Title: Stochastic Financial Models

Q: Pricing derivatives
A: Classical approach: expected discounted payoff, i.e. $\E[H/(1+r)]$.
Modern approach: arbitrage argument. Construct portfolio of cost $V_0$ such that $V-V_0=H$. Then fair value is $V_0$. Equivalently, find the cost for the seller to hedge.

Q: Pricing forward contracts
A: If we buy a share for $S_0$ and borrow $K/(1+r)$, then we can sell the share at maturity for price $K$ and pay off the loan. So the cost for the seller to hedge is $S_0-K/(1+r)$.

Q: Contingent claims
A: A contingent claim is a derivative with a payout that is dependent on the realisation of some uncertain future event.
\[
\{\text{contigent claims} \} = \{\text{derivatives} \}\setminus\{\text{forwards} \}.
\]

Q: Utility function
A: $U:\R\to [-\infty, \infty)$ is a non-decreasing concave function such that $X$ is preferred to $Y$ iff
\[
\E[U|X] \geq \E[U|Y].
\]
We write $\mathcal{D}(U)=\{x:U(x)>-\infty\}$.

Q: Absolute risk aversion
A: An agent with wealth $w$ and $C^2$ utility $U$ will accept a contingent claim $X$ provided
\[
\E[U(w+X)] > U(w).
\]
Taylor expanding
\[
U'(w)\E[X]+\frac{1}{2}U''(w)\E[X^2] >0
\]
and rearranging
\[
\frac{2\E[X]}{\E[X^2]}> - \frac{U''(w)}{U'(w)} =:\gamma(w).
\]
If the Arrow-Pratt coefficient $\gamma>0$ is constant, then this gives the Constant Absolute Risk Aversion Utility
\[
U(x)=-\exp(-\gamma x).
\]

Q: Relative Risk Aversion
A: An agent with wealth $w$ and $C^2$ utility $U$ will accept a multiplicative gamble $X$ if
\[
\E[U(w(1+X))] > U(w).
\]
Taylor expanding
\[
wU'(w)\E[X] + \frac{1}{2}w^2U''(w)\E[X^2] > 0
\]
and rearranging
\[
\frac{2\E[X]}{\E[X^2]} > -\frac{wU''(w)}{U'(w)}=:R(w).
\]
If the Arrow-Pratt coefficient $0<R\neq 1$ is constant, then this gives the Constant Relative Risk Aversion (CRRA) utility
\[
U(x)= \begin{cases}
\frac{x^{1-R}}{1-R} \quad & x\geq 0 \\
-\infty \quad & x <0.
\end{cases}
\]

Q: Chord condition for concavity
A: A function $U:\R\to\R$ is concave $\iff$ $\forall x_1<y_1\leq x_2 < y_1$:
\[
\frac{U(y_1)-U(x_1)}{y_1-x_1} \geq \frac{U(y_2)-U(x_2)}{y_2 - x_2}.
\]
\begin{proof}
Ets for $y_1=x_2$ by joining chords, which reduces to concavity:
\begin{align*}
\frac{U(z)-U(x)}{z-x} &\geq \frac{U(y)-U(z)}{y - z} \\
\iff U(z) &\geq \frac{z-x}{y-x}U(y) + \frac{y-z}{y-x} U(x).
\end{align*}
\end{proof}

Q: One-sided derivatives of concave functions
A: If $U$ is concave, then $U_-' \geq U_+'$ exist and are decreasing.
\begin{proof}
Take limits in chord condition.
\end{proof}

Q: First order maximality condition for affine spaces
A: Let $\mathcal{A}=X_0+\mathcal{V}$ be an affine admissible set and $\E[U(X^*)]=\sup_{X\in\mathcal{A}} \E[U(X)]$. Then $\forall\xi\in\mathcal{V}, t\in\R$:
\[
\E[U(X^*)] \geq \E[U(X^*+t\xi)].
\]
Differentiating wrt $t$ gives $\E[U'(X^*)\xi] =0$.

Q: Reservation prices
A: Let $\pi_b(Y)$ be the largest $\pi$ such that
\[
\sup_{X\in\mathcal{A}} \E[U(X+Y-\pi)] \geq \E[U(X^*)]
\]
and $\pi_a(Y)$ be the smallest $\pi$ such that
\[
\sup_{X\in\mathcal{A}} \E[U(X-Y+\pi)] \geq \E[U(X^*)].
\]
Clearly $\pi_a(Y)=-\pi_b(-Y)$. Moreover: $\pi_b(Y)\leq \pi_a(Y)$.
\begin{proof}
Let the suprema above be achieved at $X', X''$ resp. By concavity and monotonicity of $U$ (plus convexity of $\mathcal{A}$)
\begin{align*}
& \; \E[U(\underbrace{\tfrac{1}{2} (X'+X'')}_{\in\mathcal{A}} + \underbrace{\tfrac{1}{2}(\pi_a(Y) - \pi_b(Y))}_{\geq 0})] \\
\geq & \; \tfrac{1}{2}\E[U(X'-Y+\pi_a(Y))] +  \tfrac{1}{2}\E[U(X''+Y-\pi_b(Y))] \\
= & \; \E[U(X^*)].
\end{align*}
\end{proof}

Q: Marginal price
A: One can show that
\[
f_Y: \R\setminus\{0\} \to\R, \quad t\mapsto \frac{\pi_a(tY)}{t}
\]
is increasing so one-sided limits exist.
For $t\neq 0$, now define $X_t^*$ by
\[
X_t^* = \arg\max_{X\in\mathcal{A}} \E[U(X-tY + \pi_a(tY))].
\]
Taylor expanding and then using the first-order maximality condition:
\begin{align*}
\E[U(X^*)] &= \E[U(X_t^*-tY + \pi_a(tY))] \\
&= \E[U(X^* + (X_t^*-X^*)-tY + tf_Y(t))] \\
&= \E[U(X^*) + U'(X^*)\{\underbrace{(X_t^*-X^*)}_{\in\mathcal{V}}-tY + tf_Y(t) \}] +o(t) \\
&= \E[U(X^*) + U'(X^*)\{-tY + tf_Y(t) \}] +o(t) \\
f_Y(t) &= \frac{\E[U'(X^*Y)}{\E[U'(X^*)]} + \frac{o(t)}{t} \\
\lim_{t\to 0} \frac{\pi_a(tY)}{t} &=  \frac{\E[U'(X^*)Y]}{\E[U'(X^*)]}.
\end{align*}
This is the price per unit at which we would buy/sell an infinitesimal amount of $Y$.

Q: Mean-variance analysis of single-period model
A: Consider a portfolio $\theta$ of assets $S_0$ at $t=0$ and consider the wealth $w_1=\theta^T S_1$ at $t=1$. Let
\[
\mu=\E[S_1], \quad V = \cov(S_1)\succ 0.
\]
The Lagrangian is
\[
\mathcal{L}=\tfrac{1}{2}\theta^TV\theta + \lambda_1(m-\theta^T\mu) + \lambda_2(w_0-\theta^T S_0).
\]
Differentiating
\[
\theta = V^{-1}(\lambda_1 \mu + \lambda_2 S_0)
\]
Substituting into the mean, budget constraints
\begin{align*}
\begin{pmatrix}
m \\
w_0
\end{pmatrix} &=
\begin{pmatrix}
\mu^T \\
S_0^T
\end{pmatrix}V^{-1}(\lambda_1 \mu + \lambda_2 S_0) \\
&=
\begin{pmatrix}
\mu^TV^{-1}\mu & \mu^T V^{-1}S_0 \\
\mu^T V^{-1}S_0 & S_0^TV^{-1} S_0
\end{pmatrix}
\begin{pmatrix}
\lambda_1 \\
\lambda_2
\end{pmatrix}.
\end{align*}
For $\mu\not\in\spn S_0$, we can invert this equation to find $\lambda$. Substituting into $\theta^TV\theta$ gives a quadratic in $m$. Minimising over $m$ gives $\theta=V^{-1}S_0$.

Q: Efficient frontier
A: The mean-variance efficient frontier $\theta=V^{-1}S_0$ describes a parabola ($\subset$) on the mean/variance graph.
To the right of the curve are the feasible portfolios.
The vertex is the minimum variance portfolio.

Q: CARA Gaussian portfolio
A: Assume $S_1\sim N(\mu, V)$. We seek to maximise
\[
\E[-\exp(-\gamma \theta^TS_1)]=-\exp(-\gamma\theta^T\mu + \tfrac{1}{2}\gamma^2\theta^TV\theta).
\]
Hence we minimise the Lagrangian
\[
\mathcal{L} = -\gamma\theta^T\mu + \tfrac{1}{2}\gamma^2\theta^TV\theta + \gamma\lambda(w_0-\theta^TS_0).
\]
Solving gives a \emph{mutual fund theorem} for $\theta$, i.e. a weighted average of \emph{min-variance} portfoli $V^{-1}S_0$ and \emph{diversified} portfolio $V^{-1}\mu$.

Q: CARA Gaussian portfolio with a bond
A: Prepend $S_1^0 = (1+r)S_0^0$. As in the regular case
\[
\mathcal{L} = -\gamma\bar\theta\bar\mu + \tfrac{1}{2}\gamma^2\bar\theta\bar V\bar\theta + \gamma\lambda(w_0-\bar\theta\bar S_0).
\]
Differentiating wrt $\bar\theta$
\[
\gamma\bar V \bar\theta = \bar\mu + \lambda \bar S_0.
\]
Row 0 gives $\lambda = -(1+r)$.
As in the standard case, we have the mutual fund theorem
\[
\theta = \gamma^{-1}V^{-1}(\mu+\lambda S_0).
\]
We write $\theta=\gamma^{-1}\theta_M$ where $\theta_M$ is the market portfolio or major share index.
If $V$ is diagonal, then
\[
\theta^j = \frac{\mu^j+\lambda S_0^j}{\gamma V_{jj}}=: \frac{1}{\gamma} \frac{R_{jj}}{V_{jj}}
\]
where $R_{jj}$ is the \emph{excess mean return}, i.e. the expected returns of asset $j$ over the bond.

Q: Capital Asset Pricing Model
A: Note that $\var(S_1^i)$ captures specific/diversifiable risk. For market/systematic risk, we define the sensitivity parameter $\beta$ by
\begin{align*}
\beta_i &= \frac{\cov(S_1^i, \theta_M^T S_1)}{\var(\theta_M^TS_1)} \\
&= \frac{(V\theta_M)^i}{\theta_M^T V \theta_M} \\
&= \frac{R_i}{\theta_M^T V \theta_M}.
\end{align*}
Now consider a mkt pf investor with $w_0=\theta_M^T S_0$ and $\mu_M=\theta_M^T\mu$. Their excess mean rt is
\begin{align*}
R_M &= \mu_M - (1+r)\theta_M^T S_0 \\
&= \theta_M^T (\mu - (1+r)S_0) \\
&= \theta_M^T (V\theta_M).
\end{align*}
Then
\[
R_i = \beta_i \theta_M^T V \theta_M = \beta_i R_M.
\]
All parameters in CAPM can be estimated from mkt data, testing the theory.

