(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]
$$
v^{\prime}(t)=2\left(2 t^{2}+t f(t)+v(t)\right) .
$$

(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. $[1]$

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 $x=\ldots, y=\ldots$

(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 $\mathbb{E}\left[M_{n+1} \mid \mathcal{F}{n}\right]=M{n}$