Candidates should attempt ALL questions.
The maximum marks for the various parts of the questions are indicated.
The paper will be marked out of 100.
This copy also contains solutions, written in blue text, with marks indicated in red.

Problem 1.

Consider the SDE
$$d X_{t}=\left(t+X_{t}\right) d t+2 t d B_{t} .$$
(a) Write this SDE in integral form, and show that $f(t)=\mathbb{E}\left[X_{t}\right]$ satisfies the differential equation
$$f^{\prime}(t)=t+f(t)$$
Show that this equation is satisfied by $f(t)=C e^{t}-t-1$.
(b) Let $Y_{t}=X_{t}^{2}$. Show that
$$d Y_{t}=2\left(2 t^{2}+t X_{t}+X_{t}^{2}\right) d t+4 t X_{t} d B_{t}$$
(c) Show that $v(t)=\mathbb{E}\left[X_{t}^{2}\right]$ satisfies the differential equation
$$v^{\prime}(t)=2\left(2 t^{2}+t f(t)+v(t)\right) .$$

Proof .

(a) Writing in integral form we have
$$X_{t}=X_{0}+\int_{0}^{t}\left(u+X_{u}\right) d u+\int_{0}^{2} 2 u d B_{u} .$$
[1] Taking expectation, and recalling that Ito integrals are zero mean martingales [1],
\begin{aligned} \mathbb{E}\left[X_{t}\right] &=\mathbb{E}\left[X_{0}\right]+\mathbb{E}\left[\int_{0}^{t}\left(u+X_{u}\right) d u\right]+\mathbb{E}\left[\int_{0}^{2} 2 u d B_{u}\right] \ &=\mathbb{E}\left[X_{0}\right]+\int_{0}^{t} \mathbb{E}\left[u+X_{u}\right] d u+0 \ &=\mathbb{E}\left[X_{0}\right]+\int_{0}^{t} u+\mathbb{E}\left[X_{u}\right] d u \ f(t) &=f(0)+\int_{0}^{t} u+f(u) d u \end{aligned}
[1] Differentiating, by the fundamental theorem of calculus, [1]
$$f^{\prime}(t)=t+f(t) .$$
If we set $f(t)=C e^{t}-t-1$ then $f^{\prime}(t)=C e^{t}-1[1]$, so clearly this is a solution.

(b) Using Ito’s formula [1] we have
\begin{aligned} d Y_{t} &=\left(0+\left(t+X_{t}\right)\left(2 X_{t}\right)+\frac{1}{2}(2 t)^{2}(2)\right) d t+(2 t)\left(2 X_{t}\right) d B_{t} \ &=2\left(2 t^{2}+t X_{t}+X_{t}^{2}\right) d t+4 t X_{t} d B_{t} \end{aligned}
$[3]$
(c) Writing in integral form we have
$$Y_{t}=Y_{0}+2 \int_{0}^{t} 2 u^{2}+u X_{u}+X_{u}^{2} d u+\int_{0}^{t} 4 u X_{u} d B_{u}$$
[1] Taking expectation, and recalling that Ito integrals are zero mean martingales [1],
\begin{aligned} \mathbb{E}\left[Y_{t}\right] &=\mathbb{E}\left[Y_{0}\right]+2 \mathbb{E}\left[\int_{0}^{t} 2 u^{2}+u X_{u}+X_{u}^{2} d u\right]+\mathbb{E}\left[\int_{0}^{t} 4 u X_{u} d B_{u}\right] \ &=\mathbb{E}\left[Y_{0}\right]+\int_{0}^{t} 2 \mathbb{E}\left[2 u^{2}+u X_{u}+X_{u}^{2}\right] d u+0 \ &=\mathbb{E}\left[Y_{0}\right]+2 \int_{0}^{t} 2 u^{2}+u \mathbb{E}\left[X_{u}\right]+\mathbb{E}\left[X_{u}^{2}\right] d u \ &=\mathbb{E}\left[Y_{0}\right]+2 \int_{0}^{t} 2 u^{2}+u f(u)+v(u) d u \end{aligned}
[1] Differentiating, by the fundamental theorem of calculus, [1]
$$v^{\prime}(t)=2\left(2 t^{2}+t f(t)+v(t)\right) .$$

Problem 2.