Q: Equilibrium pricing
A: Assume $K$ agents with CARA parameters $\gamma_k$, unit net supply of $d$ assets, zero net supply of the bond.
Each agent chooses $\theta_k=\gamma_k^{-1} \theta_M$ in the risky assets. In total, after market clearing
\begin{align*}
 \mathbf{1} &= \sum_{k=1}^K \theta_k \\
&= \Gamma^{-1}\theta_M \\
&= \Gamma^{-1}V^{-1}(\mu-(1+r)S_0). \\
\intertext{where}
\Gamma^{-1}&=\sum_{k=1}^K \gamma_k^{-1}. \\
\intertext{Rearranging}
S_0 &= \frac{\mu - \Gamma V \mathbf{1}}{1+r}.
\end{align*}
Finally we can check the total demand for the bond is zero, as required.

Q: Conditional expectation for an event
A: If $B\in\mathcal{F}$ with $\P(B)>0$, then we define
\[
\E[X|B] = \frac{\E[X\mathbbm{1}_B]}{\P(B)}.
\]

Q: Conditional expectation for countable disjoint events
A: Let $(B_i)_{i\in\N}\subseteq \mathcal{F}$ partition $\Omega$. Write $\mathcal{F}_0 = \sigma(\{B_i\})$. Then we define
\[
\E[X|\mathcal{F}_0]= \sum_{i: \P(B_i)>0} \E[X|B_i] \mathbbm{1}_{B_i}.
\]
We claim this is a $\mathcal{F}_0$ msble rv and
\[
\E[X\mathbbm{1}_A] = \E[\E[X|\mathcal{F}_0] \mathbbm{1}_A].
\]
\begin{proof}
Sum of msble functions are msble.
Writing $\mathbbm{1}_A$ as a sum from $\{\mathbbm{1}_{B_i}\}$, wlog $A=B_i$ for some $i$ with $\P(B_i)>0$. Then
\begin{align*}
\E[X\mathbbm{1}_A] &= \E[X\mathbbm{1}_{B_i}]\\
&=\E[X|B_i] \P(B_i) \\
&= \E[X| B_i] \E[\mathbbm{1}_{B_i}] \\
&= \E[\E[X|B_i] \mathbbm{1}_{B_i}] \\
&= \E[\E[X|\mathcal{F}_0] \mathbbm{1}_{B_i}] \quad (\omega \in B_i).
\end{align*}
\end{proof}

Q: Conditional expectation
A: Let $X\geq 0$ be a rv, $\mathcal{F}_0\subseteq\mathcal{F}$ a $\sigma$-algebra. We write $X_0 = \E[X| \mathcal{F}_0]$ if it is an $\mathcal{F}_0$ msble rv such that $\forall A\in\mathcal{F}_0$:
\[
\E[X\mathbbm{1}_A] = \E[X_0\mathbbm{1}_A].
\]
For $X\in L^1$, then define
\[
\E[X|\mathcal{F}_0] = \E[X^+|\mathcal{F}_0] - \E[X^-|\mathcal{F}_0].
\]
Note if $\mathcal{F}_0=\sigma(\mathcal{C})$, then enough to check for $A\in\mathcal{C}$.

Q: Conditional expectation of a random variable
A: If $X,Y$ are rvs, then define
\[
\E[X|Y] = \E[X|\sigma(Y)].
\]
By the factorisation lemma, $\E[X|Y]=f(Y)$ for some $f$ so we write
\[
\E[X | Y=y ] = f(y).
\]

Q: Factorisation lemma
A: Let $T: E\to E'$, $f:E\to\R^+$. Then $f$ is msble wrt $\sigma(T)$ $\iff$ $\exists g:(E', \mathcal{A}') \to (\R, \mathcal{B})$ msble such that $f=g\circ T$.
\begin{proof}
($\Leftarrow$) Composition of msble functions is msble.
($\Rightarrow$) First assume $f=\sum_{i=1}^n \alpha_i \mathbbm{1}_{A_i}$ for $A_i=T^{-1}(A_i')\in\sigma(T)$ and $A_i'\in\mathcal{A}'$. Hence $g=\sum_{i=1}^n \alpha_i \mathbbm{1}_{A_i'}$ works.
For the general case let $(f_n)\nearrow f$ for $f_n$ simple. Then by the simple case each $f_n=g_n\circ T$ where $g_n$ is msble. Then $g_n\nearrow g$ where $g$ is as required.
\end{proof}

Q: Radon-Nikodym Theorem
A: Let $(\nu, \Omega, \mathcal{F})$ be a prob space and $\mu$ a measure. Then $\mu << \nu$ $\iff$ the Radon-Nikodym derivative $\frac{\d\mu}{\d\nu}$ exists.
Moreover $\mu\approx\nu$ iff $\frac{\d\mu}{\d\nu} > 0$ $\nu$-a.s. In this case
\[
\frac{\d\mu}{\d\nu} = \left(\frac{\d\nu}{\d\mu}\right)^{-1}.
\]

Q: Radon-Nikodym derivative
A: Let $(\nu, \Omega, \mathcal{F})$ be a prob space and $\mu$ a measure. We write $\varphi = \frac{\d\mu}{\d\nu}$ for the \emph{density} or derivative, if $\varphi\geq 0$ is $\mathcal{F}$ msble such that $\forall A \in\mathcal{F}$:
\[
\mu(A) = \int_A \varphi \; \d\nu.
\]
In particular $\E_\nu\left[\frac{\d\mu}{\d\nu}\right] = \mu(\Omega)$ and $\forall f\in L^1(\nu, \Omega, \mathcal{F})$:
\[
\mu(f) = \nu\left(f\frac{\d\mu}{\d\nu}\right).
\]

Q: Absolute continuity
A: Let $(E, \mathcal{A})$ be a measure space and $\mu,\nu$ measures. We write $\mu<< \nu$, saying $\mu$ is \emph{dominated by} or is \emph{absolutely continuous} wrt $\nu$, if $\forall A\in\mathcal{A}$:
\[
\nu(A) = 0 \implies \mu(A)=0.
\]
We say the measures are \emph{equivalent} if $\mu<<\nu << \mu$, that is $\forall A\in\mathcal{A}$:
\[
\nu(A) = 0 \iff \mu(A)=0.
\]

Q: Conditional expectation: existence
A: $X_0=\E[X|\mathcal{F}_0]$ exists.
\begin{proof}
Define
\[
\mu(A) = \int_A X \; \d\P
\]
a positive measure with $\mu<<\P$ on $\mathcal{F}$. In particular, $\mu<<\P$ on $\mathcal{F}_0$. Applying Radon-Nikodym Theorem to $(\Omega, \mathcal{F}_0)$, $\exists X_0=\frac{\d\mu}{\d\nu}$ such that $\forall A\in\mathcal{F}_0$:
\[
\E[X\mathbbm{1}_A] = \mu(A) = \int_A X_0 \; \d\P = \E[X_0 \mathbbm{1}_{A_0}].
\]
\end{proof}

Q: Conditional expectation: uniqueness
A: Let $X_0, \tilde{X}_0$ be conditional expectations of $X$ given $\mathcal{F}_0$. Then $X_0=\tilde{X}_0$ $\P$-a.s.
\begin{proof}
Write $A=\{ X_0 > \tilde{X}_0\}\in\mathcal{F}_0$ (since both are msble). Then
\[
\E[(X_0-\tilde{X}_0)]\mathbbm{1}_A ]=0.
\]
So $(X_0-\tilde{X}_0)\mathbbm{1}_A$ vanishes $\P$-a.s. and thus $\P(A)=0$.
\end{proof}

Q: Conditional expectation with a measurable factor
A: Let $Y_0\geq 0$ be $\mathcal{F}_0$ msble. Then
\[
\E[XY_0 | \mathcal{F}_0] = Y_0\E[X | \mathcal{F}_0] \quad \P \text{-a.s.}
\]
In particular
\[
\E[Y_0 | \mathcal{F}_0] = Y_0 \quad \P \text{-a.s.}
\]
\begin{proof}
By MCl, wlog $Y=\mathbbm{1}_{A_0}$ for some $A_0\in\mathcal{F}_0$. Then $\forall A\in\mathcal{F}_0$:
\begin{align*}
\E[X Y_0\mathbbm{1}_{A} ] &= \E[X \mathbbm{1}_{A\cap A_0} ] \\
&= \E[\E[X | \mathcal{F}_0]\mathbbm{1}_{A\cap A_0} ] \\
&= \E[\E[X | \mathcal{F}_0] Y_0\mathbbm{1}_{A_0} ].
\end{align*}
\end{proof}

Q: Tower property (projectivity) of conditional expectations
A: Let $\mathcal{F}_0\subseteq\mathcal{F}_1\subseteq\mathcal{F}$ be $\sigma$-algebras. Then
\[
\E[X|\mathcal{F}_0 ]  = \E[\E[X|\mathcal{F}_1] | \mathcal{F}_0] \quad \P\text{-a.s.}
\]
In particular, setting $\mathcal{F}_0$ to be trivial:
\[
\E[X ]  = \E[\E[X | \mathcal{F}_1] ] \quad \P\text{-a.s.}
\]
\begin{proof}
For any $A\in \mathcal{F}_0\subseteq\mathcal{F}_1$:
\begin{align*}
\E[X\mathbbm{1}_A ] &= \E[ \underbrace{\E[X|\mathcal{F}_1]}_{\text{dfn}} \mathbbm{1}_A ]\\
&= \E[ \underbrace{\E[\E[X|\mathcal{F}_1] | \mathcal{F}_0]}_{\text{dfn}}\mathbbm{1}_A].
\end{align*}
\end{proof}

Q: Conditional expectation of an independent RV
A: Let $X, \mathcal{F}_0$ be independent. Then
\[
\E[X|\mathcal{F}_0] = \E[X] \quad \P\text{-a.s.}
\]
\begin{proof}
For $A\in\mathcal{F}_0$
\begin{align*}
\E[X\mathbbm{1}_A] &= \E[\mathbbm{1}_A]\E[X] \qquad \text{(independence)}\\
&= \E[\E[X]\mathbbm{1}_A]  \qquad \text{(linearity of $\E$)}.
\end{align*}
\end{proof}

Q: Conditional expectation of a function of RVs
A: Let $F:\R^2\to[0,\infty)$ be msble, $X$ be indep of $\mathcal{F}_0$ and $Y$ be $\mathcal{F}_0$ msble. Write $\Phi(y)=\E[F(X,y)]$. Then
\[
\E[F(X,Y) | \mathcal{F}_0 ]  = \Phi(Y) \quad\P\text{-a.s.}
\]
\begin{proof}
By monotone class theorem, wlog $F(x,y)=f(x)g(y)$ for $f,g:\R\to [0,\infty)$ be msble. Then
\begin{align*}
\E[F(X,Y) | \mathcal{F}_0 ](\omega) &= g(Y(\omega)) \E[f(X) | \mathcal{F}_0](\omega) \\
&= g(Y(\omega)) \E[f(X)] \\
&=  \E[f(X)g(Y(\omega))] \\
&= \Phi(Y(\omega)).
\end{align*}
\end{proof}

Q: Filtration
A: A \emph{filtration} is an increasing family of $\sigma$-algebras $(\mathcal{F}_t)_{t\in I}\subseteq\mathcal{F}$.

