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 ...
12
Except in highly unusual cases, financial PDEs lack analytic solutions. The mathematical tools used are Monte Carlo, plus the usual ones for solving PDEs on grids, almost always one of the following:
Trees, for very simple cases
Explicit finite differencing, for throwaway projects or very specific cases
Implicit or Crank-Nicolson finite differencing for ...
9
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 ...
6
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:
$$Y_T=Y_S+\int_S^T e^{-\mu t} \sigma dW_t=e^{-\mu S} X_S +\sigma \int_S^T e^{-\mu t} dW_t$$
$$\Rightarrow \quad X_T=e^{\mu ...
6
You may want to look into these two open source projects:
QuantLib which is aimed at providing a comprehensive software framework for quantitative finance. This is written in C++.
JQuantLib the 100% Java implementation based on the first project.
6
You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable $W_t$ is stochastic and hence $S_t$ is as well.
So, to derive $S_t$ from $dS_t$, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it.
If you want to do it the other way round,...
5
I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform:
\begin{align*}
\mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy,
\end{align*}
...
5
As you noted, this is a Riccati type ODE and it can thus be simplified using the standard transformations for this class - see e.g. Wikipedia. We start by defining
\begin{equation}
C(t, T) = \frac{1}{2} \alpha B(t, T) \qquad \Rightarrow \qquad C_t(t, T) = \frac{1}{2} \alpha B_t(t, t)
\end{equation}
and get
\begin{eqnarray}
C_t(t, T) & = & C^2(t, T)...
4
The actual problem one solves for American options is an optimal stopping time problem, so the value of the option is
$$
V_0 = \max_\tau E_{\tau}\left[e^{-r \tau} (S_\tau-K)^+ \right]
$$
where the maximum is taken over all stopping times (exercise strategies $\tau>0$ permissible in the contract).
With a PDE operator such as you have, the instantaneous ...
4
I would start with explaining random walk (this should be fairly simple) and then making a connection to heat equation in discrete time. This paper is doing exactly this and by leaving out technicalities you should make this pretty intuitive for students.
Basically the intuition is as follows:
At each integer time unit, the heat at each point is spread ...
4
As far as I know, differential equations such as the Black-Scholes PDE are solved once analytically and then the result is used directly. If a given derivatives-pricing differential equation could not be solved analytically, it would probably be better to model it numerically using Monte Carlo methods than to derive a complicated PDE which must then be ...
4
Martingale Approach
As you noted, you need to solve
\begin{eqnarray}
F(0) & = & e^{-r T} \mathbb{E} \left[ \left( X_T - K \right)^2 \right]\\
& = & e^{-r T} \left( \mathbb{E} \left[ X_T^2 \right] - 2 K \mathbb{E} \left[ X_T \right] + K^2 \right)
\end{eqnarray}
Let $Y_t = X_t^2$. Then, by applying the Itō formula, we get
\begin{eqnarray}
\...
4
You maximize the terms in $q$ in the PDE because this is a consequence of the Bellman principle of optimality in dynamic programming. The intuition is that the global optimal strategy $\{q_t\}_{0 \leq t \leq T}$ is locally optimal such that (under the risk neutral measure because the option is dynamically hedged)
$$V_t = \max_{|q_t|\leq 1}e^{-r dt}E_t\left[...
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
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}...
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 ...
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
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(\...
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
The mathematical analysis above is correct, but to understand WHY we say that "a futures has no value" it is helpful to understand how a Futures Exchange works.
When you enter into a position (for example go long 1 crude oil contract at 45.25) you do not have to pay anything, nor does the seller of the contract receive any money from you. So it is correct ...
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
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,
$$
\mathbf{z} = \boldsymbol{\omega(T)} = \begin{bmatrix}
e^{-\alpha\kappa \...
1
The value of a futures contract is the forward value of the payment, discounted back to today -
$$
V(t,T) = e^{-r(T-t)} \mathbb{E} \left[ S(T) - F | \mathcal{F}_t \right]
$$
and the price of a futures contract is the price $F$ that gives the contract zero present value -
$$
\begin{align}
F & = \mathbb{E}\left[ S(T) | \mathcal{F}_t\right] \\
& = ...
1
Reflection principal ? Reflection principle.
It holds for the Brownian process, not the GBM. [Reflection principle is quite specific to symmetric random walks].
By chance, if $\mu-\frac{\sigma^2}{2}=0$ and $\sigma>0$, then you have :
$$\mathbb{P}(\tau^S_B<T)=\mathbb{P}(\tau^W_{\frac{1}{\sigma}\ln(B)}<T)$$ and you can apply reflection principle.
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