29
The way you do it in the first place is a discretization of the Geometric Brownian Motion (GBM) process. This method is most useful when you want to compute the path between $S_0$ and $S_t$, i.e. you want to know all the intermediary points $S_i$ for $0 \leq i \leq t$.
The second equation is a closed form solution for the GBM given $S_0$. A simple ...
20
I give you a brief outline about some key properties of Lévy processes.
Lévy processes have stationary and independent increments but do not necessarily have continuous sample paths. In fact, Brownian motion is the only Levy process with continuous sample paths. Some Lévy processes (e.g. Poisson process) have single, rare but large jumps (finite activity) ...
19
Using the Ito Formula
The general approach that often works for these kinds of question is to search for functions such that their Ito differential contains the terms that we are interested in. In your case, we are looking for a function $f(t, x)$ such that $f_t(t, x) = t x$. Let
\begin{equation}
f(t, x) = \frac{1}{2} t^2 x
\end{equation}
with
\begin{...
19
Life Without a Risk-Neutral Measure
How would we price assets without the measure $\mathbb Q$? Well, we would start with some version of the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, where $M$ is the stochastic discount factor (SDF). This equation holds under very weak assumptions (law of one price) and uses real-world probabilities. So, we take the ...
18
Here is a short list (to be edited and improved - community wiki) :
Standard brownian motion (also called Wiener process) for which:
$d\, W_t \sim \mathcal N(0, \sqrt{d t})$
Geometric brownian motion, used in the Black-Scholes model (1973):
$d\,X_t = \mu X_t\,dt + \sigma X_t\,dW_t$
Constant elasticity of variance ("CEV") model (1975):
$d\,X_t=\mu X_t dt + \...
16
Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ...
13
To complement @SRKX comment ,i'll try to explain the "simple mathematical proof" beetween both formula :
I assume you know the geometric or arithmetic brownian motion :
Geometric:
\begin{equation*}
dS = \mu S dt + \sigma Sdz
\end{equation*}
Arithmetic :
\begin{equation*}
dS = \mu dt + \sigma dz
\end{equation*}
Then another important stochastic tool you ...
13
If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list.
Pricing.
For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ...
13
Let
\begin{align*}
Y_t = e^{(a+\frac{c^2}{2})t-cW_t}.
\end{align*}
Then
\begin{align*}
dY_t = Y_t\left[\big(a+c^2\big)dt -c dW_t \right].
\end{align*}
Moreover,
\begin{align*}
d(X_tY_t) &= Y_t dX_t + X_t dY_t + d\langle X, Y\rangle_t\\
&=abY_tdt.
\end{align*}
That is,
\begin{align*}
X_t = Y_t^{-1}\left(X_0 + ab\int_0^t Y_sds\right).
\end{align*}
12
The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...
12
I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar.
The key property that a uniformly integrable martingale has is the so-called closure ...
12
Quadratic variation and variance are two different concepts.
Let $X $ be an Ito process and $t\geq 0$.
Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that you denoted by $[X,X]_t $ is a random variable.
What is confusing you is the fact that when $X $ is a martingale then
$X^2_t-[X,X]_t$ is a martingale thus you ...
12
Another approach consists in using the Fubini theorem to write that
\begin{align}
\int_0^T u W_u du &= \int_0^T \int_0^u u\, dW_v\, du \tag{$W_u = \int_0^u dW_v$} \\
&= \int_0^T \int_v^T u\, du\, dW_v \tag{Fubini}\\
&= \frac{1}{2}\int_0^T (T^2 - v^2) dW_v
\end{align}
This is an Itô integral. Since the integrand ...
11
Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter.
Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...
11
I will assume a white noise is a process $(\varepsilon_t)$ with zero mean, no autocorrelation and constant variance $\sigma^2 > 0$ while a random walk is a process $(x_t)$ defined by
$$
x_{t+1} = x_t + \varepsilon_{t+1}
$$
where $\varepsilon$ is a white noise.
1) No since $Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$ is stricly increasing while ...
11
Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed
\begin{equation}
\int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right).
\end{equation}
In your case, we have $f(t) = e^{-\lambda t}$ and thus
\begin{equation}
\int_0^t f^2(u) \mathrm{d}u = \...
11
By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$.
The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{E}[I_t]=I_0=0$.
Because $W_s\overset{d}{=}\sqrt{s}Z$, where $Z\sim N(0,1)$, we indeed have
\begin{align*}
\int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\...
11
You need to rotate them so we can find some orthogonal axes.
A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression
\begin{align}
W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2}
\end{align}
where $\tilde{...
10
I will try to answer this a bit differently.
The rigorous answer: because Ito calculus tells us that we need the second order term. Look at
$$
S_t = S_0\exp(\mu t + \sigma B_t).
$$
Assume that $S_0$ is known and fixed and look at
by Ito's formula
$$
d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt.
$$
Then with some abuse of notation:
$$
E[d(S_t/...
10
Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function.
We know that
$$
\mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t
$$
i.e. an $N$-dimensional vector $X$ of correlated Brownian motions has ...
9
There is a lot of ways to understand why stationarity allows to apply usual time series analysis. Here is one more.
Very often, the theoretical justification of what you do in time series need to be able to identify the mean formula and the expectation:
$$\frac{1}{N}\sum_{n=1}^N X_n \underset{N\rightarrow +\infty}{\longrightarrow} \mathbb{E} X, $$
where the ...
9
This is an interesting question that I have asked myself. Below is my take.
Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations:
$$\begin{align}
\text{d}S_t&=\alpha(t,S_t)\text{d}t+\...
9
I will provide some references such that you can see where the different processes are used. These papers typically motivate their models and show which effect the single paramaters have and what asset price dynamics the model intends to capture.
Geometric Brownian motion
The geometric Brownian motion is the easiest model for exponential growth with ...
8
Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future.
For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation:
$$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$
This blog post provides a closed form solution:
$$ S_t = S_0 \...
8
To provide a straight forward answer: It is not a good model. It never was, it never will be. Until we all do not come up with a better model that provides better modeling accuracy while it is equally intuitive and makes similarly simplifying assumptions the BS model with its geometric brownian motion component is here to stay.
It actually does not matter ...
8
We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so
$$
E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]
$$
and that implies that $E[W_pW_qW_r] = 0$.
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