Q: Adapted process
A: A stochastic process $Y=(Y_t)_{t\in I}$ is \emph{adapted} to a filtration $(\mathcal{F}_t)_{t\in I}$ if $Y_t$ is $\mathcal{F}_t$-msble for each $t\in I$.

Q: Martingale
A: Let $(\mathcal{F}_t)_{t\in I}$ be a filtration on $(\Omega, \mathcal{F})$. A stochastic process $M=(M_t)_{t\in I}$ on $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\in I}, \P)$ is a $\P$-\emph{martingale} if
\begin{enumerate}
\item $M$ is adapted
\item $M_t\in L^1(\Omega, \P)$ for each $t$
\item The \emph{martingale property} holds, i.e. $\forall s\leq t \in I$:
\[
\E[M_t | \mathcal{F}_s ] = M_s \quad \P\text{-a.s.}
\]
In words, the expected future profit is zero.
\end{enumerate}
If the martingale property is replaced by $\geq, \leq$, then $M$ is called a \emph{sub-martingale} or \emph{super-martingale} property respectively.

Q: Expectation of a martingale
A: If $(M_t)_{t\in I}$ is a martingale, then $\forall s < t \in I$:
\[
\E[M_s] = \E[M_t].
\]
\begin{proof}
Applying the tower property of conditional expectation to the the martingale property:
\[
\E[M_s] = \E[\E[M_t | \mathcal{F}_s]] = \E[M_t].
\]
\end{proof}

Q: Previsible
A: A stochastic process $(C_n)_{n\in\N}$ is called \emph{previsible} wrt filtration $(\mathcal{F}_n)_{n\in\N}$ if for each $n\in\N$, $C_n$ is $\mathcal{F}_{n-1}$-msble.

Q: Discrete-time integral of previsible
A: Let $M=(M_n)$ be a martingale and $C=(C_n)$ bounded previsible.
\begin{enumerate}
\item The process
\[
Y_n = \sum_{k=1}^n C_k( M_k - M_{k-1})
\]
is a martingale.
\item If $C\geq 0$, then can replace martingale with sub/super.
\end{enumerate}
\begin{proof}
\begin{enumerate}
\item $C$ bounded $\implies$ $Y_n\in\mathcal{L}^1$ for each $n$. Since $C_n$ is $\mathcal{F}_{n-1}$-msble
\[
\E[Y_n - Y_{n-1} | \mathcal{F}_{n-1}] = C_n\E[(M_n - M_{n-1}) | \mathcal{F}_{n-1}]=0. \tag{$*$}
\]
\item $C_n\geq 0$ ensures the appropriate inequality in $(*)$.
\end{enumerate}
\end{proof}

Q: Upcrossing
A: We say that an \emph{upcrossing} of $[a,b]$ occurs between times $s<t$ if
\begin{enumerate}
\item $X_s<a<b<X_t$
\item $\forall r\in (s,t): X_r\in [a,b]$.
\end{enumerate}
We write $U_N(X,[a,b])$ for the number of upcrossings of $[a,b]$ by time $N$. We can also define the monotone limit
\[
U_\infty(X,[a,b]) = \lim_{N\to\infty} U_N(X,[a,b]).
\]

Q: Upcrossing wealth process
A: Define the previsible strategy $(C_n)$ by
\[
C_1=\mathbbm{1}_{X_0<a}, \quad C_{n+1} = \mathbbm{1}_{C_n=1} \mathbbm{1}_{X_n\leq b} + \mathbbm{1}_{C_n=0} \mathbbm{1}_{X_n< a}.
\]
The associated wealth process for martingale asset $X$ is
\[
W_n = \sum_{k=1}^n C_k (X_k - X_{k-1}).
\]
In terms of upcrossings, the minimum winnings by time $N$ is
\[
W_N \geq (b-a)U_N(X,[a,b]) - |a-X_N|\mathbbm{1}_{X_N<a}.
\]

Q: Doob's upcrossing lemma
A: Let $X$ be a super-martingale. Then $\forall a<b\in\R$:
\[
\E[U_N(X,[a,b])] \leq \frac{\E[(X_N-a)^-]}{b-a}.
\]
\begin{proof}
The upcrossing wealth process $(W_n)$ is a super-martingale. Therefore $\E[W_N]\leq 0$. Taking expectations in our lower bound
\begin{align*}
W_N &\geq (b-a)U_N(X,[a,b]) - |a-X_N|\mathbbm{1}_{X_N<a} \\
0 &\geq (b-a) \E[U_N(X,[a,b])] - \E[(X_N-a)^-].
\end{align*}
\end{proof}

Q: Upcrossing bound for $\mathcal{L}^1$-bounded supermartingale
A: Let $(X_n)$ be an $\mathcal{L}^1$-bounded (e.g. non-negative) supermartingale. Then
\[
\E[U_\infty(X,[a,b])] \leq \frac{a + \sup_n \E |X_n|}{b-a} < \infty.
\]
In particular $U_\infty(X,[a,b])<\infty$ a.s.
\begin{proof}
From Doob's upcrossing lemma, $\forall N\in\N$
\[
\E[U_N(X,[a,b])] \leq \frac{\E[(X_N-a)^-]}{b-a} \leq \frac{a + \sup_N\E |X_N|}{b-a}< \infty.
\]
Done by MCT.
\end{proof}

Q: Doob's super-martingale convergence theorem
A: Let $(X_n)$ be an $\mathcal{L}^1$-bounded (e.g. non-negative) super-martingale. Then $X_\infty = \lim_{n\to\infty} X_n<\infty$ a.s.
\begin{proof}
\emph{Convergence}. Define
\[
\Lambda_{a,b} = \{ \omega: \lim\sup X_n(\omega)>b>a > \lim\inf X_n(\omega)\}
\]
such that $X_n$ converges on
\[
\Omega \setminus \bigcup_{a<b\in\Q} \Lambda_{a,b}.
\]
But
\[
\Lambda_{a,b} \subseteq \{ \omega: U_\infty(X, [a,b]) = \infty\}
\]
are null sets by Doob's upcrossing lemma.
\emph{Finite}. By Fatou's Lemma
\[
\E |X_\infty|  = \E[\lim\inf |X_n|] \leq \lim\inf \E |X_n| \leq \sup \E |X_n| < \infty.
\]
\end{proof}

Q: Stopping time
A: In discrete time, a map $\tau: \Omega \to \{0, 1, \cdots \infty\}$ is a \emph{stopping time} wrt filtration $(\mathcal{F}_n)$ for each $n$
\[
\{\tau = n \} \in\mathcal{F}_n.
\]
In cts time, a map $\tau: \Omega \to [0,\infty]$ is a \emph{stopping time} if
\[
\{\tau \leq t \} \in\mathcal{F}_t=\sigma(\{ X(s): s\leq t\}).
\]

Q: Hitting time
A: \emph{Discrete time}. Let $(X_n)$ be an adapted process and let $B\in\mathcal{B}(\R)$. Define the stopping time
\[
\tau_B(\omega) = \inf \{ n > 0: X_n(\omega) \in B\}.
\]
\emph{Continuous time}. The hitting time of $A\subseteq S$ is
\[
T_A = \inf\{ t>0: X(t)\in A\}.
\]

Q: Stopped process
A: Let $(X_n)$ be a stochastic process and $\tau$ a stopping time. Define the \emph{stopped process}
\[
X_n^\tau = X_{n\wedge \tau}.
\]
If $(X_n)$ is a (sub-) martingale, then $(X_n^\tau)$ is a (sub-) martingale.
\begin{proof}
Exercise.
\end{proof}

Q: Doob's optional stopping theorem
A: Let $(X_n)$ be a martingale and $\tau$ a stopping time. Assume any of
\begin{enumerate}
\item $\tau$ is bounded
\item $X^\tau$ is bounded and $\tau<\infty$ a.s.
\item $\E[\tau]<\infty$ and $\exists K<\infty$ such that $\forall n\in\N$:
\[
|X_n - X_{n-1} | \leq K.
\]
\end{enumerate}
Then $X_\tau\in\mathcal{L}^1$ and $\E[X_\tau] = \E[X_0]$. Similarly for sub/super- martingales.
\begin{proof}
Since $(X_n^\tau)$ is a martingale, $\forall n\in\N$
\[
\E[X_n^\tau] = \E[X_0^\tau] = \E[X_0].
\]
\begin{enumerate}
\item If $\tau \leq N$, then $\E[X_\tau] = \E[X_N^\tau] =\E[X_0]$.
\item $\tau<\infty$ so $X_\tau = \lim X_n^\tau$ a.s. Since $X^\tau$ is bounded, by DCT
\[
\E[X_0] = \lim \E[X_n^\tau] = \E[\lim X_n^\tau] = \E[X_\tau].
\]
\item By summation, we have the bound
\[
|X_n^\tau - X_0| \leq  \sum_{k=1}^{\tau \wedge n} | X_k - X_{k-1} |  \leq K\tau.
\]
So done again by DCT.
\end{enumerate}
\end{proof}

Q: Arbitrage opportunity for single-period model
A: In a single-period model, a portfolio $\bar\theta\in\R^{d+1}$ is an \emph{arbitrage opportunity} if
\begin{enumerate}
\item $\bar\theta^T\bar{S}_0 \leq 0$
\item $\bar\theta^T\bar{S}_1 \geq 0$
\item $\P(\bar\theta^T\bar{S}_1 > 0)>0$.
\end{enumerate}
N.B. $\P$ need only be specified up to equivalence.

Q: FLAP: single-period
A: Assume $S^0\equiv 1$ and write $Y=S_1-S_0$. Then TFAE
\begin{enumerate}
\item no arbitrage
\item $\exists\Q\approx\P$ such that $\E_\Q[S_1]=S_0$ and the density is
\[
\frac{\d\Q}{\d\P} = \frac{1}{Z_\theta} \exp\left( -\theta^T Y - \tfrac{1}{2}|Y|^2\right)
\]
for some $\theta\in\R^d$ where $Z_\theta$ is a normalising constant.
\end{enumerate}
\begin{proof}
By removing LD assets, assume $\forall\theta\in\R^d\setminus\{0\}: \P(\theta^TY=0)<1$.
(ii) $\implies$ (i): If $\bar\theta$ is an arb, then
\[
0 \leq \E_\Q[\bar\theta^T \bar S_1] = \bar\theta^T \bar{S}_0 \leq 0.
\]
So $\bar\theta^T \bar S_1 = 0$ $\Q$-a.s.
(i) $\implies$ (ii): Define $\varphi:\R^d\to[0,\infty)$ convex diffble given by
\[
\theta\mapsto \frac{\E[\exp(-\theta^TY -\tfrac{1}{2}|Y|^2 )]}{\E[\exp(-\tfrac{1}{2}|Y|^2 )]}.
\]
If $\theta$ is a minimum of $\varphi$, then differentiating
\[
0=\E[\exp(-\theta^TY -\tfrac{1}{2}|Y|^2 )Y] =\E_\Q[Y]
\]
as required. Otherwise, for $\alpha\geq 0$, define the compact sets
\[
F_\alpha = \{ \theta\in\R^d: \|\theta \|=1 , \varphi(\alpha\theta) \leq 1\}.
\]
Since $\inf\varphi \leq 1$ is not attained, $F_\alpha\neq\emptyset$ i.o. Since $\varphi$ is convex and $\varphi(0)=1$, $(F_\alpha)\searrow \bigcap F_\alpha$. Hence $\exists \theta\in \bigcap F_\alpha$ such that
\[
\forall\alpha\geq 0: \quad \varphi(\alpha \theta)=\frac{\E[\exp(-\alpha\theta^TY -\tfrac{1}{2}|Y|^2 )]}{\E[\exp(-\tfrac{1}{2}|Y|^2 )]} \leq 1.
\]
Hence $\theta^T Y \geq 0$ $\P$-a.s. Take $\bar\theta = (-\theta^T S_0, \theta)$.
\end{proof}

Q: FTAP for single-period model
A: Assume positive \emph{num\'eraire}, i.e. $S_t^0>0$. Then TFAE
\begin{enumerate}
\item no arbitrage
\item $\exists\Q\approx\P$ a \emph{risk-neutral} or an \emph{equivalent martingale} measure, i.e.
\[
\E_\Q\left[ \frac{S_1}{S_1^0}\right] = \frac{S_0}{S_0^0}.
\]
\end{enumerate}
\begin{proof}
$\bar\theta$ is an arb for $\bar S$ $\iff$ $\bar\theta$ is an arb for $\tilde{S}_t = S_t / S_t^0$. Done by FLAP.
\end{proof}

Q: Riskless asset
A: Money-market account, bond, bank account or riskless asset, earns simple interest at the risk-free rate $r$.
Often used as the \emph{num\'eraire} $S^0$.