Let $B_{t}$ be a standard Brownian motion.
(a) Write down the distribution of $B_{t}$, and write down $\mathbb{E}\left[B_{t}\right]$ and $\mathbb{E}\left[B_{t}^{2}\right]$.
(b) Let $0 \leq u \leq t$. Show that $\mathbb{E}\left[\left(B_{t}-B_{u}\right)^{2} \mid \mathcal{F}_{u}\right]=t-u$.

Proof .

(a) $B_{t} \sim N(0, t),[1]$ and $\mathbb{E}\left[B_{t}\right]=0,[1] \mathbb{E}\left[B_{t}^{2}\right]=t .[1]$
(b) We have
\begin{aligned} \mathbb{E}\left[\left(B_{t}-B_{u}\right)^{2} \mid \mathcal{F}{u}\right] &=\mathbb{E}\left[\left(B{t}-B_{u}\right)^{2}\right] \ &=t-u . \end{aligned}
[1] In the first line we use that, by the properties of Brownian motion, $B_{t}-B_{u}$ is independent of $\mathcal{F}{u}$. [1] Then, we use that $B{t}-B_{u} \sim N(0, t-u)$, which is the same distribution as $B_{t-u}[1]$, followed by the third formula in part (a) with $t-u$ in place of
t. $[1]$

Problem 3.

Consider the binomial model with $r=\frac{1}{11}, d=0.9, u=1.2, s=100$ and time steps $t=0,1,2$.
(a) Draw a recombining tree of the stock price process, for time $t=0,1,2$.
(b) Find the value, at time $t=0$, of a European call option that gives its holder the option to purchase one unit of stock at time $t=2$ for a strike price $K=90$. Write down the hedging strategy that replicates the value of this contract, at all nodes of your tree.
You may annotate your tree from (a) to answer (b).

Proof .

Solution. As in the lecture notes, we write the value of a unit of stock (in blue) inside the nodes of the tree, to answer (a), and write the value of the contingent claim at the various nodes, in square boxes (in green), next to the nodes themselves; the answer to the first part of
(b) appears at the root node. For the second part of (b), the replicating portfolios $h=(x, y)$ that would be held at each node are written (in orange) as $x=\ldots, y=\ldots$

Problem 4.

Let $\left(X_{n}\right)$ be a sequence of i.i.d. random variables, each with a uniform distribution on $[-1,1]$. Define
$$S_{n}=\sum_{i=1}^{n} X_{i}$$
where $S_{0}=0 .$ Let $\mathcal{F}{n}=\sigma\left(X{1}, X_{2}, \ldots, X_{n}\right)$.
(a) Show that $S_{n}$ is a martingale, with respect to the filtration $\mathcal{F}{n}$. (b) Find $\mathbb{E}\left[S{3}^{2} \mid \mathcal{F}{2}\right]$ in terms of $X{2}$ and $X_{1}$, and hence show that
$$\mathbb{E}\left[S_{3}^{2} \mid \mathcal{F}{2}\right]=S{2}^{2}+\frac{1}{3}$$
(c) Write down a deterministic function $f: \mathbb{N} \rightarrow \mathbb{R}$ such that
$$M_{n}=S_{n}^{2}-f(n)$$
is a martingale (justification is not required – make a guess!).

.

Proof .

