27
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 ...
18
From what I remember, there is no real relation between Markov and Martingale, and my intuition was confirmed by this post.
Basically, it says that you can say neither of the following:
If A is Markov, then A is a martingale.
If A is a martingale, then A is Markov.
further down the post, you can find two counter examples:
$dX_t = a dt + \sigma dW_t$ is ...
17
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 + \...
17
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{...
15
I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ...
12
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 ...
11
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 ...
11
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 ...
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:
$$
E[d(S_t/...
9
I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar.
The key property that a uniformly integrable martingale has is the so-called closure ...
9
If you want to address interesting problems that are interesting for financial mathematics, I do not believe you have the good list.
Pricing.
For instance, most of explicit formulas for pricing that are not available yet will never be. In this direction, you should have a look at simulation techniques. See for instance Nonlinear Option Pricing. Interesting ...
9
This is an interesting question that I have asked myself. Below is my take.
Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations:
$$\begin{align}
\text{d}S_t&=\alpha(t,S_t)\text{d}t+\...
8
In general, if you have a process that you can write under the form $F(B_t,t)$ where $F$ is $\mathcal{C}^{2,1}$ then Itô's lemma gives you the drift term and diffusion term of $dF$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ...
8
"Treshold Garch" or T-Garch models are designed to capture this asymmetry. See this exposition by U. Chicago's Ruey Tsay who has a terrific text on time-series models in "Analysis of Financial Time Series".
You can use the structure of the T-Garch models to simulate data with this property.
There is a package called fGarch that creates APARCH models. A T-...
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:
$$ S_t = S_0 \...
8
The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...
8
We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so
$$
E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]
$$
and that implies that $E[W_pW_qW_r] = 0$.
8
Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter.
Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...
8
I will assume a white noise is a process $(\varepsilon_t)$ with zero mean, no autocorrelation and constant variance $\sigma^2 > 0$ while a random walk is a process $(x_t)$ defined by
$$
x_{t+1} = x_t + \varepsilon_{t+1}
$$
where $\varepsilon$ is a white noise.
1) No since $Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$ is stricly increasing while ...
8
$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$.
Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process)
$$ df(t,X_t) = \frac{...
8
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 = \...
7
These patterns are of course well-known enough to have been "priced in" to the financial markets. Jump diffusions are a classic way to capture the phenomenon, and often have closed-form option pricing formulas associated with them. The implied option skew, for example, gets a lot flatter when you use a JD model.
Jump diffusions are often combined with ...
7
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 ...
7
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 ...
7
The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:
$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$
where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ...
7
I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact.
Like any other differential, this differential is defined in terms of its integral:
$$
\int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2}
$$
Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since
$$
W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\...
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