Q: Trading strategy
A: A \emph{trading strategy} is an $\R^{d+1}$-valued previsible process $\{\bar\theta_t: 0 \leq t \leq T \}$.

Q: Self-financing strategy
A: A trading strategy $\bar\theta$ is \emph{self-financing} if for $1\leq t<T$
\[
\bar\theta_t^T \bar{S}_t = \bar\theta_{t+1}^T \bar{S}_t.
\]
Price fluctuations are the only source of variation in the portfolio.

Q: Locally riskless bond
A: Assume $S_0^0\equiv 1$ with previsible spot rates $(r_t)_{1\leq t \leq T}$. Then
\[
S_t^0 = \prod_{k=1}^t (1+r_k)
\]
is previsible too.

Q: Discounted price process
A: Using the num\'eraire as our unit
\[
X_t^i = \frac{S_t^i}{S_t^0}.
\]
In particular $X_t^0\equiv 1$.

Q: Discounted value process
A: The \emph{discounted value process} of a trading strategy $\bar\theta$ is given by
\[
V_t = \begin{cases}
\bar\theta_t^T \bar{X}_t  \quad & 1\leq t\leq T \\
\bar\theta_1^T \bar{X}_0  \quad & t=0.
\end{cases}
\]

Q: Self-financing value process
A: For a trading strategy $\bar\theta$, TFAE
\begin{enumerate}
\item $\bar\theta$ is self-financing
\item $\bar\theta_t^T\bar{X}_t = \bar{\theta}_{t+1}^T\bar{X}_t$ for $1\leq t < T$
\item $V_t = V_0 + \sum_{s=1}^t \theta_s^T(X_s - X_{s-1})$ for $0\leq t \leq T$.
\end{enumerate}
\begin{proof}
(i) $\iff$ (ii): Immediate dividing by $S_t^0$.
(ii) $\iff$ (iii): For $1\leq t < T$
\[
\text{(ii)} \iff \bar\theta_{t+1}^T \bar{X}_{t+1} - \bar\theta_t^T \bar{X}_t = \bar\theta_{t+1}^T ( \bar{X}_{t+1} -  \bar{X}_t) \iff\text{(iii)}.
\]
\end{proof}

Q: Uniqueness of num\'eraire
A: If $\bar\theta$ is a self-financing trading strategy, then for $1\leq t <T$:
\[
(\bar{\theta}_{t+1}-\bar{\theta}_t)^T\bar{X}_t=0
\]
so
\[
\theta_{t+1}^0 - \theta_t^0 = -(\theta_{t+1}-\theta_t)^TX_t.
\]
Also
\[
\theta_1^0 = V_0 - \theta_1^T X_0.
\]
Hence the num\'eraire $\theta^0$ is uniquely determined by $V_0$ and $\theta$.

Q: Arbitrage opportunity for value process
A: A self-financing strategy $\bar\theta$ is an \emph{arbitrage opportunity} if the value process $V$ satisfies
\[
V_0 \leq 0, \qquad V_T \geq 0 \quad \P\text{-a.s.}, \qquad V_T >0 \quad \P\text{-p.p.}
\]

Q: Equivalent martingale measure
A: A PM $\Q$ on $(\Omega, \mathcal{F})$ is an \emph{equivalent martingale measure}, written $\Q\in\mathcal{P}$, if
\begin{enumerate}
\item $\Q\approx\P$
\item the discounted price process $X$ is a $\Q$-martingale.
\end{enumerate}

Q: FLAP: multi-period
A: Let $\Q\in\mathcal{P}$, $\bar\theta$ self-financing, $V_T\geq 0$ $\P$-a.s. Then $V$ is a $\Q$-martingale.
\begin{proof}
\textbf{Non-negativity}: $V_t \geq 0$ $\P$-a.s. By induction
\[
V_{t-1} = V_t - \theta_t \cdot (X_t - X_{t-1}) \geq - \theta_t \cdot (X_t - X_{t-1}).
\]
Define $\theta_t^c = \mathbbm{1}_{|\theta_t| \leq c}\theta_t$. Then $\theta^c\leq c$ so
\begin{align*}
V_{t-1}\mathbbm{1}_{|\theta_t| \leq c} &= \E_\Q[ V_{t-1} \mathbbm{1}_{|\theta_t| \leq c} | \mathcal{F}_{t-1}] \\
&\geq -\E_\Q [\theta_t^c \cdot (X_t - X_{t-1}) | \mathcal{F}_{t-1}] = 0.
\end{align*}
Letting $c\to\infty$ gives $V_{t-1}\geq 0$.
\textbf{Martingale property}. Since $X$ is a $\Q$-martingale
\[
\E_\Q[V_t \mathbbm{1}_{|\theta_t|\leq c} | \mathcal{F}_{t-1}] = \E_\Q[V_{t-1} \mathbbm{1}_{|\theta_t| \leq c}  +  \theta_t^c \cdot (X_t - X_{t-1}) | \mathcal{F}_{t-1} ] = V_{t-1}\mathbbm{1}_{|\theta_t|\leq c}.
\]
Letting $c\to \infty$, MCT gives $\E_\Q[V_t | \mathcal{F}_{t-1}]=V_{t-1}$.
\end{proof}

Q: FTAP for multi-period model
A: Assume $S^0_t>0$ $\P$-a.s. Then no arbitrage $\iff$ $\mathcal{P}\neq\emptyset$.
\begin{proof}
($\Leftarrow$) Let $\Q\in\mathcal{P}$, $(\bar\theta, V)$ an arb. Then FLAP gives
\[
\E_\Q[V_T] = V_0 \leq 0,
\]
contradicting $V_T\geq 0$ $\P$-a.s. and $V_T>0$ $\P$-p.p.
($\Rightarrow$) Stochastic Finance (F\"ollmer and Schied) Theorem 5.17, Section 1.6.
\end{proof}

Q: European contingent claim
A: A rv $C\geq 0$ is called a \emph{European contingent claim} or \emph{European option}.
$C$ is a \emph{derivative} of $S^0, S^1, \cdots S^d$ if it is measurable wrt the $\sigma$-algebra generated by the price process $(\bar{S}_t)_{0\leq t\leq T}$.
$T$ is called the \emph{expiration date} or \emph{maturity} of $C$.

Q: European call/put
A: \[
C = (S_T^i - K)^+, \qquad P = (K - S_T^i )^+.
\]

Q: Asian option
A: Define the TWAP
\[
S_\text{av}^i = \frac{1}{|\mathbb{T}|} \sum_{i\in\mathbb{T}} S_t^i
\]
for a predetermined averaging period $\mathbb{T}\subseteq\{0, 1, \cdots T\}$.
\begin{itemize}
\item Average price call: $(S_\text{av}^i -K)^+$
\item Average price put: $(K - S_\text{av}^i)^+$
\item Average strike call: $(S_T^i - S_\text{av}^i)^+$
\item Average strike put: $(S_\text{av}^i - S_T^i )^+$.
\end{itemize}

Q: Lookback option
A: A Lookback call is
\[
S_T^i - \min_{0\leq t \leq T}S_t^i
\]
and a Lookback put is
\[
\max_{0\leq t \leq T}S_t^i - S_T^i .
\]

Q: Attainable claim
A: A European contingent claim $C$ is \emph{attainable}, \emph{replicable} or \emph{redundant} if $\exists\bar\theta$ self-financing, called a \emph{replicating strategy}, whose terminal portfolio coincides with $C$, i.e.
\[
C = \bar{\theta}_T\cdot \bar{S}_T \quad \text{a.s.}
\]

Q: Attainable discounted claim
A: The \emph{discounted claim} associated to European claim $C$ is
\[
H = \frac{C}{S_T^0}.
\]
Then $C$ is attainable with replicating strategy $\bar\theta$ iff
\[
H = \frac{\theta_T\cdot \bar{S}_T}{S_T^0} = \theta_T\cdot \bar{X}_T = V_T = V_0 + \sum_{t=1}^T \theta_t \cdot (X_t - X_{t-1}).
\]

Q: Expectation of an attainable discounted claim
A: Let $H$ be an attainable discounted claim and $\Q\in\mathcal{P}$. Then
\[
\E_Q[H]<\infty
\]
and the value process $V$ of any replicating strategy satisfies
\[
V_t = E_\Q[H|\mathcal{F}_t] \quad \P\text{-a.s.}
\]
In particular, $V$ is a non-negative strategy-independent $\Q$-Martingale.
\begin{proof}
We know $H=V_T$ so $V_T \geq 0$. By FLAP, $V$ is a $\Q$-martingale so
\[
V_t = E_\Q[V_T | \mathcal{F}_t ] = E_\Q[H | \mathcal{F}_t ] .
\]
\end{proof}

Q: Discounted fair price
A: Let $H$ be an attainable discounted claim with replicating strategy $\bar\theta$. Then the (discounted) initial investment needed to replicate $H$
\[
\bar\theta_1 \cdot \bar{X}_0 = V_0 = \E_\Q[H]
\]
is the \emph{unique (discounted) fair price} of $H$.
\begin{proof}
Suppose the price of $H$ is $\tilde{\pi}> V_0$. Then we can
\[
\begin{cases}
\text{buy portfolio }\bar{\theta}_1, \text{ sell $H$ for $\tilde{\pi}$} \quad & t=0 \\
\text{settle $H$ using $V_T$} \quad & t=T
\end{cases}
\]
leaving profit $\tilde{\pi} - V_0>0$.
\end{proof}

Q: Second FTAP
A: A \emph{complete} market model is one in which every European contingent claim is attainable.
Equivalently, $|\mathcal{P}|=1$, i.e. there exists a unique equivalent martingale measure. (Theorem 5.38 Foellmer and Schied)

Q: CRR model
A: The Cox-Ross-Rubinstein binomial model consists of a bond $S_t^0 = (1+r)^t$ and risky asset
\[
S_t^1 = S_0^1\prod_{k=1}^t(1+R_k).
\]
where the returns in the $t$th trading period
\[
R_t = \frac{S_t^1 - S_{t-1}^1}{S_{t-1}^1}
\]
are $\{a,b\}$-valued rvs with $-1<a<b$.
Formally, let $\Omega=\{-1,1\}^T$ with projections $Y_t$. Then define
\[
R_t(\omega) = a\frac{1-Y_t(\omega)}{2} + b\frac{1+Y_t(\omega)}{2}.
\]
As filtration define
\[
\mathcal{F}_t = \sigma(S_0^1, \cdots S_t^1) = \sigma(X_0, \cdots X_t)
\]
such that
\[
\mathcal{F}_t = \begin{cases}
\{\emptyset, \Omega\} \quad & t= 0 \\
\sigma(Y_1, \cdots Y_t) = \sigma(R_1, \cdots R_t) \quad & t\geq 1 \\
\mathcal{P}(\Omega) \quad & t= T.
\end{cases}
\]
Let $\P$ be a positive PM on $(\Omega, \mathcal{F})$.

