(a) Using the triangle inequality, and monotonicity of $\mathbb{E},[1]$
$$
\mathbb{E}\left[\left|S_{n}\right|\right]=\mathbb{E}\left[\left|\sum_{i=1}^{n} X_{i}\right|\right] \leq \mathbb{E}\left[\sum_{i=1}^{n}\left|X_{i}\right|\right]=\mathbb{E}\left[\sum_{i=1}^{n} \frac{1}{i^{2}}\right]=\sum_{i=1}^{n} \frac{1}{i^{2}} \text { . }
$$(b) From part (a) we have $\mathbb{E}\left[\left|S_{n}\right|\right] \leq \sum_{i=1}^{\infty} \frac{1}{i^{2}}<\infty$, which is finite and independent of $n .[1]$ Hence $\sup {n} \mathbb{E}\left[\left|S{n}\right|\right]<\infty .[1]$
(c) We aim to use the martingale convergence theorem. $[1]$ We must check that $\left(S_{n}\right)$ is a martingale. We use the filtration $\mathcal{F}{n}=\sigma\left(X{i}: i=1, \ldots, n\right)$. Since $X_{i} \in m \mathcal{F}{n}$, we have $S{n} \in$ $m \mathcal{F}{n} \cdot[1]$ We have already shown in (a) that $\mathbb{E}\left[\left|S{n}\right|\right]<\infty$, so $S_{n} \in L^{1}$. [1] Lastly,
$$
\begin{aligned}
\mathbb{E}\left[S_{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[X{n+1}+S_{n} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1}\right]+S_{n} \
&=S_{n}
\end{aligned}
$$
[1] Here we use that $S_{n} \in m \mathcal{F}{n},[1]$ that $X{n+1}$ is independent of $\mathcal{F}{n},[1]$ and that $\mathbb{E}\left[X{n+1}\right]=0$
(d) We have
$$
\begin{aligned}
\mathbb{E}\left[\left|S_{n}\right|^{2}\right] &=\mathbb{E}\left[S_{n}^{2}\right] \
&=\mathbb{E}\left[\sum_{i=1}^{n} \sum_{j=1}^{n} X_{i} X_{j}\right] \
&=\sum_{i=1}^{n} \sum_{j=1}^{n} \mathbb{E}\left[X_{i} X_{j}\right] \
&=\sum_{i=1}^{n}\left(\mathbb{E}\left[X_{i}^{2}\right]+\sum_{j=1 \atop j \neq i}^{n} \mathbb{E}\left[X_{i} X_{j}\right]\right.\
&=\sum_{i=1}^{n}\left(\frac{1}{i^{4}}+\sum_{j=1 \atop j \neq i}^{n} 0\right) \
&=\sum_{i=1}^{n} \frac{1}{i^{4}} .
\end{aligned}
$$

(a)
$$
\begin{aligned}
d Z_{t} &=\left{\left(B_{t}\right)+(0)(t)+\frac{1}{2}(1)^{2}(1)\right} d t+(t)(1) d B_{t} \
&=B_{t} d t+t d B_{t}
\end{aligned}
$$(b)
$$
\begin{aligned}
d Z_{t} &=\left{2 t X_{t}+(\mu)\left(t^{2}\right)+\frac{1}{2}\left(\sigma B_{t}\right)^{2}(0)\right} d t+\left(\sigma B_{t}\right)\left(t^{2}\right) d B_{t} \
&=\left(2 t X_{t}+\mu t^{2}\right) d t+\sigma t^{2} B_{t} d B_{t} .
\end{aligned}
$$(c)
$$
\begin{aligned}
d Z_{t} &=\left{\left(-2 e^{-2 t} S_{t}\right)+\left(2 S_{t}\right)\left(e^{-2 t}\right)+\frac{1}{2}\left(5 S_{t}\right)^{2}(0)\right} d t+\left(5 S_{t}\right)\left(e^{-2 t}\right) d B_{t} \
&=5 e^{-2 t} S_{t} d B_{t}
\end{aligned}
$$

Case (c) is a martingale because here $d Z_{t}$ has only a $(\ldots) d B_{t}$ term, and therefore $Z_{t}=$ $Z_{0}+\int_{0}^{t} \ldots d B_{t}$ is a martingale because Ito integrals are martingales.