(a) Since $X_{i} \in \sigma\left(X_{i}\right)$ we have $X_{i} \in \sigma \mathcal{F}{n}$ for all $i \leq n[1] .$ Hence, since sums of $\mathcal{F}{n}$ measurable functions are measurable, we have also that $S_{n} \in \mathcal{F}{n}[1]$. Since $\left|X{i}\right| \leq 1$ for all $i$, we have
$$\left|S_{n}\right| \leq\left|X_{1}\right|+\left|X_{2}\right|+\ldots+\left|X_{n}\right| \leq n$$
Thus $S_{n}$ is a bounded random variable and hence $S_{n} \in L^{1}$. [1] Lastly,
\begin{aligned} \mathbb{E}\left[S_{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[X{n+1}+S_{n} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]+\mathbb{E}\left[S{n} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1}\right]+S_{n} \ &=S_{n} \end{aligned}
[1] Here, we use the linearity of conditional expectation to deduce the second line, followed by using that $X_{n+1}$ is independent of $\mathcal{F}{n}[1]$ and $S{n} \in \mathcal{F}{n}$ to deduce the third line [1]. The final line follows because $\mathbb{E}\left[X{i}\right]=0$ for all $i$. Hence $S_{n}$ is a martingale.

(b) We have
$$S_{n}^{3}=\left(X_{1}+X_{2}+X_{3}\right)^{2}=X_{1}^{2}+X_{2}^{2}+X_{3}^{2}+2 X_{1} X_{2}+2 X_{2} X_{3}+2 X_{1} X_{3}$$
[1] Hence,
\begin{aligned} \mathbb{E}\left[S_{3}^{2} \mid \mathcal{F}{2}\right]=& \mathbb{E}\left[X{1}^{2} \mid \mathcal{F}{n}\right]+\mathbb{E}\left[X{2}^{2} \mid \mathcal{F}{2}\right]+\mathbb{E}\left[X{3}^{2} \mid \mathcal{F}{2}\right] \ & \quad+2 \mathbb{E}\left[X{1} X_{2} \mid \mathcal{F}{2}\right]+2 \mathbb{E}\left[X{2} X_{3} \mid \mathcal{F}{2}\right]+2 \mathbb{E}\left[X{1} X_{3} \mid \mathcal{F}{2}\right] \ =& X{1}^{2}+X_{2}^{2}+\mathbb{E}\left[X_{3}^{2}\right]+2 X_{1} X_{2}+2 X_{2} \mathbb{E}\left[X_{3}\right]+2 X_{1} \mathbb{E}\left[X_{3}\right] \ =&\left(X_{1}+X_{2}\right)^{2}+\frac{1}{3} \ =& S_{2}^{2}+\frac{1}{3} \end{aligned}
[1]. Here, in the first line we use linearity of conditional expectation. To deduce the second line we use that $X_{3}$ is independent of $\mathcal{F}{2}[1]$, and that $X{1}, X_{2} \in m \mathcal{F}{2}$ to ‘take out what is known'[1]. We then use that $$\mathbb{E}\left[X{3}^{2}\right]=\int_{-1}^{1} x^{2} \frac{1}{2} d x=\frac{1}{3}$$
to deduce the final lines $[1]$. Pitfall: $X_{n}$ has the continuous uniform distribution on the interval $[-1,1]$.
(c) In view of (b), we take $f(n)=\frac{n}{3}$, so that $M_{n}=S_{n}-\frac{n}{3}[2]$.
To make this guess: note from (b) that $\mathbb{E}\left[S_{n}^{2}\right]$ drifts upwards by $\frac{1}{3}$ on each time step, so $\mathbb{E}\left[S_{n}^{2}-\frac{n}{3}\right]$ stays constant. This is the only way to compensate for the drift in the form $S_{n}-f(n)$, and (possibly) obtain a martingale. To see that $M_{n}$ really is a martingale: Since $S_{n} \in \mathcal{F}{n}$ we have $M{n} \in \mathcal{F}{n}$, and $\left|M{n}\right| \leq$ $\left|S_{n}^{2}\right|+\frac{2 n}{3} \leq n^{2}+\frac{n}{3}$ so $M_{n} \in L^{1} . \quad A$ similar calculation to (b) then shows that $\mathbb{E}\left[S_{n+1}^{2} \mid \mathcal{F}{n}\right]=S{n}^{2}+\frac{1}{3}$, hence $\mathbb{E}\left[M_{n+1} \mid \mathcal{F}{n}\right]=M{n}$

real analysis代写analysis 2, analysis 3请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。