Q: Arbitrage-free CRR model
A: The CRR model is arbitrage-free $\iff$ $a<r<b$. In this case $\mathcal{P}=\{\Q\}$ and $(R_t)$ are $\Q$-iid with distribution
\[
\Q(R_t = b) = p^*:= \frac{r-a}{b-a}.
\]
\begin{proof}
By definition $\Q\in\mathcal{P} \iff \forall t<T$:
\begin{align*}
 & X_t = \E_\Q[X_{t+1} | \mathcal{F}_t ] = X_t \E_\Q\left[ \left. \frac{1+R_{t+1}}{1+r}\right| \mathcal{F}_t\right] \quad \Q\text{-a.s.} \\
\iff & r = E_\Q[R_{t+1} | \mathcal{F}_t ] = b\Q(R_{t+1} = b | \mathcal{F}_t) + a(1 - \Q(R_{t+1}=b | \mathcal{F}_t)) \\
\iff & \Q(R_{t+1}=b | \mathcal{F}_t) = \frac{r-a}{b-a} = p^* \\
\iff & \Q(R_{t+1}=b)  = p^*, (R_t) \text{ iid.}
\end{align*}
($\Rightarrow$) Arbitrage-free $\implies\exists\Q\in\mathcal{P}\implies p^*\in (0,1) \implies a<r<b$.
($\Leftarrow$) If $a<r<b$, then define $\Q\approx \P$ on $(\Omega, \mathcal{F})$ by $\Q\sim\Bin (T, p^*)$. Then $(Y_t), (R_t)$ are iid with $\Q(Y_t = 1)=\Q(R_t = b) = p^*$. So $\Q\in\mathcal{P}$ and hence no arbitrage.
\end{proof}

Q: Value process CRR component form
A: The value process $V_t=\E_\Q[H|\mathcal{F}_t]$ of a replicating strategy of $H=h(X_0, \cdots X_T)$ is of the form
\[
V_t = v_t(X_0, X_1, \cdots X_t)
\]
where $v_t$ is given by
\[
v_t(x_0, \cdots x_t) = \E_Q\left[ h\left(x_0, \cdots x_t, x_{t}\frac{X_1}{X_0} \cdots x_t \frac{X_{T-t}}{X_0} \right)\right].
\]
Recursively $V_T=H$, $V_{t+1}=\E_\Q[V_t | \mathcal{F}_t]$, i.e.
\[
v_t(x_0, \cdots x_t) =p^* v_{t+1}(x_0, \cdots x_t, x_t\hat{b}) + (1-p^*) v_{t+1}(x_0, \cdots x_t, x_t\hat{a})
\]
where $\hat{a}=\frac{1+a}{1+r}, \hat{b}=\frac{1+b}{1+r}$.
\begin{proof}
By definition
\[
V_t = \E_Q\left[ h\left(X_0, \cdots X_t, X_{t}\frac{X_{t+1}}{X_t} \cdots X_t \frac{X_T}{X_t}\right)  | \mathcal{F}_t \right].
\]
Since $X_0, \cdots X_t$ are $\mathcal{F}_t$ msble and
\[
\frac{X_{t+s}}{X_t} \sim_\Q \frac{X_s}{X_0}  = \prod_{k=1}^s \frac{1+R_k}{1+r}
\]
independently of $\mathcal{F}_t$. Done by properties of $\E$.
\end{proof}

Q: Delta hedge for CRR
A: The \emph{hedging strategy} for CRR is
\begin{align*}
\theta_t &= \Delta_t (X_0, \cdots X_{t-1}) \\
\Delta_t (x_0, \cdots x_{t-1})&:= \frac{v_t(x_0, \cdots x_{t-1},x_{t-1}\hat{b} ) - v_t(x_0, \cdots x_{t-1},x_{t-1}\hat{a} )}{x_{t-1}(\hat{b}-\hat{a})}.
\end{align*}
\begin{proof}
Recall any self-financing strategy $\bar\theta$ obeys
\[
\theta_t \cdot (X_t - X_{t-1}) \equiv V_{t} - V_{t-1}.
\]
Write $\omega=(y_1, \cdots y_T)$ and define
\[
\omega^{\pm}:= (y_1, \cdots y_{t-1}, \pm 1, y_{t+1}, \cdots y_T).
\]
Substituting these into $X_t, V_t$:
\begin{align*}
\theta_t(\omega) \cdot X_{t-1}(\omega)(\hat{b} - 1) &= V_{t}(\omega^+) - V_{t-1}(\omega) \\
\theta_t(\omega) \cdot X_{t-1}(\omega)(\hat{a} - 1) &= V_{t}(\omega^-) - V_{t-1}(\omega).
\end{align*}
Taking the difference gives the result.
\end{proof}

Q: Reflection principle for simple random walk
A: Let $Z_t$ be the SSRW on $\Z$. Then $\forall k\geq 1, \ell\geq 0$:
\[
\P\left(\max_{0\leq t \leq T} Z_t \geq k, Z_T=k-\ell \right) = \P(Z_T=k+\ell)
\]
\begin{proof}
Define the hitting time and event
\begin{align*}
\tau &= \inf\{ t\geq 0: Z_t=k\} \\
A_{k,\ell} &=\{ \tau\leq T, Z_T = k-\ell\}.
\end{align*}
Define the reflection map $\phi:\Omega\to\Omega$ by
\[
\omega=(y_1, \cdots y_T) \mapsto (y_1, \cdots y_{\tau(\omega)}, -y_{\tau(\omega)+1}, \cdots -y_T)
\]
giving a bijection
\[
A_{k,\ell}\to\{ \tau\leq T, Z_T = k+\ell\} =\{ Z_T = k+\ell\}.
\]
By uniformity
\[
\P(A_{k,\ell}) = \P(Z_T = k+\ell).
\]
\end{proof}

Q: CRR density
A: Let $\P,\Q$ be the uniform and martingale measures resp. Write $Z_t$ for the SSRW. Then the density is
\[
\frac{\d\Q}{\d\P} = 2^T (p^*)^{\frac{T+Z_T}{2}} (1-p^*)^{\frac{T-Z_T}{2}}.
\]
\begin{proof}
Given $\omega\in\Omega$, for some $k\leq T$
\[
\Q(\{\omega\}) = (p^*)^k (1-p^*)^{T-k}, \qquad \P(\{\omega\})=2^{-T}.
\]
We solve for $k$:
\begin{align*}
Z_T(\omega) &= k-(T-k) \\
&=2k-T \\
k &= \frac{Z_T(\omega) + T}{2}.
\end{align*}
By countability of $\Omega=\{-1,1\}^T$:
\[
\frac{\d\Q}{\d\P} (\omega) = \frac{\Q (\{\omega \})}{\P (\{\omega\})}.
\]
\end{proof}

Q: Probability measures on countable space
A: Let $\Omega$ be countable, $\mathcal{F}=\mathcal{P}(\Omega)$, $\P$ be a strictly +ve PM on $(\Omega, \mathcal{F})$.
Then $\forall \Q$ a PM on $(\Omega, \mathcal{F})$, $\Q << \P$ and $\frac{\d\Q}{\d\P}$ is given by
\[
 \phi(\omega) := \frac{\Q(\{\omega\})}{\P(\{\omega\})}, \qquad \omega\in\Omega.
\]
\begin{proof}
For any $A\in\mathcal{F}$
\begin{align*}
\Q(A) &= \sum_{\omega\in A} \Q(\{\omega\}) \\
&= \sum_{\omega\in A} \frac{\Q(\{\omega\})}{\P(\{\omega\})} \P(\{\omega\}) \\
&= \sum_{\omega\in A} \phi(\omega) \P(\{\omega\}) \\
&= \E_\P[\phi\mathbbm{1}_A] \\
&= \int_A \phi \; \d\P.
\end{align*}
\end{proof}

Q: Brownian motion: definition
A: $(B_t)$ is \emph{Brownian motion} or \emph{Wiener process} if
\begin{enumerate}
\item $B_0=0$ a.s.
\item For each $\omega\in\Omega$, the \emph{sample path} $t\mapsto B_t(\omega)$ is cts.
\item For any $0=t_0<\cdots < t_n$:
\[
B_{t_i}-B_{t_{i-1}} \sim N(0, t_i-t_{i-1})
\]
are independent.
\end{enumerate}

Q: Brownian motion is Markovian
A: $(B_t)$ is a MC with transition probabilities
\[
p_{xy}(t) = \frac{1}{\sqrt{2\pi t}} \exp\left( - \frac{(y-x)^2}{2t} \right).
\]

Q: Brownian motion is a martingale
A: $(B_t)$ is a martingale wrt $\mathcal{F}_t=\sigma(B_s: s\leq t)$.
\begin{proof}
For $0\leq s \leq t$
\[
E[B_t | \mathcal{F}_s] = B_s + \E[B_t - B_s]=B_s.
\]
\end{proof}

Q: Gaussian process
A: A stochastic process $(X_t)_{t\geq 0}$ is called a \emph{Gaussian process} if for $0=t_0<\cdots < t_n$:
\[
(X_{t_1}, \cdots X_{t_n}) \sim \text{MVN}.
\]

Q: Brownian motion: characterisation by expectations
A: Brownian motion is the Gaussian process $(B_t)\subseteq\R$ such that
\begin{enumerate}
\item $B_0=0$ a.s.
\item For each $\omega\in\Omega$, $t\mapsto B_t(\omega)$ is cts.
\item $\E[B_t]=0$ and $\forall s,t\geq 0: \E[B_sB_t]=s\wedge t$.
\end{enumerate}
\begin{proof}
$(\Rightarrow)$ Ets (iii). For $s<t$:
\begin{align*}
\E[B_sB_t] &= \E[(B_t-B_s)B_s + B_s^2] \\
&= \E[B_t-B_s]\E[B_s] + \E[B_s^2] \\
&= s.
\end{align*}
$(\Leftarrow)$ The law of a Gaussian process is uniquely determined by mean and covariance.
\end{proof}

Q: Brownian motion in higher dimensions
A: A $d$-dimensional Brownian motion is a stochastic process $(B_t)_{t\geq 0}\subseteq\R^d$ with $B^i$ MI Brownian motions.

