^{n-1} t_j W_{t_{\void_{j+1}}} - \sum_{j=0}^{n-1} t_j W_{t_{\void_{j}}}\\&= \sum_{j=0}^{n-1} \P{t_j-t_{j+1}+t_{j+1}} W_{t_{\void_{j+1}}} - \sum_{j=0}^{n-1} t_j W_{t_{\void_{j}}}\\&= \sum_{j=0}^{n-1} \P{t_j-t_{j+1}}W_{t_{\void_{j+1}}} + \sum_{j=0}^{n-1} \P{t_{j+1} W_{t_{\void_{j+1}}} - t_j W_{t_{\void_{j}}}}\\&= -\sum_{j=0}^{n-1} \P{t_{j+1}-t_j}W_{t_{\void_{j+1}}} + tW_t - 0W_0\end{align}$$>>Let the *mesh* of this partition tend to zero, we have>>$$\int_0^t s\;\dd W_s = tW_t - \int_0^t W_s\;\dd s$$  Question 7Use Itô's Formula to show that $\d{\int_0^t s\;\dd W_s = tW_t - \int_0^t W_s\;\dd s}$>An application of Itô's Formula to $tW_t$ gives>>$$\dd \P{tW_t}= \P{W_t + 0 + 0}\dd t + t \dd W_t\Rightarrow tW_t = \int_0^t s\;\dd W_s + \int_0^t W_s\;\dd s$$  Question 8Suppose that $X$ has the stochastic differential $\dd X\P t = \alpha X\P t\dd t + \sigma\dd W\P t$ where $X\P0 = x$ and $\alpha$ is a real number, compute $\Exp\SB{X\P t}$>Denote $Y_t = X_t e^{-\alpha t}$ trying to eliminate the alpha in the SDE:>>$$\begin{align}\dd Y\P t &= \P{-\alpha\cdot e^{-\alpha t}X\P t + e^{-\alpha t}\cdot \alpha X\P t+0}\dd t + \sigma \cdot e^{-\alpha t}\dd W_t\\&= \sigma \cdot e^{-\alpha t}\dd W_t\\\end{align}$$>>Integral both sides of the equation above, with $Y_0 = x\cdot e^0 = x$, we have>>$$Y\P t=x + \sigma \int_0^t e^{-\alpha s}\;\dd W_s \Rightarrow X_t = xe^{\alpha t} + e^{\alpha t}\sigma \int_0^t e^{-\alpha s}\;\dd W_s$$>>Thus, $\Exp\SB{X\P t} = xe^{\alpha t}$ Question 8.extendedNow what if $\dd X\P t = \alpha X\P t\dd t + \sigma X\P t\dd W\P t$ with same conditions?>Easy to find that using the same method we have this is still $xe^{\alpha t}$, and here's another method, by directly solve the SDE: given the equation we have>>$$\dd \log X\P t = \P{\alpha - \ffrac{\sigma^2}{2}}\dd t + \sigma \dd W_t$$>>Take the integral, we have>>$$\log\ffrac{X\P t}{X\P0} = \P{\alpha - \ffrac{\sigma^2}{2}}t + \sigma W_t\Rightarrow X\P t = x\cdot \exp\CB{\P{\alpha - \ffrac{\sigma^2}{2}}t + \sigma W_t}$$>>$$\Exp\SB{X\P t} = x \cdot \exp\CB{\P{\alpha - \ffrac{\sigma^2}{2}}t}\cdot\exp\CB{\ffrac{\sigma^2}{2}t} = xe^{\alpha t}$$  Question 9Let $X$ and $Y$ be given as the solutions to the following system of stochastic differential equations$$\begin{align}\dd X\P{t} &= \alpha X_{t} \dd t- Y\P{t} \dd W\P{t}, X\P{0} = x_0\\\dd Y\P{t} &= \alpha Y_{t} \dd t- X\P{t} \dd W\P{t}, Y\P{0} = y_0\end{align}$$Prove that the process $R$ defined by $R\P{t} = X^2 \P{t} + Y^2 \P{t}$ is deterministic.>$$\begin{align}\dd X^2\P{t} &= \P{0+\alpha X\P{t} \cdot 2X\P{t} + \ffrac{1}{2}Y^2\P{t}\cdot 2}\dd t-Y\P{t}\cdot 2X\P{t}\dd W_t\\&= 2\alpha X^2\P{t} \dd t + Y^2\P{t} \dd t - 2X\P{t}Y\P{t} \dd W_t\\\dd Y^2\P{t} &= \P{0+\alpha Y\P{t} \cdot 2X\P{t} + \ffrac{1}{2}X^2\P{t}\cdot 2}\dd t+X\P{t}\cdot 2Y\P{t}\dd W_t\\&= 2\alpha Y^2\P{t} \dd t + X^2\P{t} \dd t + 2X\P{t}Y\P{t} \dd W_t\\\dd R &= \P{2\alpha + 1 }\P{X^2 \P{t} + Y^2 \P{t}}\dd t = \P{2\alpha + 1}R\dd t\end{align}$$  Question 10Suppose that the stock price today is $S_t = 2.00$, interest rate $r = 0$, and the time to maturity is $3$ months. Consider a contingent claim whose (Black-Scholes) price is given by the function$$V\P{t,S_t} = S_t^2 e^{2\P{T-t}}$$where the time is in annual terms.$\P{\text a}$ What is the claim price today?>$$V\P{t,S_t} = S_t^2 e^{2\P{T-t}} = 2^2 e^{2\times3/12} = 6.5949$$$\P{\text b}$ Find the volatility $\sigma$>We can use the same way to derive the Black-Scholes-Merton equation, and plug $V$ in we have>>$$\begin{align}0 &= \ffrac{\partial V}{\partial t} + \ffrac{1}{2}\sigma^2S_t^2\ffrac{\partial^2 V}{\partial S_t^2} + rS_t\ffrac{\partial V}{\partial S_t} - rV\\&= -2S_t^2 e^{2\P{T-t}} + \ffrac{1}{2}\sigma^2 S_t^2 \times 2e^{2\P{T-t}} + 0 - 0\\\sigma^2 &= 2 \Rightarrow \sigma = \sqrt 2\end{align}$$>>Or using risk neutral valuation, since the terminal payoff is $V\P{T,S_T} = S_T^2$, we have>>$$\begin{align}V\P{t} &= \Exp_t^\QQ\SB{S_T^2 e^{-r\P{T-t}}}\\&= \Exp_t^\QQ\SB{S_T^2}\\&= \Exp_t^\QQ\SB{ S_t^2 \exp\CB{2\P{r-\ffrac{\sigma^2}{2}}\P{T-t} + 2\sigma\P{W_T-W_t}} }\\&= S_t^2 e^{-\sigma^2\P{T-t}}\Exp^\QQ_t\SB{e^{2\sigma\P{W_T-W_t}}}\\&= S_t^2 e^{-\sigma^2\P{T-t}}\exp\CB{\ffrac{\P{2\sigma}^2}{2}\P{T-t}}\\&= S_t^2 e^{\sigma^2\P{T-t}}\end{align}$$>>Thus, $\sigma^2 = 2$$\P{\text c}$ If the stock at maturity is $S(T) = 2.00$, what is the payoff of the claim at maturity?>$$V\P{T,S_T} = 2^2 e^0 = 4$$ ***<jupyter_code>import math
4*math.exp(0.5)<jupyter_output><empty_output>
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