(a) i. Using the explicit formula for geometric Brownian motion (see the formula sheet) we obtain
$$
\begin{aligned}
e^{-r(T-t)} \mathbb{E}^{\mathbb{Q}}\left[3 S_{T}+5 \mid \mathcal{F}{t}\right] &=e^{-r(T-t)} \mathbb{E}^{\mathbb{Q}}\left[3 S{t} e^{\left(r-\frac{1}{2} \sigma^{2}\right)(T-t)+\sigma\left(B_{T}-B_{t}\right)}+5 \mid \mathcal{F}{t}\right] \ &=e^{-r(T-t)}\left(3 S{t} e^{\left(r-\frac{1}{2} \sigma^{2}\right)(T-t)} \mathbb{E}^{\mathbb{Q}}\left[e^{\sigma\left(B_{T}-B_{t}\right)} \mid \mathcal{F}{t}\right]+5\right) \ &=e^{-r(T-t)}\left(3 S{t} e^{\left(r-\frac{1}{2} \sigma^{2}\right)(T-t)} \mathbb{E}^{Q}\left[e^{\sigma\left(B_{T}-B_{t}\right)}\right]+5\right) \
&=e^{-r(T-t)}\left(3 S_{t} e^{\left(r-\frac{1}{2} \sigma^{2}\right)(T-t)+\frac{1}{2} \sigma^{2}(T-t)}+5\right) \
&=e^{-r(T-t)}\left(3 S_{t} e^{r(T-t)}+5\right) \
&=3 S_{t}+5 e^{-r(T-t)}
\end{aligned}
$$
$[4]$ Here, we use that $S_{t}$ is $\mathcal{F}{t}$ measurable, $[1]$ and that $Z=\sigma\left(B{T}-B_{t}\right) \sim N\left(0, \sigma^{2}(T-\right.$
$t)$ is independent of $\mathcal{F}{t}$. [1] We use the formula sheet to provide an explicit formula for $\mathbb{E}\left[e^{Z}\right]$ ii. Assuming $1 \leq t \leq T$, we have $$ \begin{aligned} e^{-r(T-t)} \mathbb{E}^{\mathbb{Q}}\left[S{1} S_{T}+1 \mid \mathcal{F}{t}\right] &=e^{-r(T-t)}\left(S{1} \mathbb{E}^{\mathbb{Q}}\left[S_{T} \mathcal{F}{t}\right]+1\right) \ &=S{1} e^{r t} e^{-r T} \mathbb{E}^{Q}\left[S_{T} \mathcal{F}{t}\right]+e^{-r(T-t)} \ &=S{1} e^{r t} e^{-r t} S_{t}+e^{-r(T-t)} \
&=S_{1} S_{t}+e^{-r(T-t)}
\end{aligned}
$$
[2] Here we use that $S_{1} \in \mathcal{F}{t}$ for $t \geq 1,[1]$ and the fact (from Lemma 14.4.1 in lectures) that $M{t}=e^{-r t} S_{t}$ is a martingale in the risk-neutral world. $[1]$
(b) i. At time 0, we buy three units of stock [1] and $5 e^{-r T}$ in cash. [1] It’s value at time is then $3 S_{t}+5 e^{-r T} e^{r t}=\Phi\left(S_{T}\right)$
Therefore, this portfolio replicates $\Phi\left(S_{T}\right)$ for all $t \in[0, T] .[1]$
ii. It isn’t possible to replicate $\Psi\left(S_{T}\right)$ with a constant portfolio. [1] The replicating portfolio provided by Theorem $14.3 .1$ is unique, and contains a stock component $y_{t}=\frac{\partial F}{\partial s}\left(t, S_{t}\right)$ where $F(t, s)$ is the pricing formula obtained in (a.ii); in this case for $t \geq 1$ we have $F(t, s)=S_{1} s+e^{-r(T-t)}$ so $y_{t}$ is non-constant. $[1]$
(c) i. The value of our portfolio at time $t$ is given by $F\left(t, S_{t}\right)$, where $F$ is as in part
(a). If we add an amount $\alpha$ of stock into our portfolio then its new value will be $V\left(t, S_{t}\right)=F\left(t, S_{t}\right)+\alpha S_{t} .[1]$ To achieve delta neutrality, we want to choose $\alpha$ such that
$$
0=\frac{\partial V}{\partial s}\left(0, S_{0}\right)=3+\alpha
$$
$[1]$ Hence $\alpha=-3 .[1]$
ii. Our new portfolio has value $V\left(t, S_{t}\right)=F\left(t, S_{t}\right)-3 S_{t}=5 e^{-r(T-t)}$, and hence $\frac{\partial V}{\partial s}=0$ for all time. Hence, in this case our portfolio will stay delta neutral for all time.
(d) A delta neutral portfolio is advantageous because its value is, typically, less sensitive so sudden changes in the stock price. [1]