Q: Brownian motion: existence
A: Let $(Y_i)_{i\in\N}$ be iid with $\E[Y_i]=0, \var(Y_i)=1$.
Define $X_k=\sum_{i=1}^k Y_i$ and
\[
X_t^{(n)} = \frac{1}{n} X_{\lfloor n^2t \rfloor} + \frac{tn^2-\lfloor tn^2 \rfloor}{n}(X_{\lfloor n^2t \rfloor + 1} - X_{\lfloor n^2t \rfloor}).
\]
That is, $X^{(n)}_{k/n^2}=X_k/n$ with linear interpolation between these times.
By Donsker's invariance principle / Functional Central Limit Theorem $\lim X^{(n)}$ is a Brownian motion.

Q: Brownian motion: H\"older continuity
A: The Brownian motion is locally H\"older cts of exponent $\alpha$ iff $\alpha<\frac{1}{2}$.
In particular, Brownian motion is nowhere differentiable.
\begin{proof}
See Norris Advanced Probability Theorem 7.7.2.
\end{proof}

Q: Brownian motion transformations
A: If $B$ is a BM, then so are
\begin{enumerate}
\item $-B_t$
\item $cB_{t/c^2}$ (\emph{scale invariance})
\item $tB_{1/t}$ (\emph{time-inversion})
\item $B_{T+t}-B_T$ ($\forall T\geq 0$).
\end{enumerate}
\begin{proof}
Use the fact that the law of a Gaussian process is determined by its mean and covariance.
For cty at 0 of (iii):
\[
\{\lim_{t\to 0}\tilde{B}_t= 0 \} = \bigcap_n \bigcup_m \bigcap_{0<q<1/m \in\Q} \{ |\tilde{B}_q|<\frac{1}{n}\}.
\]
is a.s. since $\tilde{B}_t:=tB_{1/t}\sim B_t$ by the above.
\end{proof}

Q: Brownian motion law of large numbers
A: Let $B$ be a Brownian motion. Then
\[
\lim_{t\to\infty} \frac{B_t}{t} = 0 \quad\text{a.s.}
\]
\begin{proof}
Immediate by cty of $\tilde{B}_t = tB_{1/t}$ at $t=0$.
\end{proof}

Q: Brownian motion explodes
A: For $B$ a BM
\[
\P(\sup B_t=\infty)=1.
\]
\begin{proof}
Write $Z=\sup B_t$, $\tilde{B}_t=c^{-1}B_{c^2t}$ and $B_t'=B_{t+1}-B_1$. By \emph{scale invariance}
\[
Z =\sup B_{c^2 t} \sim c \sup B_t = cZ.
\]
Hence the cdf obeys
\[
\P(Z\leq z) = \P(cZ\leq z)
\]
so is constant on $(0,\infty)$. Hence $Z\in\{0,\infty\}$ a.s.
By \emph{translation invariance}  $Z'=\sup B_t'\sim Z\in\{0,\infty\}$ a.s. so
\begin{align*}
\P(Z=0) &= \P(Z=0, Z'=0) \\
&\leq \P(B_1\leq 0, Z'=0) \\
&= \P(B_1\leq 0)\P(Z'=0) \\
&= \frac{1}{2}\P(Z=0) \\
\P(Z=0) &= 0.
\end{align*}
\end{proof}

Q: Brownian motion: Strong Markov Property
A: Let $(B_t)$ be an $(\mathcal{F}_t)$-BM and $\tau$ an a.s. finite stopping time. Define the filtration
\[
\mathcal{F}_\tau = \{ A\in\mathcal{F}: A \cap \{\tau\leq t\}\in\mathcal{F}_t (\forall t \geq 0) \}.
\]
Then $(B_{\tau + t})$ is an $(\mathcal{F}_{\tau +t})$ BM.
In particular, $(B_{\tau + t} - B_\tau)$ is a BM independent of $\mathcal{F}_\tau$.
\begin{proof}
Norris Advanced Probability Thm 7.5.1.
\end{proof}

Q: Brownian motion reflected
A: Let $(B_t)$ be a BM, $B_0=0$ and $a>0$. Then
\[
\tilde{B}_t = \begin{cases}
B_t \quad & t< \tau_a \\
2a - B_t \quad & t\geq \tau_a
\end{cases}
\]
is an identically distributed BM.
\begin{proof}
Write $Y_t=B_t$ for $t\leq \tau_a$ and $Z_t=B_{\tau_a+t}-B_{\tau_a}$.
By SMP $\pm Z$ are BMs independent of $Y$ with $(Y,Z)\sim (Y,-Z)$.
For $t_0>0$, define the composition of paths $\Phi_{t_0}: C([0,t_0])\times C([0, \infty))\to C([0,\infty))$ by
\[
\Phi_{t_0}(y,z)(t) = \begin{cases}
y_t \quad & t\leq t_0 \\
y_{t_0}+z_{t-t_0} \quad & t>t_0
\end{cases}
\]
Then $B=\Phi_{\tau_a}(Y,Z) \sim \Phi_{\tau_a}(Y,-Z)=\tilde{B}$.
\end{proof}

Q: Brownian motion reflection principle
A: For any $a,y\geq 0$ and $t\geq 0$:
\[
\P(\max_{s\leq t} B_s\geq a, B_t\leq a-y)=\P(B_t \geq a+y).
\]
\begin{proof}
Writing $\tilde{B}$ for the reflected BM about $a$. Then
\begin{align*}
\P(\max_{s\leq t} B_s\geq a, B_t\leq a-y) &= \P(\max_{s\leq t} \tilde{B}_s\geq a, \tilde{B}_t\leq a-y) \\
&= \P(\max_{s\leq t} B_s\geq a, B_t\geq a+y) \\
&= \P(B_t \geq a+y).
\end{align*}
\end{proof}

Q: Laplace transform of hitting time by reflection principle
A: For any $a>0, \lambda >0$
\[
E[e^{-\lambda\tau_a}] = e^{-a\sqrt{2\lambda}}.
\]
\begin{proof}
Using the reflection principle
\begin{align*}
\P(\tau_a\leq t) &= \P(\max_{s\leq t} B_s \geq a) \\
&= \P(\max_{s\leq t} B_s \geq a, B_t\leq a) + \P(\max_{s\leq t} B_s \geq a, B_t\geq a)\\
&= 2\P(B_t\geq a) \\
&= 2\left[1-\Phi\left(\frac{a}{\sqrt{t}}\right)\right] \\
\P(\tau_a \in \d t) &= -2\Phi'(at^{-1/2})\left(-\tfrac{1}{2} at^{-3/2} \right) \\
&= \frac{a}{\sqrt{2\pi t^3}} e^{-a^2/2t} \\
E[e^{-\lambda\tau_a}] &= \int_0^\infty e^{-\lambda t}  \left[\frac{a}{\sqrt{2\pi t^3}} e^{-a^2/2t}\right] \; \d t \\
&= e^{-a\sqrt{2\lambda}}.
\end{align*}
\end{proof}

Q: Laplace transform of hitting time by OST
A: For any $a>0, \lambda >0$
\[
E[e^{-\lambda\tau_a}] = e^{-a\sqrt{2\lambda}}.
\]
\begin{proof}
Define the martingale
\[
M_t^\theta = \exp(\theta B_t - \tfrac{1}{2}\theta^2 t) \qquad (\theta>0).
\]
Applying OST and DCT
\begin{align*}
1&=\E[M_{\tau_a\wedge n}^\theta]\\
& \to e^{\theta a} \E[e^{-\tfrac{1}{2}\theta^2 \tau_a }] \qquad\text{(DCT)} \\
\E[e^{-\tfrac{1}{2}\theta^2 \tau_a }] &= e^{-\theta a} \\
\E[e^{-\lambda \tau_a }] &= e^{-a\sqrt{2\lambda}} \qquad (\lambda=\theta^2 / 2).
\end{align*}
\end{proof}

Q: Wiener measure
A: Let $\Omega=C([0,T])$ be the \emph{path space}, $B$ the coordinate process and $\mathcal{F}=\sigma(B_t: 0\leq t \leq T)$ coinciding with the Borel $\sigma$-algebra on $C([0,T]), \|\ph\|_\infty$.
By Donsker's invariance principle, $\exists\P$ a PM on $(\Omega, \mathcal{F})$ called the \emph{Wiener measure} which makes $B$ a BM.

Q: Cameron-Martin Theorem
A: Write $\tilde{B}_t = B_t+ct$ for the drifted BM, $c\in\R, t\leq T$. Then $\forall F:C([0,T])\to \R$ bounded msble
\[
\E[F(\tilde{B})] = \E[F(B)\exp(cB_T- \tfrac{c^2}{2} T)].
\]
In particular, $\tilde{B}$ has PM $\P^{(c)}$ given by density
\[
\frac{\d\P^{(c)}}{\d\P} = \exp(cB_T - \tfrac{c^2}{2}T).
\]
\begin{proof}
By linearity and density, wlog $F=\mathbbm{1}_{\{\forall i: B_{t_i} +ct_i\in A_i\}}$ for some $0=t_0 < \cdots < t_n=T$ and $A_1, \cdots A_n \in\mathcal{B}(\R)$ (we say $F$ is a \emph{cylindrical functional}):
\begin{align*}
 &\E[\mathbbm{1}_{\{\forall i: B_{t_i} +ct_i\in A_i\}} ] \\
=&\P(\forall i: B_{t_i}+ct_i\in A_i)  \\
=& \int \cdots \int_{A_i-ct_i} \prod_{i=1}^n p_{t_i-t_{i-1}} (x_i - x_{i-1}) \d x_n \cdots \d x_1 \qquad (x_0:=0) \\
=& \int \cdots \int_{A_i} p_{t_i-t_{i-1}} (y_i - y_{i-1}-c(t_i-t_{i-1})) \d y_n \cdots \d y_1 \qquad (y_i = x_i+ct_i)\\
=& \int \cdots \int_{A_i} \prod_{i=1}^n p_{t_i-t_{i-1}}(y_i-y_{i-1}) \exp(c(y_i-y_{i-1}) - \tfrac{c^2}{2}(t_i-t_{i-1})) \d y_n \cdots \d y_1  \\
=& \int \cdots \int_{A_i} \left[\prod_{i=1}^n p_{t_i-t_{i-1}}(y_i-y_{i-1})\right] \exp(cy_n-\tfrac{c^2}{2}t_n) \d y_n \cdots \d y_1 \\
=& \E[\mathbbm{1}_{\{\forall i: B_{t_i}\in A_i\}} \exp(cB_{t_n}-\tfrac{c^2}{2}t_n) ].
\end{align*}
\end{proof}

