14
Martingales + Markovian
Here is the motivation. Conditional expectations are martingales by the tower property of conditional expectations (an easy exercise to show). Suppose $r=0$, by the risk neutral pricing theorem $E^\star\left[h(X_T)\bigg|\mathscr{F}_t,\,X_t=x\right]$ is the price of any derivative security with $X$ as the underlying asset and payoff ...
11
Let's skip to the stochastic differential equation (SDE):
$$ dF=\left[\frac{\partial F}{\partial t}+\mu \frac{\partial F}{\partial x}+\frac{1}{2}\sigma^2 \frac{\partial^2 F}{\partial x^2} \right]dt + \sigma \frac{\partial F}{\partial x}dW $$
What does this equation actually represent? It suggests that a change in $F$ (represented by $\Delta F$) equals a ...
9
The Feynman-Kac theorem primarily makes sense in a pricing context. If you know that some function solves the Feynman-Kac equation you can represent it's soluation as an Expectation with respect to the process. (confer this document)
On the other hand a pricing function solves the FK-PDE. Thus often one would try solving the PDE to get a closed form ...
7
To solve the SDE you should use the so called variation of constant method.
Define a process $Y_t=e^{-\mu t}X_t$, so that using Itô we obtain:
$$dY_t=-\mu Y_t dt+ e^{-\mu t}X_t=e^{-\mu t} \sigma dW_t $$
Therefore by integrating we have:
$$Y_T=Y_S+\int_S^T e^{-\mu t} \sigma dW_t=e^{-\mu S} X_S +\sigma \int_S^T e^{-\mu t} dW_t$$
$$\Rightarrow \quad X_T=e^{\mu ...
6
You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable $W_t$ is stochastic and hence $S_t$ is as well.
So, to derive $S_t$ from $dS_t$, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it.
If you want to do it the other way round,...
5
I would start with explaining random walk (this should be fairly simple) and then making a connection to heat equation in discrete time. This paper is doing exactly this and by leaving out technicalities you should make this pretty intuitive for students.
Basically the intuition is as follows:
At each integer time unit, the heat at each point is spread ...
5
I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform:
\begin{align*}
\mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy,
\end{align*}
...
5
As you noted, this is a Riccati type ODE and it can thus be simplified using the standard transformations for this class - see e.g. Wikipedia. We start by defining
\begin{equation}
C(t, T) = \frac{1}{2} \alpha B(t, T) \qquad \Rightarrow \qquad C_t(t, T) = \frac{1}{2} \alpha B_t(t, t)
\end{equation}
and get
\begin{eqnarray}
C_t(t, T) & = & C^2(t, T)...
5
Black and Scholes (1973) were not the first ones to use the geometric Brownian motion as a model for stock prices. For example, Samuelson did it before them.
It all started with a Brownian motion as simplest time continuous stock price model. However, then the stock price is normally distributed and can be negative. Not a great property! So, Samuelson ...
4
The actual problem one solves for American options is an optimal stopping time problem, so the value of the option is
$$
V_0 = \max_\tau E_{\tau}\left[e^{-r \tau} (S_\tau-K)^+ \right]
$$
where the maximum is taken over all stopping times (exercise strategies $\tau>0$ permissible in the contract).
With a PDE operator such as you have, the instantaneous ...
4
We consider the case where the Novikov condition is satisfied, that is,
\begin{align*}
E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty.
\end{align*}
Then $\{L_t \mid t \ge 0\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale.
On $\mathscr{F}_T$, we define the probability measure $Q$ by
\begin{align*}
\frac{dQ}{dP}\big|_{\mathscr{F}...
4
You maximize the terms in $q$ in the PDE because this is a consequence of the Bellman principle of optimality in dynamic programming. The intuition is that the global optimal strategy $\{q_t\}_{0 \leq t \leq T}$ is locally optimal such that (under the risk neutral measure because the option is dynamically hedged)
$$V_t = \max_{|q_t|\leq 1}e^{-r dt}E_t\left[...
4
Martingale Approach
As you noted, you need to solve
\begin{eqnarray}
F(0) & = & e^{-r T} \mathbb{E} \left[ \left( X_T - K \right)^2 \right]\\
& = & e^{-r T} \left( \mathbb{E} \left[ X_T^2 \right] - 2 K \mathbb{E} \left[ X_T \right] + K^2 \right)
\end{eqnarray}
Let $Y_t = X_t^2$. Then, by applying the Itō formula, we get
\begin{eqnarray}
\...
4
Using Itô's Lemma, notice that:
$$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$
Hence:
$$X_t=tS_t-\int S_udu$$
Using independence of Brownian increments, $E(S_udW_u)=E(S_u)E(dW_u)=0$, and the chain rule for the 4th step:
$$\begin{align}
E(X_t)&=E\left(\int dX_u\right)
\\
&=\int uE(dS_u)
\\
&=\int u\mu E(S_u)du
\\
&=S_0\int u\mu e^{\mu u}du
\\
&=S_0\...
3
It must be a typo for the equation in the book. That is, the equation for a caplet is of the form
\begin{align*}
\frac{\partial V}{\partial t} + LV - r_t V +\max(r_t-r^*, 0) = 0,
\end{align*}
which can also be derived using the martingale approach.
Specifically, note that the accumulated payments from time $t$ up to maturity $T$ is given by
\begin{align*}
\...
3
I think you misunderstood the underlying idea of the risk-neutrality and the market price of risk.