Q: Reflection principle for drifted BM
A: Write $\tilde{B}_t = B_t + ct$ for $c\in\R, t\leq T$. Then $\forall y>a$:
\[
\P(\tilde{B}_T\in\d y, \min \tilde{B} \geq a) = e^{cy-\tfrac{c^2}{2}T}(p_T(y) - p_T(2a-y)).
\]
Differentiating wrt $y$ and changing signs
\[
\P(\tilde{B}_T\in\d y, \min \tilde{B} \in\d a) = \frac{2(y-2a)}{\sqrt{2\pi T^3}} \exp\left( -\frac{(2a-y)^2}{2T} + cy - \frac{c^2}{2}T \right).
\]
\begin{proof}
By the reflection principle (with $B\mapsto 2a-B, y\mapsto a+y$)
\begin{align*}
\P(B_T\geq y, \min B < a) &= \P_x(B_T\leq 2a - y)\\
\P(B_T\geq y, \min B \geq a) &= \P(B_T\geq y) - \P(B_T\leq 2a - y) \\
\P(B_T\in \d y, \min B \geq a) &= (p_T(y) - p_T(2a-y)) \d y.
\end{align*}
Done by Cameron-Martin.
\end{proof}

Q: Brownian motion: associated martingales
A: Let $f\in C_b^{1,2}([0,\infty) \times\R)$, $\mathcal{L}=\partial_t + \tfrac{1}{2}\partial_{xx}^2$ and $B$ a BM. Then
\[
M_t = f(t,B_t) - f(0,B_0) - \int_0^t \mathcal{L}f(s,B_s) \; \d s
\]
is a martingale.
\begin{proof}
Cts, adapted and $L^1$ clear. Now fix $s\geq 0$ and set
\[
\tilde{f}(t,\ph):= f(s+t, \ph), \quad \tilde{B}_t=B_{s+t}, \quad \tilde{\mathcal{F}}_t:=\mathcal{F}_{s+t}.
\]
Then $\tilde{B}$ is an $\tilde{\mathcal{F}}$-BM and
\begin{align*}
M_{s+t} - M_s &= f(s+t, B_{s+t}) - f(s, B_s) - \int_s^{s+t} \mathcal{L} f(r, B_r) \; \d r \\
&= \tilde{f}(t,\tilde{B}_t) - \tilde{f}(0,\tilde{B}_0) - \int_0^t \mathcal{L}\tilde{f}(r,\tilde{B}_r) \; \d s \\
&=:\tilde{M}_t.
\end{align*}
Rtp $\E[ \tilde{M}_t| \tilde{\mathcal{F}}_0]=0$ so by similarity ets $\E[ M_t| \mathcal{F}_0]=0$ a.s. Hence ets $\forall x: \E_x[ M_t]=0$. But $M_0=0$ so ets $\E_x[M_t-M_s]=0$ for $0<s<t$. Indeed
\begin{align*}
\E_x[M_t-M_s] &= \E_x[f(t,B_t) - f(s,B_s) + \int_s^t \mathcal{L}f(r, B_r) \; \d t] \\
&= [\E_x f(u,B_u)]_s^t  - \int_\R \int_s^t  p_r(x,y) (\partial_r + \tfrac{1}{2} \partial^2_{yy})f(r, y) \; \d r \d y \\
&= \int_\R \int_s^t f(r, y) (\partial_r - \tfrac{1}{2} \partial^2_{yy})p_r(x,y)  \; \d r \d y \qquad \text{(parts)}
\end{align*}
which vanishes as $p$ satisfies the heat equation.
\end{proof}

Q: Black-Scholes martingale measure
A: By Cameron-Martin, $W_t := B_t-ct$ is a $\Q$-BM where
\[
\d\Q: = \exp(cB_T - \tfrac{c^2}{2}T) \; \d\P.
\]
Now pick $c\in\R$ such that
\[
\sigma c + \mu - r = -\frac{\sigma^2}{2} \iff c = \frac{r-\tfrac{\sigma^2}{2} -\mu}{\sigma}.
\]
Then the discounted price process becomes
\begin{align*}
X_t &= e^{-rt} S_t \\
&= S_0\exp(\sigma B_t + (\mu - r)t) \\
&= S_0 \exp(\sigma W_t + (\sigma c + \mu - r)t) \\
&= S_0 \exp(\sigma W_t - \tfrac{\sigma^2}{2}t )
\end{align*}
which is a $\Q$-martingale. In fact, this is the unique martingale measure.

Q: Samuelson market model
A: Assume riskless bond
\[
S_t^0 = e^{rt}
\]
and risky asset
\[
S_t = S_0 \exp(\sigma B_t + \mu t)
\]
with interest rate $r\geq 0$ and drift $\mu\in\R$.

Q: Black-Scholes as limit of binomial model
A: Split $[0,T]$ into $NT$ trading periods. Define CRR models
\[
r_N = \frac{r}{N}, \quad a_N = -\frac{\sigma}{\sqrt N}, \quad b_N = \frac{\sigma}{\sqrt N}, \quad p_N = \frac{1}{2} +  \frac{1}{2}\frac{\mu}{\sigma\sqrt{N}}.
\]
Assume $S_{0,N}^0=1, S_{0,N}^1=S_0^1>0$. Hence
\begin{align*}
S_{NT, N}^0 &= (1+\tfrac{r}{N})^{NT} \to e^{rT} \\
S_{NT, N}^1 &= S_0^1\prod_{i=1}^{NT} (1+R_i^{(N)}) \\
\log\left( \frac{S_{NT, N}^1}{S_0^1} \right) &=\sum_{i=1}^{NT} \left[ R_i^{(N)} - \frac{1}{2} (R_i^{(N)})^2 \right] + O(N^{-1/2})
\end{align*}
where the returns $R_i^{(N)}$ have mean and variance
\[
\E[R_i^{(N)}] = \frac{\mu}{N}, \quad \var(R_i^{(N)}) = \frac{\sigma^2}{N} - \frac{\mu^2}{N^2}.
\]
Hence the RHS summand has mean and variance
\[
\hspace{-2cm}\E[R_i^{(N)} - \tfrac{1}{2} (R_i^{(N)})^2 ] = \frac{\mu-\tfrac{1}{2}\sigma^2}{N}, \quad \var(R_i^{(N)} - \tfrac{1}{2} (R_i^{(N)})^2)  = \frac{\sigma^2}{N}.
\]
By non-constant mean/variance CLT
\[
\log\left( \frac{S_{NT, N}^1}{S_0^1} \right) \xrightarrow{d} N((\mu - \tfrac{1}{2}\sigma^2) T, \sigma^2 T).
\]
The equivalent martingale measure $\Q_N$ is given by
\[
p_N^* = \frac{r_N-a_N}{b_N - a_N} = \frac{1}{2} + \frac{1}{2} \frac{r}{\sigma\sqrt{N}}
\]
so letting $\mu\mapsto r$ the same result holds in $\Q_N$.

Q: Central limit theorem with non-constant mean/variance
A: For each $N\in\N$, let $Y_1^{(N)}, \cdots Y_N^{(N)}$ be independent RVs and define
\[
Z_N= \sum_{i=1}^N Y_i^{(N)}.
\]
Assume
\begin{enumerate}
\item $|Y_i^{(N)}| \leq \gamma_N \to 0$ a.s.
\item $\E[Z_N]  \to m\in\R$
\item $\var(Z_N) \to \sigma^2 > 0$.
\end{enumerate}
Then $Z_n\wto N(m, \sigma^2)$.
\begin{proof}
 Chung. A course in probability theory. Cor of Thm 7.1.2
\end{proof}

Q: Pricing terminal-value based assets in binomial limit
A: Let $f$ be cts. Then limit of the arbitrage-free prices $C^{(N)}=f(S^1_{NT,N})$ is given by the B-S price $v(T,S_0^1)$ where
\begin{align*}
v(T,x) &=e^{-rT}\E_\Q\left[ f\left( x\exp\left\{\sigma B_T + (r-\tfrac{1}{2}\sigma^2) T\right\} \right)\right]  \\
&=  e^{-rT} \int_{-\infty}^\infty f\left(x\exp (\sigma\sqrt{T}y + (r-\tfrac{1}{2}\sigma^2) T)\right) \frac{1}{\sqrt{2\pi}} e^{-y^2/2} \; \d y.
\end{align*}
\begin{proof}
The special case $f\in C_b(\R)$ is immediate from
\[
S_{NT,N}^1 \xrightarrow{d} S_0^1 \exp \left( \sigma B_T + (r - \tfrac{1}{2}\sigma^2)T\right).
\]
\end{proof}

Q: Delta-hedging in Black-Scholes
A: The hedging/replicating strategy $\bar\theta = (\theta^0, \theta)$ for claim $C=f(S_T)$ is given by
\[
\theta_t = \Delta(T-t, S_t), \qquad \theta_t^0 = v(T-t, S_t) - \theta_t e^{-rt}S_t,
\]
where $\Delta = \partial_xv$, and has discounted value process $V_t=v(T-t, S_t)$.
\begin{proof}
Follows from B-S as binomial limit and similar result for CRR.
\end{proof}

Q: Black-Scholes price for European Call
A: The B-S price for a European call is $v(T,S_0)$ where
\[
v(T,x) = x\Phi(d_+) - Ke^{-rT}\Phi(d_-)
\]
and
\[
d_\pm(T,x) := \frac{ \log(x/K) + (r\pm\tfrac{1}{2}\sigma^2)T } { \sigma\sqrt{T} }.
\]
\begin{proof}
Apply terminal-value formula with $f(x) = (x-K)^+$, defining $\tilde\sigma=\sigma\sqrt{T}$ and $\tilde{K}=e^{-rT}K/x$, to obtain
\begin{align*}
v(T,x) &= e^{-rT}\E\left[ \left\{x\exp\left(\sigma \sqrt{T}Z + (r-\tfrac{1}{2}\sigma^2) T\right) - K\right\}^+\right] \\
&= x\int_{-\infty}^\infty \left( e^{\tilde{\sigma}z - \tfrac{1}{2}\tilde\sigma^2} - \tilde{K} \right)^+ \frac{e^{-z^2/2}}{\sqrt{2\pi}} \; \d z \\
&= x\int_{-d_-}^\infty \left( e^{\tilde{\sigma}z - \tfrac{1}{2}\tilde\sigma^2} - \tilde{K} \right) \frac{e^{-z^2/2}}{\sqrt{2\pi}} \; \d z \\
&= \frac{x}{\sqrt{2\pi}}\int_{-d_-}^\infty  e^{-\tfrac{(z-\tilde{\sigma})^2}{2}}   \; \d z - x\tilde{K}\left(1 - \Phi(-d_-) \right) \\
&= x(1-\Phi(-d_+)) - Ke^{-rT}\left(1 - \Phi(-d_-) \right)  \\
&= x\Phi(d_+) - Ke^{-rT}\Phi(d_-).
\end{align*}
where $-d_- = \frac{\log\tilde K + \tfrac{1}{2}\tilde\sigma^2}{\tilde\sigma}$ and $d_+=d_- + \tilde\sigma$.
\end{proof}

Q: Put-call parity
A: The pay-off of European options obeys
\[
C^{\text{call}} -  C^{\text{put}}= (S_T-K)^+ - (K-S_T)^+ = S_T - K = F.
\]
Discounting gives the price relationship
\[
\pi(C^{\text{call}}) -  \pi(C^{\text{put}})=  S_0 - e^{-rT}K.
\]

Q: Black-Scholes price for European Put
A: The B-S price of a European put is
\[
\pi(C^{\text{put}}) = e^{-rT}K\Phi(-d_-) - S_0\Phi(-d_+).
\]
\begin{proof}
From put-call parity
\begin{align*}
\pi(C^{\text{put}}) &=\pi(C^{\text{call}}) - (S_0-e^{-rT}K) \\
&=S_0\Phi(d_+) - Ke^{-rT}\Phi(d_-) - S_0 + e^{-rT}K \\
&= e^{-rT}K\Phi(-d_-) - S_0 \Phi(-d_+).
\end{align*}
\end{proof}

Q: Delta of a European call
A: The Delta-hedging strategy is $\theta_t=\Delta(T-t, S_t)$ where
\[
\Delta(t, x) = \Phi(d_+).
\]
In particular, $\theta_t\in (0,1)$ so we are \emph{long stock}.
\begin{proof}
Differentiate wrt $x$ the value function
\[
v(t, x) =  x\Phi(d_+) - Ke^{-rt} \Phi(d_-).
\]
\end{proof}

Q: Gamma of European call
A: The \emph{Gamma} of the European call is
\[
\Gamma(t, x) = \frac{\Phi'(d_+)}{x\sigma\sqrt{t}}.
\]
In particular $x\mapsto v(t, x)$ is strictly convex.
\begin{proof}
Differentiate the value function twice wrt $x$:
\[
\Gamma(t, x) = \frac{\partial}{\partial x} \Phi(d_+).
\]
\end{proof}

Q: Theta of European call
A: The \emph{Theta} of a European call option is
\[
\Theta(t, x) = \frac{\partial v}{\partial t} = \frac{x\sigma}{2\sqrt{t}} \Phi'(d_+) + Kre^{-rt}\Phi(d_-).
\]
Note $\Theta >0$ due to increasing optionality.

Q: Rho of a European call
A: The value of a European call increases with interest rates:
\[
\rho(t, x):= \frac{\partial v}{\partial r} = Kte^{-rt}\Phi(d_-)>0.
\]

Q: Vega of a European call
A: The price of a European call increases with volatility:
\[
\vega (t,x):= \frac{\partial v}{\partial \sigma} = x\sqrt{t}\Phi'(d_+)>0.
\]
Hence $\sigma\mapsto v(t,x)$ is injective (in fact strictly increasing), allowing calculation of \emph{implied volatility}.

Q: Black-Scholes PDE
A: Given $u\in C_b^{1,2}([0,\infty) \times\R)$, define $f(t,x)=u(t,s)$ where
\[
s(t,x):= S_0e^{\sigma x + (r-\tfrac{1}{2}\sigma^2)t}.
\]
Writing $\mathcal{G}u:= \mathcal{L}f=\partial_t f + \tfrac{1}{2}\partial_{xx}^2 f$, the chain rule gives
\begin{align*}
\partial_t f &= \partial_t u +(r-\tfrac{1}{2}\sigma^2) s\partial_s u \\
\partial_x f &= \sigma s \partial_s u \\
\partial_{xx} f &= \sigma^2 s  \partial_s u + \sigma^2 s^2 \partial_{ss}u  \\
\mathcal{G}u&= \partial_t u + rs\partial_s u + \tfrac{1}{2}\sigma^2s^2\partial_{ss}u.
\end{align*}
Therefore
\[
u(t, S_t) - u(0,S_0) - \int_0^t \mathcal{G}u(r,S_r) \; \d r = f(t, W_t) - f(0,W_0) - \int_0^t \mathcal{L}f(r,W_r) \; \d r
\]
is a cts $\Q$-martingale. In particular, if $\mathcal{G}u=0$ with BC $u(T,\ph)=\phi$, then $(u(t,S_t))$ is a martingale. Hence
\[
\E_\Q[\phi(S_T) | S_t=s] = u(t,s)
\]
is the (undiscounted) value of a European option $C=\phi(S_T)$ (it can be checked that this solves the PDE).
Differentiating the discounted value $V(t,s):=e^{-r(T-t)}u(t,s)$ gives the PDE
\begin{align*}
\partial_t V &= r V + e^{-r(T-t)} (-rs\partial_s u - \tfrac{1}{2}\sigma^2 s^2 \partial_{ss} u) \\
&= r V - rs\partial_s V - \tfrac{1}{2}\sigma^2 s^2 \partial_{ss} V
\end{align*}
with BC $V(T,\ph)=\phi$.

Q: Black-Scholes PDE with dividends
A: Replacing $r\mapsto r-D$ gives the undiscounted value process $u\mapsto\tilde u$. Discounting as before, $\tilde{V}:=e^{-r(T-t)}\tilde{u}$ obeys the PDE
\[
0=\partial_t \tilde{V} + (r-D)s\partial_s \tilde{V} + \frac{1}{2}\sigma^2 s^2 \partial_{ss} \tilde{V} - r \tilde{V}.
\]

Q: Numerical solution of Black-Scholes PDE
A: Reduce to heat equation $\partial_t u = \partial_{xx}u$ on $[0,T]\times [0,L]$ with initial value $u(0,\ph)=g$ and spatial BCs $u(\ph, 0)=a$, $u(\ph, L)=b$.
Define \emph{increments} $h=L/N_x$, $k=T/N_t$, \emph{grid values} $U_j^i=u(ik, jh)$ and \emph{grid vectors} $U^i=(U_j^i)_{1\leq j \leq N_x - 1}$. The BCs are
\[
U_0^{i}= a(ik), U_{N_x}^{i}= b(ik) .
\]
Define \emph{ratio} $\nu:=k/h^2$.

Q: FTCS-method
A: Forward-difference-in-Time and Central-difference-in-Space method:
\begin{align*}
\frac{U_j^{i+1} - U_j^i }{k} &= \frac{U_{j+1}^i - 2U_j^i + U_{j-1}^i} {h^2} \\
U_j^{i+1}&= U_j^i + \nu ( U_{j+1}^i  -2U_j^i + U_{j-1}^i ) \\
U^{i+1} = FU^i + p^i
\end{align*}
for suitable $F, p^i$.

Q: BTCS Method
A: Backward-difference-in-Time and Central-difference-in-Space: we must solve the system of linear equations
\begin{align*}
\frac{U_j^{i} - U_j^{i-1} }{k} &= \frac{U_{j+1}^i - 2U_j^i + U_{j-1}^i} {h^2} \\
U_j^{i-1} &= U_j^{i}- \nu (U_{j+1}^{i}  -2 U_j^{i} +  U_{j-1}^{i} ) \\
BU^{i+1} &= U^i + q^i
\end{align*}
for suitable $B, q^i$.

Q: Crank-Nicolson method
A: We use FT, \emph{half-central} apx in space:
\begin{align*}
\frac{U_j^{i+1} - U_j^i }{k} &= \frac{1}{2} \left(
\frac{U_{j+1}^{i+1} - 2U_j^{i+1} + U_{j-1}^{i+1}} {h^2} +
\frac{U_{j+1}^i - 2U_j^i + U_{j-1}^i} {h^2} \right) \\
2(1+\nu) U_j^{i+1} &=2U_j^i + \nu(U_{j+1}^{i+1} + U_{j-1}^{i+1} + U_{j+1}^{i} - 2U_{j}^{i} + U_{j-1}^{i}) \\
(I+B)U^{i+1} &= (F+I)U^i + p^i + q^i
\end{align*}
which is FTCS $+$ BTCS.

Q: FTCS: local accuracy
A: Taylor expanding with $U_j^i=u(ik,jh)$ gives
\begin{align*}
R_j^i &:=\frac{U_j^{i+1} - U_j^i }{k} - \frac{U_{j+1}^i - 2U_j^i + U_{j-1}^i} {h^2} \\
&= \frac{1}{k} \left[k\partial_t u + \frac{1}{2}k^2\partial_t^2 u + O(k^3) \right] \\
& \quad - \frac{1}{h^2}\Bigg[ h\partial_x u + \frac{1}{2} h^2 \partial_x^2 u + \frac{1}{6} h^3 \partial_x^3 u + O(h^4) \\
& \qquad\quad\!  -h\partial_x u + \frac{1}{2} h^2 \partial_x^2 u - \frac{1}{6} h^3 \partial_x^3 u + O(h^4) \Bigg] \\
&= \left(\partial_t u -  \partial_x^2 u \right) +\frac{1}{2}k\partial_t^2 u + O(k^2) + O(h^2) \\
&= \frac{1}{2}k\partial_t^2 u + O(k^2) + O(h^2).
\end{align*}

Q: BTCS: local error
A: Taylor expanding with $U_j^i=u(ik,jh)$ gives
\begin{align*}
R_j^i &:=\frac{U_j^{i} - U_j^{i-1} }{k} - \frac{U_{j+1}^i - 2U_j^i + U_{j-1}^i} {h^2} \\
&= -\frac{1}{k} \left[-k\partial_t u + \frac{1}{2}k^2\partial_t^2 u + O(k^3) \right] \\
& \quad - \frac{1}{h^2}\Bigg[ h\partial_x u + \frac{1}{2} h^2 \partial_x^2 u + \frac{1}{6} h^3 \partial_x^3 u + O(h^4) \\
& \qquad\quad\!  -h\partial_x u + \frac{1}{2} h^2 \partial_x^2 u - \frac{1}{6} h^3 \partial_x^3 u + O(h^4) \Bigg] \\
&= \left(\partial_t u -  \partial_x^2 u \right) -\frac{1}{2}k\partial_t^2 u + O(k^2) + O(h^2) \\
&= -\frac{1}{2}k\partial_t^2 u + O(k^2) + O(h^2).
\end{align*}

Q: Crank-Nicolson: local error
A: Simple averaging of FTCS/BTCS gives local accuracy $O(k^2) + O(h^2)$.

Q: Reflection equation for SRW
A: Let $Z_t$ be the SSRW on $\Z$. Then $\forall k\geq 1, \ell\geq 0$:
\[
\P\left(\max_{0\leq t \leq T} Z_t = k, Z_T=k-\ell \right) = \frac{2(k+\ell+1)}{T+1} \P(Z_{T+1}=1+k+\ell ) .
\]
\begin{proof}
By parity, both sides are trivial if $T+k+\ell$ is odd. Otherwise write $T+k+\ell =2j$ and compute using the reflection principle twice
\begin{align*}
\text{LHS}(2) &= \P\left(\max_{0\leq t \leq T} Z_t \geq k, Z_T=k-\ell \right) -  \P\left(\max_{0\leq t \leq T} Z_t \geq k+1, Z_T=k-\ell \right) \\
&= \P(Z_T=k+\ell) -  \P(Z_T=k+\ell+2) \\
&= 2^{-T} \left({T \choose j} - {T \choose j+1} \right) \\
&= 2^{-T} {T+1 \choose j+1}\frac{2j+1-T}{T+1} \\
&= \frac{2(k+\ell +1)}{T+1} \P(Z_{T+1}=1+k+\ell).
\end{align*}
\end{proof}

Q: Forward start options
A: For $0<t<T$, define
\[
\text{FS}(T) = (S_T-KS_t)^+ = S_t\left(\frac{S_T}{S_t}-K\right)^+.
\]
Taking discounted expectations
\[
\text{FS}(0) = \E_\Q[e^{-rt}S_t] \E_\Q\left[e^{-r(T-t)}\left(\frac{S_T}{S_t}-K\right)^+\right]=S_0 \text{EC}(T-t, 1, K) .
\]