The basic idea is to price the option with a portfolio consisting of the underlying asset $S$ and another option. In order to make this portfolio risk-free and because of no-arbitrage arguments, the change in the portfolio should correspond to the change of ...
3
This is a corollary of Feynman-Kac theorem. For self-containedness,
I re-produce the proof as follows.
Assume that there exists a $C^{1,2}$-function $F=F(t,x)$ defined
on $[0,T]\times\mathbb{R}$ that satisfies the PDE on the
interior
$$
F_{t}+\beta F_{x}+\frac{1}{2}\sigma^{2}F_{xx}=0,
$$
and the boundary condition: $F(T,x)=g(x)$. Consider the process $\left(...
3
About the integration problem: Your integrand is highly oscillatory, and the adaptive quadrature of Matlab doesn't handle such integrands very well. In general, I would recommend Mathematica when Matlab's standard procedures don't perform well. In this case, a Levin-type method would perform much better. The reason that quadv produces NaN values is because ...
3
The way I think of it is that the PDE describes the flow of a time dependent probability distribution. The stochastic process describes individual realisations (random walks with a drift), but if you ran a large number of them you'd build up a distribution.
The PDE says how that distribution changes in time (first term) due to deterministic drift (the ...
3
At the terminal time $T$, the terminal condition is $g(T, q) = -\alpha q^2$, this implies,
$$
\begin{aligned}
g(T, q) &= \frac{1}{\kappa} \log{\omega(T, q)} = -\alpha q^2\\
\Rightarrow \omega(T,q) &= e^{-\kappa\alpha q^2}
\end{aligned}
$$
Therefore, $\mathbf{z}$ is given by,
$$
\mathbf{z} = \boldsymbol{\omega(T)} = \begin{bmatrix}
e^{-\alpha\kappa \...
3
Edits have been made, but the original question asked about $\frac{\mathrm{d}t}{W_t}$.
The random variable does exist
In your [original] question you ask about $\frac{\mathrm{d}t}{W_t}$, although I think you really meant $\frac{\mathrm{d}t}{\mathrm{d}W_t}$. The quantity $\frac{\mathrm{d}t}{W_t}$ does exist and is just a random variable. Be careful though, ...
3
The SDE you are describing is called the Geometric Brownian Motion. In the end its just a model, which underlies certain assumptions, which are usually not met in the real world scenarios. There are many further extensions and variation of SDEs for modelling prices f.e. including a jump component (jump diffusion models), mean reversion (f.e. Ornstein-...
3
Take the analogy of equations modelling something in physics.
Just because you write down an equation, it does not mean it has to be connected to anything in reality. It only do so to the extent you have adapted the equation and it's parameters to fit reality.
In finance things are a bit more complicated when it comes to the predicting power though.
...
2
I am not sure any of the other answers mentioned this but the main reason you should not use an option model to buy/sell the underlying (BS or other) is that the option models are more about market-making in options and hedging using the underlying rather than forecasting the underlying.
The layman way to understand this is that: using an option model, you ...
2
I didn't work out the explicit details, but you can reproduce Black&Scholes methodology using the Ito's formula for Jump Diffusions. See for example, the sectio about Poisson jump processes in http://en.wikipedia.org/wiki/Itō's_lemma
In general every Markov process admits some kind of Ito's formula, known as Dynkin formula, which says that for a markov ...
2
As you state in your comment, you only have trouble with the second partial derivative w.r.t. the spot. So you understand how the first partial derivative is obtained
\begin{equation}
\frac{\partial C}{\partial S} = e^{-x} \frac{\partial U}{\partial x}.
\end{equation}
Then you just carefully apply the chain rule again. In ALL details:
\begin{eqnarray}
\...
2
By Bayes' rule for conditional expectation (or here),
$$E_{\mathbb Q}[X_t | \mathscr F_u] E[L_T| \mathscr F_u] = E[X_tL_T| \mathscr F_u]$$
$$ \to E_{\mathbb Q}[X_t | \mathscr F_u] L_u = E[X_tL_T| \mathscr F_u]$$
$$\to E_{\mathbb Q}[X_t | \mathscr F_u] = E[\frac{X_tL_t}{L_u}| \mathscr F_u]$$
$$= \frac{1}{L_u} E[ \frac{X_tL_t}{1} | \mathscr F_u]$$
$$= \...
2
In convertible bond pricing there is something similar called a "soft call" with similar properties so you might want to search for literature on them. The main difference is that soft calls are an exercise condition rather than an exercise price.
One key point is that, if expiration $t$ is distant, very little error is introduced by ignoring the softness. ...
2
Note that
\begin{align*}
M(r_t, t) &\equiv Q(r_t, t) e^{-\int_0^t r_u du} \\
&=E\left(e^{-\int_0^T r_u du} h(r_T, T) \mid \mathscr{F}_t \right)
\end{align*}
is a martingale. Moreover,
\begin{align*}
dM &= \frac{\partial M}{\partial t}dt + \frac{\partial M}{\partial r} dr_t + \frac{1}{2}\frac{\partial^2 M}{\partial r^2}d\langle r, r\rangle_t\\
&...
2
May be I have overlooked something, but I believe that
\begin{align*}
Q(t, S) = \mathbb{P}\left(\tau_{B} \le T \mid \mathcal{F}_t\right).
\end{align*}
Then $\{Q(t, S), \, 0<t < T\}$ is a martingale, and the PDE follows immediately, by noting that
\begin{align*}
dQ &= Q_t dt + Q_S dS + \frac{1}{2}Q_{SS} d\langle S, S\rangle_t\\
&=\Big(\...
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